UBC Wiki - User contributions [en]

Course:MATH110/Archive/2010-2011/003/Teams/Zurich/Homework 13 Logarithmic Scale - pH

2011-02-02T03:38:33Z

EllenTsang:

==What is pH?==

[[File:Ph Environment Canada elmhurts.edu.gif|right|Courtesy of http://www.elmhurst.edu/~chm/vchembook/184ph.html]]

'''The mathematical definition of pH (power of Hydrogen) is the negative logarithmic value of the Hydrogen ion (H+) concentration.'''

The acidity or alkalinity of a substance (ph) can be written as the equation:

'''<math>pH = -log [H+]</math>'''

When <math>H+</math> is the hydrogen ion solution in moles per litre.

The pH graph looks like this, courtesy of WolframAlpha:

[[File: Log_Graph.gif]]

Here is a table showing the correlation of pH values and Hydronium ion concentrations:

[[File:Ph - wwwchem.csustan.edu.gif]]

Courtesy of http://wwwchem.csustan.edu/chem3070/3070m07.htm]

==Example==

Suppose we test tomato juice and we find that the <math>H+</math> is equal to 0.0001.
How do we find the pH value and determine whether the tomato juice is acidic or alkaline?

'''1) Substitute into our logarithmic substance formula'''

In this case:
<math>pH = -log[0.0001] (base 10)
10^-4=0.0001</math>
Hence:
pH = 4

'''This means that the tomato juice is acidic with a pH of 4.'''

Since the negative log of H+ is used in the pH scale, we will usually get positive values (as long as H+ is below 1). Additionally, the larger our pH, the smaller our H+.

The pH scale turns a concise scale into a large scale between the values 0 and 1.

We know that there is an infinite number of values between 0 and 1 which shows that using a log scale allows us to covert small quantities into large quantities.

The pH scale is logarithmic and as a result, each whole pH value below 7 is ten times more acidic than the next higher value. For example, pH 4 is ten times more acidic than pH 5 and 100 times (10 times 10) more acidic than pH 6. The same holds true for pH values above 7, each of which is ten times more alkaline (another way to say basic) than the next lower whole value. For example, pH 10 is ten times more alkaline than pH 9 and 100 times (10 times 10) more alkaline than pH 8.

Thread:Course talk:MATH110/003/Math Forum/Webwork 13 Q5 Bug?/reply (2)

2011-02-02T00:42:06Z

EllenTsang: Reply to Webwork 13 Q5,6 Bug?

Okay thanks :)

Thread:Course talk:MATH110/003/Math Forum/Webwork 13 Q5 Bug?

2011-02-01T23:34:46Z

EllenTsang:

I've tried to do Q5 and I got the part a (the equation) correct but when I plug in the numbers for Part b it doesn't work.

Here's a printscreen of my problem:

[[File:Webworkbugq5.jpg]]

I've also discussed with my group members and we all couldn't get Q6, although it seems correct.

[[File:Webworkbugq6a.JPG]]

Thanks.

Thread:Course talk:MATH110/003/Math Forum/Webwork 13 Q5 Bug?

2011-02-01T23:32:46Z

EllenTsang:

I've tried to do Q5 and I got the part a (the equation) correct but when I plug in the numbers for Part b it doesn't work.

Here's a printscreen of my problem:

[[File:Webworkbugq5.jpg]]

Thanks.

I've also discussed with my group members and we all couldn't get Q6, although it seems correct.

[[File:Webworkbugq6a.JPG]]

File:Webworkbugq6a.JPG

2011-02-01T23:31:26Z

EllenTsang:

Thread:Course talk:MATH110/003/Math Forum/Webwork 13 Q5 Bug?

2011-02-01T23:20:29Z

EllenTsang: New thread: Webwork 13 Q5 Bug?

I've tried to do Q5 and I got the part a (the equation) correct but when I plug in the numbers for Part b it doesn't work.

Here's a printscreen of my problem:

[[File:Webworkbugq5.jpg]]

Thanks.

File:Webworkbugq5.jpg

2011-02-01T23:20:04Z

EllenTsang:

Course:MATH110/Archive/2010-2011/003/Teams/Zurich/Homework 13 Logarithmic Scale - pH

2011-02-01T22:38:01Z

EllenTsang:

====Links====
http://en.wikipedia.org/wiki/Logarithmic_scale

----

The mathematical definition of pH (power of Hydrogen) is the negative logarithmic value of the Hydrogen ion (H+) concentration.

The acidity or alkalinity of a substance (ph) can be written as the equation:

<math>pH = -log [H+]</math>

When <math>H+</math> is the hydrogen ion solution in moles per litre.

==Example==

Suppose we test tomato juice and we find that the <math>H+</math> is equal to 0.0001.
How do we find the pH value and determine whether the tomato juice is acidic or alkaline?

'''1) Substitute into our logarithmic substance formula'''

In this case:
<math>pH = -log[0.0001] (base 10)
10^-4=0.0001</math>
Hence:
pH = 4

'''This means that the tomato juice is acidic with a pH of 4.'''

Since the negative log of H+ is used in the pH scale, we will usually get positive values (as long as H+ is below 1). Additionally, the larger our pH, the smaller our H+.

The pH scale turns a concise scale into a large scale between the values 0 and 1.

We know that there is an infinite number of values between 0 and 1 which shows that using a log scale allows us to covert small quantities into large quantities.

The pH scale is logarithmic and as a result, each whole pH value below 7 is ten times more acidic than the next higher value. For example, pH 4 is ten times more acidic than pH 5 and 100 times (10 times 10) more acidic than pH 6. The same holds true for pH values above 7, each of which is ten times more alkaline (another way to say basic) than the next lower whole value. For example, pH 10 is ten times more alkaline than pH 9 and 100 times (10 times 10) more alkaline than pH 8.

[[File:Ph Environment Canada elmhurts.edu.gif]]

http://www.elmhurst.edu/~chm/vchembook/184ph.html

Pure (neutral) water has a pH around 7 at 25 °C (77 °F); this value varies with temperature. When an acid is dissolved in water the pH will be less than 7 (if at 25 °C (77 °F)) and when a base, or alkali is dissolved in water the pH will be greater than 7 (if at 25 °C (77 °F)). A solution of a strong acid, such as hydrochloric acid, at concentration 1 mol dm−3 has a pH of 0. A solution of a strong alkali, such as sodium hydroxide, at concentration 1 mol dm−3 has a pH of 14. Thus, measured pH values will mostly lie in the range 0 to 14.

Since pH is a logarithmic scale a difference of one pH unit is equivalent to a tenfold difference in hydrogen ion concentration.

http://en.wikipedia.org/wiki/PH

[[File:Ph - wwwchem.csustan.edu.gif]]

http://wwwchem.csustan.edu/chem3070/3070m07.htm

Course:MATH110/Archive/2010-2011/003/Teams/Zurich/Homework 13 Logarithmic Scale - pH

2011-02-01T10:17:17Z

EllenTsang:

====Links====
http://en.wikipedia.org/wiki/Logarithmic_scale

----

The pH scale is logarithmic and as a result, each whole pH value below 7 is ten times more acidic than the next higher value. For example, pH 4 is ten times more acidic than pH 5 and 100 times (10 times 10) more acidic than pH 6. The same holds true for pH values above 7, each of which is ten times more alkaline (another way to say basic) than the next lower whole value. For example, pH 10 is ten times more alkaline than pH 9 and 100 times (10 times 10) more alkaline than pH 8.

[[File:Ph Environment Canada elmhurts.edu.gif]]

http://www.elmhurst.edu/~chm/vchembook/184ph.html

Pure (neutral) water has a pH around 7 at 25 °C (77 °F); this value varies with temperature. When an acid is dissolved in water the pH will be less than 7 (if at 25 °C (77 °F)) and when a base, or alkali is dissolved in water the pH will be greater than 7 (if at 25 °C (77 °F)). A solution of a strong acid, such as hydrochloric acid, at concentration 1 mol dm−3 has a pH of 0. A solution of a strong alkali, such as sodium hydroxide, at concentration 1 mol dm−3 has a pH of 14. Thus, measured pH values will mostly lie in the range 0 to 14.

Since pH is a logarithmic scale a difference of one pH unit is equivalent to a tenfold difference in hydrogen ion concentration.

http://en.wikipedia.org/wiki/PH

[[File:Ph - wwwchem.csustan.edu.gif]]

http://wwwchem.csustan.edu/chem3070/3070m07.htm

Course:MATH110/Archive/2010-2011/003/Teams/Zurich/Homework 13 Logarithmic Scale - pH

2011-02-01T10:12:54Z

EllenTsang:

The pH scale is logarithmic and as a result, each whole pH value below 7 is ten times more acidic than the next higher value. For example, pH 4 is ten times more acidic than pH 5 and 100 times (10 times 10) more acidic than pH 6. The same holds true for pH values above 7, each of which is ten times more alkaline (another way to say basic) than the next lower whole value. For example, pH 10 is ten times more alkaline than pH 9 and 100 times (10 times 10) more alkaline than pH 8.

[[File:Ph Environment Canada elmhurts.edu.gif]]

http://www.elmhurst.edu/~chm/vchembook/184ph.html

Pure (neutral) water has a pH around 7 at 25 °C (77 °F); this value varies with temperature. When an acid is dissolved in water the pH will be less than 7 (if at 25 °C (77 °F)) and when a base, or alkali is dissolved in water the pH will be greater than 7 (if at 25 °C (77 °F)). A solution of a strong acid, such as hydrochloric acid, at concentration 1 mol dm−3 has a pH of 0. A solution of a strong alkali, such as sodium hydroxide, at concentration 1 mol dm−3 has a pH of 14. Thus, measured pH values will mostly lie in the range 0 to 14.

Since pH is a logarithmic scale a difference of one pH unit is equivalent to a tenfold difference in hydrogen ion concentration.

http://en.wikipedia.org/wiki/PH

[[File:Ph - wwwchem.csustan.edu.gif]]

http://wwwchem.csustan.edu/chem3070/3070m07.htm

File:Ph Environment Canada elmhurts.edu.gif

2011-02-01T10:11:53Z

EllenTsang: Courtesy of Environment Canada http://www.ns.ec.gc.ca/

Courtesy of Environment Canada
http://www.ns.ec.gc.ca/

Course:MATH110/Archive/2010-2011/003/Teams/Zurich/Homework 13 Logarithmic Scale - pH

2011-02-01T10:04:44Z

EllenTsang: Created page with "The pH scale is logarithmic and as a result, each whole pH value below 7 is ten times more acidic than the next higher value. For example, pH 4 is ten times more acidic than pH 5..."

The pH scale is logarithmic and as a result, each whole pH value below 7 is ten times more acidic than the next higher value. For example, pH 4 is ten times more acidic than pH 5 and 100 times (10 times 10) more acidic than pH 6. The same holds true for pH values above 7, each of which is ten times more alkaline (another way to say basic) than the next lower whole value. For example, pH 10 is ten times more alkaline than pH 9 and 100 times (10 times 10) more alkaline than pH 8.

http://www.elmhurst.edu/~chm/vchembook/184ph.html

Pure (neutral) water has a pH around 7 at 25 °C (77 °F); this value varies with temperature. When an acid is dissolved in water the pH will be less than 7 (if at 25 °C (77 °F)) and when a base, or alkali is dissolved in water the pH will be greater than 7 (if at 25 °C (77 °F)). A solution of a strong acid, such as hydrochloric acid, at concentration 1 mol dm−3 has a pH of 0. A solution of a strong alkali, such as sodium hydroxide, at concentration 1 mol dm−3 has a pH of 14. Thus, measured pH values will mostly lie in the range 0 to 14.

Since pH is a logarithmic scale a difference of one pH unit is equivalent to a tenfold difference in hydrogen ion concentration.

[[File:PH Scale.svg|thumb|right|Some typical pH values]]

http://en.wikipedia.org/wiki/PH

[[File:Ph - wwwchem.csustan.edu.gif]]

http://wwwchem.csustan.edu/chem3070/3070m07.htm

File:Ph - wwwchem.csustan.edu.gif

2011-02-01T10:04:17Z

EllenTsang: Courtesy of California State University Stanislaus Department of Chemistry http://wwwchem.csustan.edu/chem3070/3070m07.htm

Courtesy of California State University Stanislaus Department of Chemistry
http://wwwchem.csustan.edu/chem3070/3070m07.htm

User:EllenTsang

2011-01-28T01:54:43Z

EllenTsang:

Hi I'm Ellen and I am in 1st year Commerce.

I prefer to say I am from Hong Kong although I do have Canadian citizenship. I am planning to major in Marketing, with a minor in International Business.
My favourite hobby is singing karaoke and I love Korean and Japanese music.
I like Statistics but this is unfortunately not in our MATH 110 course, aww.

I like pi(e).

Main Links:

[[Course:MATH110/003]]

[[Course:MATH110/003/Teams/Zurich]]

[[Course:MATH110/003/Groups/Group 09]]

[[Help:Contents]]

----

== Pythagorean Theorem ==

Here is shown the animated proof of the Pythagorean Theorem.

To prove that, when a and b are the two sides of a right-angled triangle and c is the hypotenuse:

==='''<math>a^2 + b^2 = c^2</math>'''===

[[File:Pythprf.gif|left|frame|400px|Uploaded on the United States Naval Academy]] [[File:PythagoreanTheorem16c.gif|300px]]

The area of the middle-sized square, <math>b^2</math>, is cut into congruent quadrilaterals, as shown in the animation.

Then the quadrilaterals are rearranged and shifted to the largest square, <math>c^2</math>.

Lastly, the smallest square, <math>a^2</math>, is translated to fit into the remaining middle area of <math>c^2</math>.

Hence, this proves that the sum of the squares of the two smaller sides equals the square of the largest side.

==Applications of Calculus==

Calculus is essential for a wide variety of careers. For example, differential calculus can be used in Biology to find the rate of growth of Bacteria over time given different constraints and variables such as food and temperature. Integral calculus could be used to calculate the amount of material required to construct a curved dome. At the present, since I am taking a Commerce degree, I want to focus on the applications of calculus in the Business world. The uses of calculus in Business are extensive, for instance, optimization i.e. how to maximize profits by optimizing operations to attain maximum sales and minimum costs. Calculus could also be used to find marginal figures such as marginal cost and marginal revenues.

===An Example of an Application of Calculus in Business===

A factory is capable of producing 60,000 phones in a day and the total daily cost of producing x phones in a day is given by,

[[File:Calculusapplication1.jpg]]

The question is:
<pre>How many phones per day should they produce in order to minimize production costs?</pre>

====Solution====

Here we need to minimize the cost subject to the constraint that x must be in the range:

[[File:Calculusapplication2.jpg]]

In this case, the cost function is not continuous at the left endpoint and so we won’t be able to just plug critical points and endpoints into the cost function to find the minimum value.

First, we must find the first couple of derivatives of the cost function.

[[File:Calculusapplication3.jpg]]

The critical points of the cost function are:

[[File:Calculusapplication4.jpg]]

In this case, the negative value doesn’t make any sense so we can ignore it.

Hence, we have a single critical point in the range of possible solutions: 50,000.

Now, as long as <math>x>0</math>, the second derivative is positive and so, in the range of possible solutions the function is always concave up and so producing 50,000 phones will yield the absolute minimum production cost.

I believe that no matter in which occupational field, calculus plays an indispensable role in the success of its operations.

File:Calculusapplication4.jpg

2011-01-28T01:38:12Z

EllenTsang:

File:Calculusapplication3.jpg

2011-01-28T01:37:02Z

EllenTsang:

File:Calculusapplication2.jpg

2011-01-28T01:34:27Z

EllenTsang:

File:Calculusapplication1.jpg

2011-01-28T01:31:38Z

EllenTsang:

User:EllenTsang

2011-01-28T00:00:25Z

EllenTsang:

Hi I'm Ellen and I am in 1st year Commerce.

I am from Hong Kong with Canadian citizenship and am planning to major in Marketing, with a minor in International Business.
My favourite hobby is singing karaoke and I love Korean and Japanese music.
I like Statistics but this is unfortunately not in our MATH 110 course, aww.

I like pi(e).

Main Links:

[[Course:MATH110/003]]

[[Course:MATH110/003/Teams/Zurich]]

[[Course:MATH110/003/Groups/Group 09]]

[[Help:Contents]]

----

== Pythagorean Theorem ==

Here is shown the animated proof of the Pythagorean Theorem.

To prove that, when a and b are the two sides of a right-angled triangle and c is the hypotenuse:

==='''<math>a^2 + b^2 = c^2</math>'''===

[[File:Pythprf.gif|left|frame|400px|Uploaded on the United States Naval Academy]] [[File:PythagoreanTheorem16c.gif|300px]]

The area of the middle-sized square, <math>b^2</math>, is cut into congruent quadrilaterals, as shown in the animation.

Then the quadrilaterals are rearranged and shifted to the largest square, <math>c^2</math>.

Lastly, the smallest square, <math>a^2</math>, is translated to fit into the remaining middle area of <math>c^2</math>.

Hence, this proves that the sum of the squares of the two smaller sides equals the square of the largest side.

Thread:Course talk:MATH110/003/Teams/Zurich/Homework 1/Deletion/reply (3)

2011-01-27T23:59:46Z

EllenTsang: Reply to Deletion

thanks, did so :)

Course:MATH110/Archive/2010-2011/003/Teams/Zurich/Homework 12

2011-01-27T23:31:39Z

EllenTsang: /* Modelling to a Real Life Problem */

==Problem==

{| class="wikitable" width="800px"
!style="background: #777777; color: #D0D0D0; text-align: left; padding:3px;" width="100%"|Team Problem
|-
|style="background: #FFFFFF; padding:12px;"|
Starting with the function
:<math>P(t) = \frac{1}{1+e^{-t}}</math>
Your goal is to modify the function so that we can use it to model a real-life problem. We want to be able to control the following things:
* Change the height of the horizontal asymptote on the right, we'll denote it by <math>K</math>.
* Change the <math>y</math>-intercept to any number between <math>0</math> and <math>K</math>
BONUS (just the point below, not what comes after)
* Change the slope of the curved part. Find a way so that the slope can go from very close to zero to almost vertical. (If you graph it, it should be quite clear).

Once you've played with the function enough, try to find an application of the graph to model something. It can be anything which starts at a value and then goes to another one (think for a population, it goes from 0 to it's carrying capacity). Explain what you are modelling and how you decide to attribute a numerical value to each of the 2 or 3 parameters that you researched just above. Then use the model to make a prediction. For example, if your model is suppose to describe a population for which you have its initial population and carrying capacity (potentially its rate of increase if you solved the bonus part), then use that data to make a prediction for the population in 20 years, or use the model to predict when will the population reach 95% of its carrying capacity).

When doing this last part, explain well where you're taking your data from (real data or imagined data), what it is that you're modelling and how you are doing the math to answer a predictive question.

==Solution==

===Changing the Height of the Horizontal Asymptote===

<pre>Change the height of the horizontal asymptote on the right, we'll denote it by K.</pre>

This is the graph for:

<math>P(t) = \frac{1}{1+e^{-t}}</math>

[[File:12graphoriginal.gif]]

We have found that if we change the '1' in the denominator to 'k', any constant, we can change the height of the horizontal asymptote to the right.

The smaller the number we put for k, with k > 0, the higher the height of the horizontal asymptote to the right, as shown in the graph below.

[[File:12graph1.gif]]

===Changing the y-intercept===

<pre> Change the y-intercept to any number between 0 and K</pre>

We found that to change the y-intercept, we manipulate the value of t by adding to it any constant, 'k'. The bigger the number k we add to t, the smaller the value of the y-intercept.

[[File:12graph2.gif]]

===Modelling to a Real Life Problem===

4 people were stranded on a shipwrecked island.

We used our model:

<math>P(t) = \frac{1}{0.005+e^{-t-1.5}}</math>

to model the maximum carrying capacity of the island and how many years it takes for the population to reach the maximum capacity, assuming the population growth is constant.

[[File:Rate of change2.gif]]

From the equation and the graph above, we can predict that after 5 years the population will rise to 153 people.

The derivative of P(t) is [[File:P'(t).gif]] . As the function P(t) travels toward the horizontal asymptotes [[File:Limit on graph 9.gif]], the P'(t) goes to zero [[File:Limit on P'(t).gif]], indicating that horizontal asymptotes has slope zero.

Course:MATH110/Archive/2010-2011/003/Teams/Zurich/Homework 12

2011-01-25T23:11:11Z

EllenTsang: /* Solution */

==Problem==

{| class="wikitable" width="800px"
!style="background: #777777; color: #D0D0D0; text-align: left; padding:3px;" width="100%"|Team Problem
|-
|style="background: #FFFFFF; padding:12px;"|
Starting with the function
:<math>P(t) = \frac{1}{1+e^{-t}}</math>
Your goal is to modify the function so that we can use it to model a real-life problem. We want to be able to control the following things:
* Change the height of the horizontal asymptote on the right, we'll denote it by <math>K</math>.
* Change the <math>y</math>-intercept to any number between <math>0</math> and <math>K</math>
BONUS (just the point below, not what comes after)
* Change the slope of the curved part. Find a way so that the slope can go from very close to zero to almost vertical. (If you graph it, it should be quite clear).

Once you've played with the function enough, try to find an application of the graph to model something. It can be anything which starts at a value and then goes to another one (think for a population, it goes from 0 to it's carrying capacity). Explain what you are modelling and how you decide to attribute a numerical value to each of the 2 or 3 parameters that you researched just above. Then use the model to make a prediction. For example, if your model is suppose to describe a population for which you have its initial population and carrying capacity (potentially its rate of increase if you solved the bonus part), then use that data to make a prediction for the population in 20 years, or use the model to predict when will the population reach 95% of its carrying capacity).

When doing this last part, explain well where you're taking your data from (real data or imagined data), what it is that you're modelling and how you are doing the math to answer a predictive question.

==Solution==

===Changing the Height of the Horizontal Asymptote===

<pre>Change the height of the horizontal asymptote on the right, we'll denote it by K.</pre>

This is the graph for:

<math>P(t) = \frac{1}{1+e^{-t}}</math>

[[File:12graphoriginal.gif]]

We have found that if we change the '1' in the denominator to 'k', any constant, we can change the height of the horizontal asymptote to the right.

The smaller the number we put for k, with k > 0, the higher the height of the horizontal asymptote to the right, as shown in the graph below.

[[File:12graph1.gif]]

===Changing the y-intercept===

<pre> Change the y-intercept to any number between 0 and K</pre>

We found that to change the y-intercept, we manipulate the value of t by adding to it any constant, 'k'. The bigger the number k we add to t, the smaller the value of the y-intercept.

[[File:12graph2.gif]]

===Modelling to a Real Life Problem===

4 people were stranded on a shipwrecked island.

We used our model:

<math>P(t) = \frac{1}{0.005+e^{-t}}</math>

to model the maximum carrying capacity of the island and how many years it takes for the population to reach the maximum capacity, assuming the population growth is constant.

File:12graph2.gif

2011-01-25T23:03:22Z

EllenTsang:

Course:MATH110/Archive/2010-2011/003/Teams/Zurich/Homework 12

2011-01-25T22:46:55Z

EllenTsang: /* Changing the Height of the Horizontal Asymptote */

Course:MATH110/Archive/2010-2011/003/Teams/Zurich/Homework 12

2011-01-25T22:44:23Z

EllenTsang: /* Solution */

Course:MATH110/Archive/2010-2011/003/Teams/Zurich/Homework 12

2011-01-25T22:38:05Z

EllenTsang: /* Solution */

File:12graphoriginal.gif

2011-01-25T22:34:25Z

EllenTsang: uploaded a new version of "File:12graphoriginal.gif"

Course:MATH110/Archive/2010-2011/003/Teams/Zurich/Homework 12

2011-01-25T22:32:14Z

EllenTsang: /* Solution */

File:12graph1.gif

2011-01-25T22:27:33Z

EllenTsang:

Course:MATH110/Archive/2010-2011/003/Teams/Zurich/Homework 12

2011-01-25T22:15:43Z

EllenTsang:

File:12graphoriginal.gif

2011-01-25T22:09:34Z

EllenTsang:

Course:MATH110/Archive/2010-2011/003/Teams/Zurich/Homework 12

2011-01-25T22:04:26Z

EllenTsang: Blanked the page

Course:MATH110/Archive/2010-2011/003/Teams/Zurich/Homework 1

2011-01-25T21:59:47Z

EllenTsang: moved Course:MATH110/003/Teams/Zurich/Homework 1 to Course:MATH110/003/Teams/Zurich/Homework 12: wrong name

#REDIRECT [[Course:MATH110/003/Teams/Zurich/Homework 12]]

Course:MATH110/Archive/2010-2011/003/Teams/Zurich/Homework 12

2011-01-25T21:59:47Z

EllenTsang: moved Course:MATH110/003/Teams/Zurich/Homework 1 to Course:MATH110/003/Teams/Zurich/Homework 12: wrong name

Thread:Course talk:MATH110/003/Teams/Zurich/Homework 1/Deletion/reply

2011-01-25T04:04:01Z

EllenTsang: Reply to Deletion

Yes haha, but we're not sure how to delete it.

Course:MATH110/Archive/2010-2011/003/Teams/Zurich/Homework 11

2011-01-19T04:49:47Z

EllenTsang:

To predict the cost of producing flags of our team’s Canton under the assumptions that the marginal cost is $7 per unit and that at the current production level of 20 items, the cost is $100, we wrote a linear model based on previous knowledge of revenue formulas:

$7 is the marginal cost, hence the slope = 7

The information above gives us the point (20,100)

To find the linear equation, we substitute this point (20,100) into the equation y = 7x + b

100 = 7(20) + b
100 = 140 + b
b = 140 – 100 = 40

Therefore we get the equation:

y = 7x - 40, where x is number of units produced, and y is the cost

We can demonstrate this with the chart:

[[File:1.png]]

Based on this model, we can predict what it would cost for 150 items to be produced:

7(150) - 40 = 1010

It would cost $1010 for the production of 150 items.

Thus, according to this model, the average cost per item increases as production levels increase.

This is shown as 1010/150 = 6.73, whereas initially 100/20 = 5

Here are some other models related to this question:

The x-axis is Cost and the y-axis is the amount produced.

1) The average cost remains constant as production increases.

Y=((40x^4)+(4x^2))/((10x^4+8x^2))

[[File: The_average_cost_remains_constant_as_production_increases.1.gif‎]]

2) The average cost diminishes as production increases.

Y=((10x^4+8x^2))/ ((40x^4)+(4x^2))

[[File:The_average_cost_diminishes_as_production_increases..gif]]

3) The average cost increases as production increases.

Y=x

[[File:The_average_cost_increases_as_production_increases..gif]]

4) The average cost increases as production increases. (2nd Example)

Y=e^x

[[File:The_average_cost_increases_as_production_increases2.gif]]

5)We obtain an economy of scale. This means that starting at some specific production level, the marginal cost is always less than the average cost.
This can be displayed in the following model:

[[File:6.png]]

As we can see in the model, y=x^2 represents the marginal cost, and y=x^6 represents the average cost. After production of 1, the marginal cost is always less than the average cost.

Course:MATH110/Archive/2010-2011/003/Teams/Zurich/Homework 11

2011-01-19T04:49:08Z

EllenTsang: Created page with "To predict the cost of producing flags of our team’s Canton under the assumptions that the marginal cost is $7 per unit and that at the current production level of 20 items, th..."

To predict the cost of producing flags of our team’s Canton under the assumptions that the marginal cost is $7 per unit and that at the current production level of 20 items, the cost is $100, we wrote a linear model based on previous knowledge of revenue formulas:

$7 is the marginal cost, hence the slope = 7

The information above gives us the point (20,100)

To find the linear equation, we substitute this point (20,100) into the equation y = 7x + b

100 = 7(20) + b
100 = 140 + b
b = 140 – 100 = 40

Therefore we get the equation:

y = 7x - 40, where x is number of units produced, and y is the cost

We can demonstrate this with the chart:

[[File:1.png]]

Based on this model, we can predict what it would cost for 150 items to be produced:

7(150) - 40 = 1010

It would cost $1010 for the production of 150 items.

Thus, according to this model, the average cost per item increases as production levels increase.

This is shown as 1010/150 = 6.73, whereas initially 100/20 = 5

----
Here are some other models related to this question:

The x-axis is Cost and the y-axis is the amount produced.

1) The average cost remains constant as production increases.

Y=((40x^4)+(4x^2))/((10x^4+8x^2))

[[File: The_average_cost_remains_constant_as_production_increases.1.gif‎]]

2) The average cost diminishes as production increases.

Y=((10x^4+8x^2))/ ((40x^4)+(4x^2))

[[File:The_average_cost_diminishes_as_production_increases..gif]]

3) The average cost increases as production increases.

Y=x

[[File:The_average_cost_increases_as_production_increases..gif]]

4) The average cost increases as production increases. (2nd Example)

Y=e^x

[[File:The_average_cost_increases_as_production_increases2.gif]]

5)We obtain an economy of scale. This means that starting at some specific production level, the marginal cost is always less than the average cost.
This can be displayed in the following model:

[[File:6.png]]

As we can see in the model, y=x^2 represents the marginal cost, and y=x^6 represents the average cost. After production of 1, the marginal cost is always less than the average cost.

Course:MATH110/Archive/2010-2011/003/Teams/Zurich

2011-01-19T04:41:41Z

EllenTsang:

{{
Infobox MATH110 Teams
| team name = Zurich
| member 1 = [[User:EllenTsang|Ellen Tsang]]
| member 2 = [[User:EmilyOates|Emily Oates]]
| member 3 = [[User:PhilipLauFaiWong|Philip Wong]]
| member 4 = [[User:RaphaelTan|Raphael Tan]]
}}

==Subpages==

<dpl>
titlematch={{PAGENAME}}/%
namespace={{NAMESPACE}}
shownamespace=false
</dpl>

Course:MATH110/Archive/2010-2011/003/Teams/Zurich

2011-01-19T04:41:02Z

EllenTsang:

{{
Infobox MATH110 Teams
| team name = Zurich
| member 1 = [[User:EllenTsang|Ellen Tsang]]
| member 2 = [[User:EmilyOates|Emily Oates]]
| member 3 = [[User:PhilipLauFaiWong|Philip Wong]]
| member 4 = [[User:RaphaelTan|Raphael Tan]]
}}
In workshop J.

Homework 11 project

==Subpages==

<dpl>
titlematch={{PAGENAME}}/%
namespace={{NAMESPACE}}
shownamespace=false
</dpl>

Course:MATH110/Archive/2010-2011/003/Teams/Zurich

2011-01-19T04:39:47Z

EllenTsang:

{{
Infobox MATH110 Teams
| team name = Zurich
| member 1 = [[User:EllenTsang|EllenTsang]]
| member 2 = Emily Oates
| member 3 = [[User:PhilipLauFaiWong|Philip Wong]]
| member 4 = Raphael Tan
}}
In workshop J.

Homework 11 project

==Subpages==

<dpl>
titlematch={{PAGENAME}}/%
namespace={{NAMESPACE}}
shownamespace=false
</dpl>

Course:MATH110/Archive/2010-2011/003/Groups/Group 09/Basic Skills Project

2010-12-15T05:51:27Z

EllenTsang:

For the Basic Skills Project, Group 9 plans on focusing on Inequalities.

We will give several worked out examples to cover all cases of questions concerning this topic.

Also, we will include tips & tricks for how to solve more difficult problems and possible references related to the topic.

<pre>Let's work with Group 10 for the group project :D
Thanks to everyone who contributed & thanks Michelle Gutmanis from Group 10 who replied.
Good luck with finals guys! & Have a great holiday! -- Ellen</pre>

=What is an inequality?=

It basically means when:

* An equation includes < or > or ≤ or ≥.
** E.g. x + 1 ≤ 3

The symbols and meanings of the inequalities are as follows:

> means greater than

< means less than

≥ means greater than or equal to

≤ means less than or equal to

=When to change the sign in an inequality problem?=

This is where most of the students face problems while doing problems regarding inequality. This occurs when there is ambiguity whether to change the sign of the inequality.

To solve the confusions, we ONLY change the sign of the inequality to its corresponding opposite when we '''multiply or divide with a negative number'''

Equations can be thought of as balanced scales, where the total weight
on the left balances what is on the right. Inequalities are like
unbalanced scales, where all you know is which side is "down"
(heavier). So for example

3x > 6

can be thought of as 3 unknown weights labeled "X" on the left,
heavier than a 6-gram weight on the right.

Negative numbers complicate it a bit. A negative constant can be
thought of as a helium balloon (barely) able to lift a certain weight.
A negative number times a variable might mean there is an antigravity
machine under it so it pulls up as hard as it would normally push
down!

So the equation

-4y < -36

would be 4 Y's on the left with an antigravity machine, and a "-36"
gram balloon on the right. The right side is "heavier," which in this
case means not that it is pushing down more, but that it is pulling up
less!

To solve it, let us first replace the -36 gram balloon with a 36-gram
weight and an antigravity machine:

-(4y) < -(36)

Now turn off the antigravity machines:

4y > 36

Why did I reverse the direction of the "<"? That is the key to this
whole thing: antigravity machines are like turning the whole world
upside down, so the side that was down is now up:

-4y -36
^ ^
\ | |
\ |
\ |
\ |
\
\

becomes

4y 36
| |
| v /
| /
| /
v /
/
/

because 4y, which pulled up harder before, now pushes down harder.

Now we can work with positive numbers, and divide both weights by 4 to
get

y > 9

So any number BIGGER than 9 will work. For example, for y = 10,

-4y = -40 < -36

Do you see how the larger number, 10 > 9, becomes the smaller number
(-40 < -36) when it is multiplied by a negative number? That is the
key. The rule is that when you multiply an inequality by a negative
number, you have to reverse the direction. Or if you prefer, you can
do this:

-4y < -36 Add 4y
0 < 4y - 36 Add 36
36 < 4y Divide by 4
9 < y Reverse the whole inequality
y > 9

By avoiding multiplication by a negative number, I avoided the need to
reverse signs until the end.

Now that we have seen with our imagination what is going on, let us try
to prove the rule that if

a > b

then

-a < -b.

Start with the original inequality and subtract a from both sides:

0 > b - a

Now subtract b from both sides:

-b > -a

But that is the same as

-a < -b

which we were looking for. It is really pretty simple - so simple it
does not grab your attention the way helium balloons and antigravity
machines do! That is why I like to start the way I did.

'''The solution above was taken from Dr. Math. For all the visual learners, i tried to find a video.. but failed to do so...'''

Note- http://mathforum.org/dr/math/ is a very good website to find help for any math problems. You should check it out!

=Basic Examples=
====Example 1====

Solve -2x > 2

we start by dividing both sides by -2 to solve the inequality

x < -1

====Example 2====
Solving linear inequalities is almost exactly like solving linear equations.

* Solve x + 3 < 0.

If they'd given "x + 3 = 0", we would know how to solve: we would have subtracted 3 from both sides. The same applies here.

x < -3

Then the solution is:

x < –3

====Example 3====

* Solve x – 4 > 0.

If they'd given "x – 4 = 0", then we would can solve by adding four to each side. The same applies here.
x >= 4

Then the solution is: x > 4

====Example 4====

* Solve 2x < 9.

If they had given "2x = 9", we would have divided the 2 from each side.

x <= 9/2

Then the solution is: x < 9/2

====Example 5====

* Solve (2x – 3)/4 < 2.
First, multiply through by 4. Since the "4" is positive, we don't have to flip the inequality sign:

(2x – 3)/4 < 2
(4) × (2x – 3)/4 < (4)(2)
2x – 3 < 8
2x < 11
x < 11/2 = 5.5

====Example 6 - Compound Inequality====

* Solve 10 < 3x + 4 < 19.

This is what is called a "compound inequality". It works just like regular inequalities, except that it has three "sides". So, for instance, when we go to subtract the 4, I will have to subtract it from all three "sides".

=10 < 3x + 4 < 19
=6 < 3x < 15
=2 < x < 5

====Example 7====

* Solve 5x + 7 < 3(x + 1).

First we multiply through on the right-hand side, and then solve as usual:

5x + 7 < 3(x + 1)

5x + 7 < 3x + 3

2x + 7 < 3

2x < –4

x < –2

====Example 8====

* Solve <math>5(x-3)/2 < 2(3x+4)</math>

First, simplify the equation then proceed to isolate x.

* (5x-15)/2 < 6x+8
5x-15 > 2(6x+8)
5x > 12x+16+15
5x > 12x+31
5x-12x > 31
-7x > 31 ;; Dividing and multiplying by negative numbers switches the sign
<math>x<-31/7</math>

====Example 9====

* Solve <math>3 < |3x+4| </math>

Given an inequality with absolute values, say |x| > a then we assume it is –a > x > a.

For better and more in-depth information, please visit here: http://www.nipissingu.ca/calculus/tutorials/absolutevalue.html

* <math>3 < 3x+4 < -3 </math>
<math>3 -4 < 3x+4 -4 < -3 -4</math> ;; We minus 4 on both sides
<math>-1 < 3x < -7 </math> ;; Divide 3 on both sides
<math>-1/3 < x < -7/3 </math>

So the solution for <math>3 < |3x+4| </math> is <math>-1/3 < x < -7/3 </math>.

=Quadratic Inequalities=

To solve a quadratic inequality, follow these steps:

1. Solve the inequality as though it were an equation. The real solutions to the equation become boundary points for the solution to the inequality.

2. Make the boundary points solid circles if the original inequality includes equality; otherwise, make the boundary points open circles.

3. Select points from each of the regions created by the boundary points. Replace these “test points” in the original inequality.

4. If a test point satisfies the original inequality, then the region that contains that test point is part of the solutions.

5. Represent the solution in graphic form and in solution test form.

====Example 1====

Solve <math>x^2-2x-15>0</math>

Treat it like a normal quadratic equation and find the zeroes
* <math>x^2-2x-15=0</math>
<math>(x-5)(x+3)=0</math>
x-5=0 or x+3=0
x=5 or -3

The zeroes divide the number line into three regions

[[File:Crappy_number_line_thing_1.png]]

For region x>5, check for a number greater than 5 (eg.6).
*<math>(6)^2-2(6)-15>0</math>
9>0
All x-values greater than 5 will work.

For region x<-3, check for a number less than -3 (eg.-4).
*<math>(-4)^2-2(-4)-15>0</math>
9>0
All x-values less than -3 will work.

*Show answer using interval notation
<math>(-infinity,-3)U(5,infinity)</math>

Then, x>5 and x<-3.

====Example 2====

Solve <math>3x^2>-x+4</math>

Make one side equal to zero
*<math>3x^2+x-4>0</math>

Find the zeroes using the quadratic formula and factor

*(3x+4)(x-1)>0
3x+4=0 or x-1=0
x=<math>-(4/3)</math> or 1

For region x><math>-(4/3)</math>, check for a number less than <math>-(4/3)</math> (eg.-2).
*<math>3(-2)^2+(-2)-4>0</math>
6>0
All x-values less than <math>-4/3</math> will work.

For region x>1, check for a number greater than 1 (eg.4).
*<math>3(4)^2+(4)-4>0</math>
48>0
All x-values greater than 1 will work.

*Show values that produce an answer greater than 0 using interval notation
<math>(-infinity, -4/3)U(1,+infinity)</math>

====Example 3====

Solve for, <math>9x+9+3x^2>-x^2+7</math>

Move terms to one side so one side is zero
*<math>x+9+3x^2+x^2-7>0</math>
<math>4x^2+9x+2>0</math>

Then find the zeroes
*(x+2)(4x+1)>0
x+2=0 or 4x+1=0
x=<math>-2</math> or <math>-(1/4)</math>

To verify, check the to see if the numbers for region x<-2 are greater than zero by plugging in some number.

*<math>4x^2+9x+2>0</math>
<math>4(-3)^2+9(-3)+2>0</math>
<math>11>0</math>

Likewise, for the region x><math>-1/4</math>

*<math>4x^2+9x+2>0</math>
<math>4(1)^2+9(1)+2>0</math>
<math>15>0</math>

And lastly, some number between -2 and <math>-1/4</math>.

*<math>4x^2+9x+2>0</math>
<math>4(-1)^2+9(-1)+2>0</math>
<math>-3>0</math>
As you can see, -3 is not greater than zero.

Then in the end, we are assume that the solution of this question is, x<-2 and x><math>-1/4</math>.

====Example 4====

Solve for, <math>-9x^2+7+22x>11x+2-15x^2</math>

Move terms to one side so one side is zero
*<math>-9x^2+7+22x-11x-2+15x^2>0</math>
<math>6x+11x+5>0</math>

Then find the zeroes
*(6x+5)(x+1)>0
6x+5=0 or x+1=0
x=<math>-5/6</math> or -1

To verify, check the to see if the numbers for region x<<math>-5/6</math> are greater than zero by plugging in some number.

*<math>6x+11x+5>0</math>
<math>6(-4)+11(-4)+5>0</math>
<math>30>0</math>

Likewise, for the region x>-1

*<math>6x+11x+5>0</math>
<math>6(6)+11(6)+5>0</math>
<math>200>0</math>

And lastly, some number between <math>-5/6</math> and -1.

*<math>6x+11x+5>0</math>
<math>6(-.9)+11(-.9)+5>0</math>
<math>-2.86>0</math>
As you can see, it is not greater than zero.

Then in the end, we are assume that the solution of this question is, x<<math>-5/6</math> and x>-1.

====References====

http://www.purplemath.com/modules/ineqquad.htm

http://www.analyzemath.com/Inequalities_Polynomial/quadratic_inequalities.html

=Graphing Inequalities=

Number Line

1. Simplify the inequality you are going to graph.

<pre>E.g.

-2x2 + 5x < -6(x + 1)

-2x2 + 5x < -6x – 6</pre>

2. Move all terms to one side so the other is zero. ''(It will be easiest if the highest power variable is positive.)''

<pre>E.g.

0 < 2x2 -6x - 5x - 6

0 < 2x2 -11x – 6</pre>

3. Pretend that the inequality sign is an equal sign and find all values of the variable.

<pre>E.g.

0 = 2x2 -11x - 6

0 = (2x + 1)(x - 6)

2x + 1 = 0, x - 6 = 0

2x = -1, x = 6

x = -1/2</pre>

4. Draw a number line including the variable solutions (in order).

[[File:Inequalitiesnumberline1.jpg]]

5. Draw a circle on the points. If the inequality symbol means less than or more than (> or <), draw an empty circle over the variable solution(s). If it means less/more than and equal to (≤ or ≥) fill in that circle.

*In this case our equation was greater than zero, so use open circles.

[[File:Inequalitynumberlineshizz2.jpg]]

6. Take a number from each of the resulting intervals and plug it back into the equality. If you get a true statement once solved, shade this region of the number line.

In the interval from (-∞,-1/2) we will take -1 and plug it into the original inequality.

<pre>0 < 2x2 -11x - 6

0 < 2(-1)2 -11(-1) - 6

0 < 2(1) + 11 - 6

0 < 7

Zero is less than 7 is correct, so shade (-∞, -1/2) on the number line.</pre>

7. Next, on the interval from (-1/2, 6) we will use zero.

<pre>0 < 2(0)2 -11(0) - 6

0 < 0 + 0 - 6

0 < -6

Zero is not less than negative six, so do not shade (-1/2,6).</pre>

Lastly, we will take 10 from the interval (6,∞).

<pre>0 < 2(10)2 - 11(10) + 6

0 < 2(100) - 110 + 6

0 < 200 - 110 + 6

0 < 96

Zero is less than 96 is correct, so shade (6,∞) as well.</pre>

Use arrows on the end of shading to indicate that the interval continues into infinity. The completed number line:

[[File:Inequalitynumberlineblah3.jpg]]

=Tips=
If x ≥ y then 1/x ≤ 1/y

=Videos teaching Inequality=

===Solving Linear Inequalities===

{{#ev:youtube | VgDe_D8ojxw | 400}}

===Other Videos===

Video 1. http://www.khanacademy.org/video/inequalities?playlist=Algebra%20I%20Worked%20Examples
- This video touches upon the concept of inequality and has a basic word problem solved.

Video 2. http://www.khanacademy.org/video/interpreting-inequalities?playlist=Algebra%20I%20Worked%20Examples
- This video is about interpreting inequalities in word problems.

Video 3. http://www.khanacademy.org/video/solving-inequalities?playlist=Algebra%20I%20Worked%20Examples
- This video is about solving basic problems regarding inequalities.

Video 4. http://www.khanacademy.org/video/inequalities-using-addition-and-subtraction?playlist=ck12.org%20Algebra%201%20Examples
- This video solves random question about inequalities with addition and subtraction.

Video 5. http://www.khanacademy.org/video/inequalities-using-multiplication-and-division?playlist=ck12.org%20Algebra%201%20Examples
- This video solves random question about inequalities with multiplication and division

Video 6. http://www.khanacademy.org/video/quadratic-inequalities?playlist=Algebra
- This video explains Quadratic Inequalities.

=Useful Links=

* http://www.purplemath.com/modules/ineqsolv.htm
* http://www.mathsisfun.com/algebra/inequality-solving.html
* http://webmath.com/solverineq.html - Allows you to check your answers after solving an Inequality
* http://www.youtube.com/user/khanacademy - Youtube channel with helpful videos on many areas of Math including Inequalities

=Group 10=

[[Course:MATH110/003/Groups/Group 10/Basic skills project]]

Course:MATH110/Archive/2010-2011/003/Groups/Group 09/Basic Skills Project

2010-12-03T10:12:38Z

EllenTsang:

For the Basic Skills Project, Group 9 plans on focusing on Inequalities.

We will give several worked out examples to cover all cases of questions concerning this topic.

Also, we will include tips & tricks for how to solve more difficult problems and possible references related to the topic.

<pre>Let's work with Group 10 for the group project :D
Thanks to everyone who contributed & thanks Micha the only person from Group 10 who replied.
Good luck with finals guys! & Have a great holiday! -- Ellen</pre>

=What is an inequality?=

It basically means when:

* An equation includes < or > or ≤ or ≥.
** E.g. x + 1 ≤ 3

The symbols and meanings of the inequalities are as follows:

> means greater than

< means less than

≥ means greater than or equal to

≤ means less than or equal to

=When to change the sign in an inequality problem?=

This is where most of the students face problems while doing problems regarding inequality. This occurs when there is ambiguity whether to change the sign of the inequality.

To solve the confusions, we ONLY change the sign of the inequality to its corresponding opposite when we '''multiply or divide with a negative number'''

Equations can be thought of as balanced scales, where the total weight
on the left balances what is on the right. Inequalities are like
unbalanced scales, where all you know is which side is "down"
(heavier). So for example

3x > 6

can be thought of as 3 unknown weights labeled "X" on the left,
heavier than a 6-gram weight on the right.

Negative numbers complicate it a bit. A negative constant can be
thought of as a helium balloon (barely) able to lift a certain weight.
A negative number times a variable might mean there is an antigravity
machine under it so it pulls up as hard as it would normally push
down!

So the equation

-4y < -36

would be 4 Y's on the left with an antigravity machine, and a "-36"
gram balloon on the right. The right side is "heavier," which in this
case means not that it is pushing down more, but that it is pulling up
less!

To solve it, let us first replace the -36 gram balloon with a 36-gram
weight and an antigravity machine:

-(4y) < -(36)

Now turn off the antigravity machines:

4y > 36

Why did I reverse the direction of the "<"? That is the key to this
whole thing: antigravity machines are like turning the whole world
upside down, so the side that was down is now up:

-4y -36
^ ^
\ | |
\ |
\ |
\ |
\
\

becomes

4y 36
| |
| v /
| /
| /
v /
/
/

because 4y, which pulled up harder before, now pushes down harder.

Now we can work with positive numbers, and divide both weights by 4 to
get

y > 9

So any number BIGGER than 9 will work. For example, for y = 10,

-4y = -40 < -36

Do you see how the larger number, 10 > 9, becomes the smaller number
(-40 < -36) when it is multiplied by a negative number? That is the
key. The rule is that when you multiply an inequality by a negative
number, you have to reverse the direction. Or if you prefer, you can
do this:

-4y < -36 Add 4y
0 < 4y - 36 Add 36
36 < 4y Divide by 4
9 < y Reverse the whole inequality
y > 9

By avoiding multiplication by a negative number, I avoided the need to
reverse signs until the end.

Now that we have seen with our imagination what is going on, let us try
to prove the rule that if

a > b

then

-a < -b.

Start with the original inequality and subtract a from both sides:

0 > b - a

Now subtract b from both sides:

-b > -a

But that is the same as

-a < -b

which we were looking for. It is really pretty simple - so simple it
does not grab your attention the way helium balloons and antigravity
machines do! That is why I like to start the way I did.

'''The solution above was taken from Dr. Math. For all the visual learners, i tried to find a video.. but failed to do so...'''

Note- http://mathforum.org/dr/math/ is a very good website to find help for any math problems. You should check it out!

=Basic Examples=
====Example 1====

Solve -2x > 2

we start by dividing both sides by -2 to solve the inequality

x < -1

====Example 2====
Solving linear inequalities is almost exactly like solving linear equations.

* Solve x + 3 < 0.

If they'd given "x + 3 = 0", we would know how to solve: we would have subtracted 3 from both sides. The same applies here.

x < -3

Then the solution is:

x < –3

====Example 3====

* Solve x – 4 > 0.

If they'd given "x – 4 = 0", then we would can solve by adding four to each side. The same applies here.
x >= 4

Then the solution is: x > 4

====Example 4====

* Solve 2x < 9.

If they had given "2x = 9", we would have divided the 2 from each side.

x <= 9/2

Then the solution is: x < 9/2

====Example 5====

* Solve (2x – 3)/4 < 2.
First, multiply through by 4. Since the "4" is positive, we don't have to flip the inequality sign:

(2x – 3)/4 < 2
(4) × (2x – 3)/4 < (4)(2)
2x – 3 < 8
2x < 11
x < 11/2 = 5.5

====Example 6 - Compound Inequality====

* Solve 10 < 3x + 4 < 19.

This is what is called a "compound inequality". It works just like regular inequalities, except that it has three "sides". So, for instance, when we go to subtract the 4, I will have to subtract it from all three "sides".

=10 < 3x + 4 < 19
=6 < 3x < 15
=2 < x < 5

====Example 7====

* Solve 5x + 7 < 3(x + 1).

First we multiply through on the right-hand side, and then solve as usual:

5x + 7 < 3(x + 1)

5x + 7 < 3x + 3

2x + 7 < 3

2x < –4

x < –2

====Example 8====

* Solve <math>5(x-3)/2 < 2(3x+4)</math>

First, simplify the equation then proceed to isolate x.

* (5x-15)/2 < 6x+8
5x-15 > 2(6x+8)
5x > 12x+16+15
5x > 12x+31
5x-12x > 31
-7x > 31 ;; Dividing and multiplying by negative numbers switches the sign
<math>x<-31/7</math>

====Example 9====

* Solve <math>3 < |3x+4| </math>

Given an inequality with absolute values, say |x| > a then we assume it is –a > x > a.

For better and more in-depth information, please visit here: http://www.nipissingu.ca/calculus/tutorials/absolutevalue.html

* <math>3 < 3x+4 < -3 </math>
<math>3 -4 < 3x+4 -4 < -3 -4</math> ;; We minus 4 on both sides
<math>-1 < 3x < -7 </math> ;; Divide 3 on both sides
<math>-1/3 < x < -7/3 </math>

So the solution for <math>3 < |3x+4| </math> is <math>-1/3 < x < -7/3 </math>.

=Quadratic Inequalities=

To solve a quadratic inequality, follow these steps:

1. Solve the inequality as though it were an equation. The real solutions to the equation become boundary points for the solution to the inequality.

2. Make the boundary points solid circles if the original inequality includes equality; otherwise, make the boundary points open circles.

3. Select points from each of the regions created by the boundary points. Replace these “test points” in the original inequality.

4. If a test point satisfies the original inequality, then the region that contains that test point is part of the solutions.

5. Represent the solution in graphic form and in solution test form.

====Example 1====

Solve <math>x^2-2x-15>0</math>

Treat it like a normal quadratic equation and find the zeroes
* <math>x^2-2x-15=0</math>
<math>(x-5)(x+3)=0</math>
x-5=0 or x+3=0
x=5 or -3

The zeroes divide the number line into three regions

[[File:Crappy_number_line_thing_1.png]]

For region x>5, check for a number greater than 5 (eg.6).
*<math>(6)^2-2(6)-15>0</math>
9>0
All x-values greater than 5 will work.

For region x<-3, check for a number less than -3 (eg.-4).
*<math>(-4)^2-2(-4)-15>0</math>
9>0
All x-values less than -3 will work.

*Show answer using interval notation
<math>(-infinity,-3)U(5,infinity)</math>

Then, x>5 and x<-3.

====Example 2====

Solve <math>3x^2>-x+4</math>

Make one side equal to zero
*<math>3x^2+x-4>0</math>

Find the zeroes using the quadratic formula and factor

*(3x+4)(x-1)>0
3x+4=0 or x-1=0
x=<math>-(4/3)</math> or 1

For region x><math>-(4/3)</math>, check for a number less than <math>-(4/3)</math> (eg.-2).
*<math>3(-2)^2+(-2)-4>0</math>
6>0
All x-values less than <math>-4/3</math> will work.

For region x>1, check for a number greater than 1 (eg.4).
*<math>3(4)^2+(4)-4>0</math>
48>0
All x-values greater than 1 will work.

*Show values that produce an answer greater than 0 using interval notation
<math>(-infinity, -4/3)U(1,+infinity)</math>

====Example 3====

Solve for, <math>9x+9+3x^2>-x^2+7</math>

Move terms to one side so one side is zero
*<math>x+9+3x^2+x^2-7>0</math>
<math>4x^2+9x+2>0</math>

Then find the zeroes
*(x+2)(4x+1)>0
x+2=0 or 4x+1=0
x=<math>-2</math> or <math>-(1/4)</math>

To verify, check the to see if the numbers for region x<-2 are greater than zero by plugging in some number.

*<math>4x^2+9x+2>0</math>
<math>4(-3)^2+9(-3)+2>0</math>
<math>11>0</math>

Likewise, for the region x><math>-1/4</math>

*<math>4x^2+9x+2>0</math>
<math>4(1)^2+9(1)+2>0</math>
<math>15>0</math>

And lastly, some number between -2 and <math>-1/4</math>.

*<math>4x^2+9x+2>0</math>
<math>4(-1)^2+9(-1)+2>0</math>
<math>-3>0</math>
As you can see, -3 is not greater than zero.

Then in the end, we are assume that the solution of this question is, x<-2 and x><math>-1/4</math>.

====Example 4====

Solve for, <math>-9x^2+7+22x>11x+2-15x^2</math>

Move terms to one side so one side is zero
*<math>-9x^2+7+22x-11x-2+15x^2>0</math>
<math>6x+11x+5>0</math>

Then find the zeroes
*(6x+5)(x+1)>0
6x+5=0 or x+1=0
x=<math>-5/6</math> or -1

To verify, check the to see if the numbers for region x<<math>-5/6</math> are greater than zero by plugging in some number.

*<math>6x+11x+5>0</math>
<math>6(-4)+11(-4)+5>0</math>
<math>30>0</math>

Likewise, for the region x>-1

*<math>6x+11x+5>0</math>
<math>6(6)+11(6)+5>0</math>
<math>200>0</math>

And lastly, some number between <math>-5/6</math> and -1.

*<math>6x+11x+5>0</math>
<math>6(-.9)+11(-.9)+5>0</math>
<math>-2.86>0</math>
As you can see, it is not greater than zero.

Then in the end, we are assume that the solution of this question is, x<<math>-5/6</math> and x>-1.

====References====

http://www.purplemath.com/modules/ineqquad.htm

http://www.analyzemath.com/Inequalities_Polynomial/quadratic_inequalities.html

=Graphing Inequalities=

Number Line

1. Simplify the inequality you are going to graph.

<pre>E.g.

-2x2 + 5x < -6(x + 1)

-2x2 + 5x < -6x – 6</pre>

2. Move all terms to one side so the other is zero. ''(It will be easiest if the highest power variable is positive.)''

<pre>E.g.

0 < 2x2 -6x - 5x - 6

0 < 2x2 -11x – 6</pre>

3. Pretend that the inequality sign is an equal sign and find all values of the variable.

<pre>E.g.

0 = 2x2 -11x - 6

0 = (2x + 1)(x - 6)

2x + 1 = 0, x - 6 = 0

2x = -1, x = 6

x = -1/2</pre>

4. Draw a number line including the variable solutions (in order).

[[File:Inequalitiesnumberline1.jpg]]

5. Draw a circle on the points. If the inequality symbol means less than or more than (> or <), draw an empty circle over the variable solution(s). If it means less/more than and equal to (≤ or ≥) fill in that circle.

*In this case our equation was greater than zero, so use open circles.

[[File:Inequalitynumberlineshizz2.jpg]]

6. Take a number from each of the resulting intervals and plug it back into the equality. If you get a true statement once solved, shade this region of the number line.

In the interval from (-∞,-1/2) we will take -1 and plug it into the original inequality.

<pre>0 < 2x2 -11x - 6

0 < 2(-1)2 -11(-1) - 6

0 < 2(1) + 11 - 6

0 < 7

Zero is less than 7 is correct, so shade (-∞, -1/2) on the number line.</pre>

7. Next, on the interval from (-1/2, 6) we will use zero.

<pre>0 < 2(0)2 -11(0) - 6

0 < 0 + 0 - 6

0 < -6

Zero is not less than negative six, so do not shade (-1/2,6).</pre>

Lastly, we will take 10 from the interval (6,∞).

<pre>0 < 2(10)2 - 11(10) + 6

0 < 2(100) - 110 + 6

0 < 200 - 110 + 6

0 < 96

Zero is less than 96 is correct, so shade (6,∞) as well.</pre>

Use arrows on the end of shading to indicate that the interval continues into infinity. The completed number line:

[[File:Inequalitynumberlineblah3.jpg]]

=Tips=
If x ≥ y then 1/x ≤ 1/y

=Videos teaching Inequality=

===Solving Linear Inequalities===

{{#ev:youtube | VgDe_D8ojxw | 400}}

===Other Videos===

Video 1. http://www.khanacademy.org/video/inequalities?playlist=Algebra%20I%20Worked%20Examples
- This video touches upon the concept of inequality and has a basic word problem solved.

Video 2. http://www.khanacademy.org/video/interpreting-inequalities?playlist=Algebra%20I%20Worked%20Examples
- This video is about interpreting inequalities in word problems.

Video 3. http://www.khanacademy.org/video/solving-inequalities?playlist=Algebra%20I%20Worked%20Examples
- This video is about solving basic problems regarding inequalities.

Video 4. http://www.khanacademy.org/video/inequalities-using-addition-and-subtraction?playlist=ck12.org%20Algebra%201%20Examples
- This video solves random question about inequalities with addition and subtraction.

Video 5. http://www.khanacademy.org/video/inequalities-using-multiplication-and-division?playlist=ck12.org%20Algebra%201%20Examples
- This video solves random question about inequalities with multiplication and division

Video 6. http://www.khanacademy.org/video/quadratic-inequalities?playlist=Algebra
- This video explains Quadratic Inequalities.

=Useful Links=

* http://www.purplemath.com/modules/ineqsolv.htm
* http://www.mathsisfun.com/algebra/inequality-solving.html
* http://webmath.com/solverineq.html - Allows you to check your answers after solving an Inequality
* http://www.youtube.com/user/khanacademy - Youtube channel with helpful videos on many areas of Math including Inequalities

=Group 10=

[[Course:MATH110/003/Groups/Group 10/Basic skills project]]

Course:MATH110/Archive/2010-2011/003/Groups/Group 09/Basic Skills Project

2010-12-03T10:05:57Z

EllenTsang: /* Example 6 */

For the Basic Skills Project, Group 9 plans on focusing on Inequalities.

We will give several worked out examples to cover all cases of questions concerning this topic.

Also, we will include tips & tricks for how to solve more difficult problems and possible references related to the topic.

<pre>Let's work with Group 10 for the group project :D
Thanks Micha the only person from Group 10 who replied. -- Ellen</pre>

=What is an inequality?=

It basically means when:

* An equation includes < or > or ≤ or ≥.
** E.g. x + 1 ≤ 3

The symbols and meanings of the inequalities are as follows:

> means greater than

< means less than

≥ means greater than or equal to

≤ means less than or equal to

=When to change the sign in an inequality problem?=

This is where most of the students face problems while doing problems regarding inequality. This occurs when there is ambiguity whether to change the sign of the inequality.

To solve the confusions, we ONLY change the sign of the inequality to its corresponding opposite when we '''multiply or divide with a negative number'''

Equations can be thought of as balanced scales, where the total weight
on the left balances what is on the right. Inequalities are like
unbalanced scales, where all you know is which side is "down"
(heavier). So for example

3x > 6

can be thought of as 3 unknown weights labeled "X" on the left,
heavier than a 6-gram weight on the right.

Negative numbers complicate it a bit. A negative constant can be
thought of as a helium balloon (barely) able to lift a certain weight.
A negative number times a variable might mean there is an antigravity
machine under it so it pulls up as hard as it would normally push
down!

So the equation

-4y < -36

would be 4 Y's on the left with an antigravity machine, and a "-36"
gram balloon on the right. The right side is "heavier," which in this
case means not that it is pushing down more, but that it is pulling up
less!

To solve it, let us first replace the -36 gram balloon with a 36-gram
weight and an antigravity machine:

-(4y) < -(36)

Now turn off the antigravity machines:

4y > 36

Why did I reverse the direction of the "<"? That is the key to this
whole thing: antigravity machines are like turning the whole world
upside down, so the side that was down is now up:

-4y -36
^ ^
\ | |
\ |
\ |
\ |
\
\

becomes

4y 36
| |
| v /
| /
| /
v /
/
/

because 4y, which pulled up harder before, now pushes down harder.

Now we can work with positive numbers, and divide both weights by 4 to
get

y > 9

So any number BIGGER than 9 will work. For example, for y = 10,

-4y = -40 < -36

Do you see how the larger number, 10 > 9, becomes the smaller number
(-40 < -36) when it is multiplied by a negative number? That is the
key. The rule is that when you multiply an inequality by a negative
number, you have to reverse the direction. Or if you prefer, you can
do this:

-4y < -36 Add 4y
0 < 4y - 36 Add 36
36 < 4y Divide by 4
9 < y Reverse the whole inequality
y > 9

By avoiding multiplication by a negative number, I avoided the need to
reverse signs until the end.

Now that we have seen with our imagination what is going on, let us try
to prove the rule that if

a > b

then

-a < -b.

Start with the original inequality and subtract a from both sides:

0 > b - a

Now subtract b from both sides:

-b > -a

But that is the same as

-a < -b

which we were looking for. It is really pretty simple - so simple it
does not grab your attention the way helium balloons and antigravity
machines do! That is why I like to start the way I did.

'''The solution above was taken from Dr. Math. For all the visual learners, i tried to find a video.. but failed to do so...'''

Note- http://mathforum.org/dr/math/ is a very good website to find help for any math problems. You should check it out!

=Basic Examples=
====Example 1====

Solve -2x > 2

we start by dividing both sides by -2 to solve the inequality

x < -1

====Example 2====
Solving linear inequalities is almost exactly like solving linear equations.

* Solve x + 3 < 0.

If they'd given "x + 3 = 0", we would know how to solve: we would have subtracted 3 from both sides. The same applies here.

x < -3

Then the solution is:

x < –3

====Example 3====

* Solve x – 4 > 0.

If they'd given "x – 4 = 0", then we would can solve by adding four to each side. The same applies here.
x >= 4

Then the solution is: x > 4

====Example 4====

* Solve 2x < 9.

If they had given "2x = 9", we would have divided the 2 from each side.

x <= 9/2

Then the solution is: x < 9/2

====Example 5====

* Solve (2x – 3)/4 < 2.
First, multiply through by 4. Since the "4" is positive, we don't have to flip the inequality sign:

(2x – 3)/4 < 2
(4) × (2x – 3)/4 < (4)(2)
2x – 3 < 8
2x < 11
x < 11/2 = 5.5

====Example 6 - Compound Inequality====

* Solve 10 < 3x + 4 < 19.

This is what is called a "compound inequality". It works just like regular inequalities, except that it has three "sides". So, for instance, when we go to subtract the 4, I will have to subtract it from all three "sides".

=10 < 3x + 4 < 19
=6 < 3x < 15
=2 < x < 5

====Example 7====

* Solve 5x + 7 < 3(x + 1).

First we multiply through on the right-hand side, and then solve as usual:

5x + 7 < 3(x + 1)

5x + 7 < 3x + 3

2x + 7 < 3

2x < –4

x < –2

====Example 8====

* Solve <math>5(x-3)/2 < 2(3x+4)</math>

First, simplify the equation then proceed to isolate x.

* (5x-15)/2 < 6x+8
5x-15 > 2(6x+8)
5x > 12x+16+15
5x > 12x+31
5x-12x > 31
-7x > 31 ;; Dividing and multiplying by negative numbers switches the sign
<math>x<-31/7</math>

====Example 9====

* Solve <math>3 < |3x+4| </math>

Given an inequality with absolute values, say |x| > a then we assume it is –a > x > a.

For better and more in-depth information, please visit here: http://www.nipissingu.ca/calculus/tutorials/absolutevalue.html

* <math>3 < 3x+4 < -3 </math>
<math>3 -4 < 3x+4 -4 < -3 -4</math> ;; We minus 4 on both sides
<math>-1 < 3x < -7 </math> ;; Divide 3 on both sides
<math>-1/3 < x < -7/3 </math>

So the solution for <math>3 < |3x+4| </math> is <math>-1/3 < x < -7/3 </math>.

=Quadratic Inequalities=

To solve a quadratic inequality, follow these steps:

1. Solve the inequality as though it were an equation. The real solutions to the equation become boundary points for the solution to the inequality.

2. Make the boundary points solid circles if the original inequality includes equality; otherwise, make the boundary points open circles.

3. Select points from each of the regions created by the boundary points. Replace these “test points” in the original inequality.

4. If a test point satisfies the original inequality, then the region that contains that test point is part of the solutions.

5. Represent the solution in graphic form and in solution test form.

====Example 1====

Solve <math>x^2-2x-15>0</math>

Treat it like a normal quadratic equation and find the zeroes
* <math>x^2-2x-15=0</math>
<math>(x-5)(x+3)=0</math>
x-5=0 or x+3=0
x=5 or -3

The zeroes divide the number line into three regions

[[File:Crappy_number_line_thing_1.png]]

For region x>5, check for a number greater than 5 (eg.6).
*<math>(6)^2-2(6)-15>0</math>
9>0
All x-values greater than 5 will work.

For region x<-3, check for a number less than -3 (eg.-4).
*<math>(-4)^2-2(-4)-15>0</math>
9>0
All x-values less than -3 will work.

*Show answer using interval notation
<math>(-infinity,-3)U(5,infinity)</math>

Then, x>5 and x<-3.

====Example 2====

Solve <math>3x^2>-x+4</math>

Make one side equal to zero
*<math>3x^2+x-4>0</math>

Find the zeroes using the quadratic formula and factor

*(3x+4)(x-1)>0
3x+4=0 or x-1=0
x=<math>-(4/3)</math> or 1

For region x><math>-(4/3)</math>, check for a number less than <math>-(4/3)</math> (eg.-2).
*<math>3(-2)^2+(-2)-4>0</math>
6>0
All x-values less than <math>-4/3</math> will work.

For region x>1, check for a number greater than 1 (eg.4).
*<math>3(4)^2+(4)-4>0</math>
48>0
All x-values greater than 1 will work.

*Show values that produce an answer greater than 0 using interval notation
<math>(-infinity, -4/3)U(1,+infinity)</math>

====Example 3====

Solve for, <math>9x+9+3x^2>-x^2+7</math>

Move terms to one side so one side is zero
*<math>x+9+3x^2+x^2-7>0</math>
<math>4x^2+9x+2>0</math>

Then find the zeroes
*(x+2)(4x+1)>0
x+2=0 or 4x+1=0
x=<math>-2</math> or <math>-(1/4)</math>

To verify, check the to see if the numbers for region x<-2 are greater than zero by plugging in some number.

*<math>4x^2+9x+2>0</math>
<math>4(-3)^2+9(-3)+2>0</math>
<math>11>0</math>

Likewise, for the region x><math>-1/4</math>

*<math>4x^2+9x+2>0</math>
<math>4(1)^2+9(1)+2>0</math>
<math>15>0</math>

And lastly, some number between -2 and <math>-1/4</math>.

*<math>4x^2+9x+2>0</math>
<math>4(-1)^2+9(-1)+2>0</math>
<math>-3>0</math>
As you can see, -3 is not greater than zero.

Then in the end, we are assume that the solution of this question is, x<-2 and x><math>-1/4</math>.

====Example 4====

Solve for, <math>-9x^2+7+22x>11x+2-15x^2</math>

Move terms to one side so one side is zero
*<math>-9x^2+7+22x-11x-2+15x^2>0</math>
<math>6x+11x+5>0</math>

Then find the zeroes
*(6x+5)(x+1)>0
6x+5=0 or x+1=0
x=<math>-5/6</math> or -1

To verify, check the to see if the numbers for region x<<math>-5/6</math> are greater than zero by plugging in some number.

*<math>6x+11x+5>0</math>
<math>6(-4)+11(-4)+5>0</math>
<math>30>0</math>

Likewise, for the region x>-1

*<math>6x+11x+5>0</math>
<math>6(6)+11(6)+5>0</math>
<math>200>0</math>

And lastly, some number between <math>-5/6</math> and -1.

*<math>6x+11x+5>0</math>
<math>6(-.9)+11(-.9)+5>0</math>
<math>-2.86>0</math>
As you can see, it is not greater than zero.

Then in the end, we are assume that the solution of this question is, x<<math>-5/6</math> and x>-1.

====References====

http://www.purplemath.com/modules/ineqquad.htm

http://www.analyzemath.com/Inequalities_Polynomial/quadratic_inequalities.html

=Graphing Inequalities=

Number Line

1. Simplify the inequality you are going to graph.

<pre>E.g.

-2x2 + 5x < -6(x + 1)

-2x2 + 5x < -6x – 6</pre>

2. Move all terms to one side so the other is zero. ''(It will be easiest if the highest power variable is positive.)''

<pre>E.g.

0 < 2x2 -6x - 5x - 6

0 < 2x2 -11x – 6</pre>

3. Pretend that the inequality sign is an equal sign and find all values of the variable.

<pre>E.g.

0 = 2x2 -11x - 6

0 = (2x + 1)(x - 6)

2x + 1 = 0, x - 6 = 0

2x = -1, x = 6

x = -1/2</pre>

4. Draw a number line including the variable solutions (in order).

[[File:Inequalitiesnumberline1.jpg]]

5. Draw a circle on the points. If the inequality symbol means less than or more than (> or <), draw an empty circle over the variable solution(s). If it means less/more than and equal to (≤ or ≥) fill in that circle.

*In this case our equation was greater than zero, so use open circles.

[[File:Inequalitynumberlineshizz2.jpg]]

6. Take a number from each of the resulting intervals and plug it back into the equality. If you get a true statement once solved, shade this region of the number line.

In the interval from (-∞,-1/2) we will take -1 and plug it into the original inequality.

<pre>0 < 2x2 -11x - 6

0 < 2(-1)2 -11(-1) - 6

0 < 2(1) + 11 - 6

0 < 7

Zero is less than 7 is correct, so shade (-∞, -1/2) on the number line.</pre>

7. Next, on the interval from (-1/2, 6) we will use zero.

<pre>0 < 2(0)2 -11(0) - 6

0 < 0 + 0 - 6

0 < -6

Zero is not less than negative six, so do not shade (-1/2,6).</pre>

Lastly, we will take 10 from the interval (6,∞).

<pre>0 < 2(10)2 - 11(10) + 6

0 < 2(100) - 110 + 6

0 < 200 - 110 + 6

0 < 96

Zero is less than 96 is correct, so shade (6,∞) as well.</pre>

Use arrows on the end of shading to indicate that the interval continues into infinity. The completed number line:

[[File:Inequalitynumberlineblah3.jpg]]

=Tips=
If x ≥ y then 1/x ≤ 1/y

=Videos teaching Inequality=

===Solving Linear Inequalities===

{{#ev:youtube | VgDe_D8ojxw | 400}}

===Other Videos===

Video 1. http://www.khanacademy.org/video/inequalities?playlist=Algebra%20I%20Worked%20Examples
- This video touches upon the concept of inequality and has a basic word problem solved.

Video 2. http://www.khanacademy.org/video/interpreting-inequalities?playlist=Algebra%20I%20Worked%20Examples
- This video is about interpreting inequalities in word problems.

Video 3. http://www.khanacademy.org/video/solving-inequalities?playlist=Algebra%20I%20Worked%20Examples
- This video is about solving basic problems regarding inequalities.

Video 4. http://www.khanacademy.org/video/inequalities-using-addition-and-subtraction?playlist=ck12.org%20Algebra%201%20Examples
- This video solves random question about inequalities with addition and subtraction.

Video 5. http://www.khanacademy.org/video/inequalities-using-multiplication-and-division?playlist=ck12.org%20Algebra%201%20Examples
- This video solves random question about inequalities with multiplication and division

Video 6. http://www.khanacademy.org/video/quadratic-inequalities?playlist=Algebra
- This video explains Quadratic Inequalities.

=Useful Links=

* http://www.purplemath.com/modules/ineqsolv.htm
* http://www.mathsisfun.com/algebra/inequality-solving.html
* http://webmath.com/solverineq.html - Allows you to check your answers after solving an Inequality
* http://www.youtube.com/user/khanacademy - Youtube channel with helpful videos on many areas of Math including Inequalities

=Group 10=

[[Course:MATH110/003/Groups/Group 10/Basic skills project]]

Course:MATH110/Archive/2010-2011/003/Groups/Group 09/Basic Skills Project

2010-12-03T09:52:31Z

EllenTsang: /* Solving Linear Inequalities */

For the Basic Skills Project, Group 9 plans on focusing on Inequalities.

We will give several worked out examples to cover all cases of questions concerning this topic.

Also, we will include tips & tricks for how to solve more difficult problems and possible references related to the topic.

<pre>Let's work with Group 10 for the group project :D
Thanks Micha the only person from Group 10 who replied. -- Ellen</pre>

=What is an inequality?=

It basically means when:

* An equation includes < or > or ≤ or ≥.
** E.g. x + 1 ≤ 3

The symbols and meanings of the inequalities are as follows:

> means greater than

< means less than

≥ means greater than or equal to

≤ means less than or equal to

=When to change the sign in an inequality problem?=

This is where most of the students face problems while doing problems regarding inequality. This occurs when there is ambiguity whether to change the sign of the inequality.

To solve the confusions, we ONLY change the sign of the inequality to its corresponding opposite when we '''multiply or divide with a negative number'''

Equations can be thought of as balanced scales, where the total weight
on the left balances what is on the right. Inequalities are like
unbalanced scales, where all you know is which side is "down"
(heavier). So for example

3x > 6

can be thought of as 3 unknown weights labeled "X" on the left,
heavier than a 6-gram weight on the right.

Negative numbers complicate it a bit. A negative constant can be
thought of as a helium balloon (barely) able to lift a certain weight.
A negative number times a variable might mean there is an antigravity
machine under it so it pulls up as hard as it would normally push
down!

So the equation

-4y < -36

would be 4 Y's on the left with an antigravity machine, and a "-36"
gram balloon on the right. The right side is "heavier," which in this
case means not that it is pushing down more, but that it is pulling up
less!

To solve it, let us first replace the -36 gram balloon with a 36-gram
weight and an antigravity machine:

-(4y) < -(36)

Now turn off the antigravity machines:

4y > 36

Why did I reverse the direction of the "<"? That is the key to this
whole thing: antigravity machines are like turning the whole world
upside down, so the side that was down is now up:

-4y -36
^ ^
\ | |
\ |
\ |
\ |
\
\

becomes

4y 36
| |
| v /
| /
| /
v /
/
/

because 4y, which pulled up harder before, now pushes down harder.

Now we can work with positive numbers, and divide both weights by 4 to
get

y > 9

So any number BIGGER than 9 will work. For example, for y = 10,

-4y = -40 < -36

Do you see how the larger number, 10 > 9, becomes the smaller number
(-40 < -36) when it is multiplied by a negative number? That is the
key. The rule is that when you multiply an inequality by a negative
number, you have to reverse the direction. Or if you prefer, you can
do this:

-4y < -36 Add 4y
0 < 4y - 36 Add 36
36 < 4y Divide by 4
9 < y Reverse the whole inequality
y > 9

By avoiding multiplication by a negative number, I avoided the need to
reverse signs until the end.

Now that we have seen with our imagination what is going on, let us try
to prove the rule that if

a > b

then

-a < -b.

Start with the original inequality and subtract a from both sides:

0 > b - a

Now subtract b from both sides:

-b > -a

But that is the same as

-a < -b

which we were looking for. It is really pretty simple - so simple it
does not grab your attention the way helium balloons and antigravity
machines do! That is why I like to start the way I did.

'''The solution above was taken from Dr. Math. For all the visual learners, i tried to find a video.. but failed to do so...'''

Note- http://mathforum.org/dr/math/ is a very good website to find help for any math problems. You should check it out!

=Basic Examples=
====Example 1====

Solve -2x > 2

we start by dividing both sides by -2 to solve the inequality

x < -1

====Example 2====
Solving linear inequalities is almost exactly like solving linear equations.

* Solve x + 3 < 0.

If they'd given "x + 3 = 0", we would know how to solve: we would have subtracted 3 from both sides. The same applies here.

x < -3

Then the solution is:

x < –3

====Example 3====

* Solve x – 4 > 0.

If they'd given "x – 4 = 0", then we would can solve by adding four to each side. The same applies here.
x >= 4

Then the solution is: x > 4

====Example 4====

* Solve 2x < 9.

If they had given "2x = 9", we would have divided the 2 from each side.

x <= 9/2

Then the solution is: x < 9/2

====Example 5====

* Solve (2x – 3)/4 < 2.
First, multiply through by 4. Since the "4" is positive, we don't have to flip the inequality sign:

(2x – 3)/4 < 2
(4) × (2x – 3)/4 < (4)(2)
2x – 3 < 8
2x < 11
x < 11/2 = 5.5

====Example 6====

* Solve 10 < 3x + 4 < 19.

This is what is called a "compound inequality". It works just like regular inequalities, except that it has three "sides". So, for instance, when we go to subtract the 4, I will have to subtract it from all three "sides".

=10 < 3x + 4 < 19
=6 < 3x < 15
=2 < x < 5

====Example 7====

* Solve 5x + 7 < 3(x + 1).

First we multiply through on the right-hand side, and then solve as usual:

5x + 7 < 3(x + 1)

5x + 7 < 3x + 3

2x + 7 < 3

2x < –4

x < –2

====Example 8====

* Solve <math>5(x-3)/2 < 2(3x+4)</math>

First, simplify the equation then proceed to isolate x.

* (5x-15)/2 < 6x+8
5x-15 > 2(6x+8)
5x > 12x+16+15
5x > 12x+31
5x-12x > 31
-7x > 31 ;; Dividing and multiplying by negative numbers switches the sign
<math>x<-31/7</math>

====Example 9====

* Solve <math>3 < |3x+4| </math>

Given an inequality with absolute values, say |x| > a then we assume it is –a > x > a.

For better and more in-depth information, please visit here: http://www.nipissingu.ca/calculus/tutorials/absolutevalue.html

* <math>3 < 3x+4 < -3 </math>
<math>3 -4 < 3x+4 -4 < -3 -4</math> ;; We minus 4 on both sides
<math>-1 < 3x < -7 </math> ;; Divide 3 on both sides
<math>-1/3 < x < -7/3 </math>

So the solution for <math>3 < |3x+4| </math> is <math>-1/3 < x < -7/3 </math>.

=Quadratic Inequalities=

To solve a quadratic inequality, follow these steps:

1. Solve the inequality as though it were an equation. The real solutions to the equation become boundary points for the solution to the inequality.

2. Make the boundary points solid circles if the original inequality includes equality; otherwise, make the boundary points open circles.

3. Select points from each of the regions created by the boundary points. Replace these “test points” in the original inequality.

4. If a test point satisfies the original inequality, then the region that contains that test point is part of the solutions.

5. Represent the solution in graphic form and in solution test form.

====Example 1====

Solve <math>x^2-2x-15>0</math>

Treat it like a normal quadratic equation and find the zeroes
* <math>x^2-2x-15=0</math>
<math>(x-5)(x+3)=0</math>
x-5=0 or x+3=0
x=5 or -3

The zeroes divide the number line into three regions

[[File:Crappy_number_line_thing_1.png]]

For region x>5, check for a number greater than 5 (eg.6).
*<math>(6)^2-2(6)-15>0</math>
9>0
All x-values greater than 5 will work.

For region x<-3, check for a number less than -3 (eg.-4).
*<math>(-4)^2-2(-4)-15>0</math>
9>0
All x-values less than -3 will work.

*Show answer using interval notation
<math>(-infinity,-3)U(5,infinity)</math>

Then, x>5 and x<-3.

====Example 2====

Solve <math>3x^2>-x+4</math>

Make one side equal to zero
*<math>3x^2+x-4>0</math>

Find the zeroes using the quadratic formula and factor

*(3x+4)(x-1)>0
3x+4=0 or x-1=0
x=<math>-(4/3)</math> or 1

For region x><math>-(4/3)</math>, check for a number less than <math>-(4/3)</math> (eg.-2).
*<math>3(-2)^2+(-2)-4>0</math>
6>0
All x-values less than <math>-4/3</math> will work.

For region x>1, check for a number greater than 1 (eg.4).
*<math>3(4)^2+(4)-4>0</math>
48>0
All x-values greater than 1 will work.

*Show values that produce an answer greater than 0 using interval notation
<math>(-infinity, -4/3)U(1,+infinity)</math>

====Example 3====

Solve for, <math>9x+9+3x^2>-x^2+7</math>

Move terms to one side so one side is zero
*<math>x+9+3x^2+x^2-7>0</math>
<math>4x^2+9x+2>0</math>

Then find the zeroes
*(x+2)(4x+1)>0
x+2=0 or 4x+1=0
x=<math>-2</math> or <math>-(1/4)</math>

To verify, check the to see if the numbers for region x<-2 are greater than zero by plugging in some number.

*<math>4x^2+9x+2>0</math>
<math>4(-3)^2+9(-3)+2>0</math>
<math>11>0</math>

Likewise, for the region x><math>-1/4</math>

*<math>4x^2+9x+2>0</math>
<math>4(1)^2+9(1)+2>0</math>
<math>15>0</math>

And lastly, some number between -2 and <math>-1/4</math>.

*<math>4x^2+9x+2>0</math>
<math>4(-1)^2+9(-1)+2>0</math>
<math>-3>0</math>
As you can see, -3 is not greater than zero.

Then in the end, we are assume that the solution of this question is, x<-2 and x><math>-1/4</math>.

====Example 4====

Solve for, <math>-9x^2+7+22x>11x+2-15x^2</math>

Move terms to one side so one side is zero
*<math>-9x^2+7+22x-11x-2+15x^2>0</math>
<math>6x+11x+5>0</math>

Then find the zeroes
*(6x+5)(x+1)>0
6x+5=0 or x+1=0
x=<math>-5/6</math> or -1

To verify, check the to see if the numbers for region x<<math>-5/6</math> are greater than zero by plugging in some number.

*<math>6x+11x+5>0</math>
<math>6(-4)+11(-4)+5>0</math>
<math>30>0</math>

Likewise, for the region x>-1

*<math>6x+11x+5>0</math>
<math>6(6)+11(6)+5>0</math>
<math>200>0</math>

And lastly, some number between <math>-5/6</math> and -1.

*<math>6x+11x+5>0</math>
<math>6(-.9)+11(-.9)+5>0</math>
<math>-2.86>0</math>
As you can see, it is not greater than zero.

Then in the end, we are assume that the solution of this question is, x<<math>-5/6</math> and x>-1.

====References====

http://www.purplemath.com/modules/ineqquad.htm

http://www.analyzemath.com/Inequalities_Polynomial/quadratic_inequalities.html

=Graphing Inequalities=

Number Line

1. Simplify the inequality you are going to graph.

<pre>E.g.

-2x2 + 5x < -6(x + 1)

-2x2 + 5x < -6x – 6</pre>

2. Move all terms to one side so the other is zero. ''(It will be easiest if the highest power variable is positive.)''

<pre>E.g.

0 < 2x2 -6x - 5x - 6

0 < 2x2 -11x – 6</pre>

3. Pretend that the inequality sign is an equal sign and find all values of the variable.

<pre>E.g.

0 = 2x2 -11x - 6

0 = (2x + 1)(x - 6)

2x + 1 = 0, x - 6 = 0

2x = -1, x = 6

x = -1/2</pre>

4. Draw a number line including the variable solutions (in order).

[[File:Inequalitiesnumberline1.jpg]]

5. Draw a circle on the points. If the inequality symbol means less than or more than (> or <), draw an empty circle over the variable solution(s). If it means less/more than and equal to (≤ or ≥) fill in that circle.

*In this case our equation was greater than zero, so use open circles.

[[File:Inequalitynumberlineshizz2.jpg]]

6. Take a number from each of the resulting intervals and plug it back into the equality. If you get a true statement once solved, shade this region of the number line.

In the interval from (-∞,-1/2) we will take -1 and plug it into the original inequality.

<pre>0 < 2x2 -11x - 6

0 < 2(-1)2 -11(-1) - 6

0 < 2(1) + 11 - 6

0 < 7

Zero is less than 7 is correct, so shade (-∞, -1/2) on the number line.</pre>

7. Next, on the interval from (-1/2, 6) we will use zero.

<pre>0 < 2(0)2 -11(0) - 6

0 < 0 + 0 - 6

0 < -6

Zero is not less than negative six, so do not shade (-1/2,6).</pre>

Lastly, we will take 10 from the interval (6,∞).

<pre>0 < 2(10)2 - 11(10) + 6

0 < 2(100) - 110 + 6

0 < 200 - 110 + 6

0 < 96

Zero is less than 96 is correct, so shade (6,∞) as well.</pre>

Use arrows on the end of shading to indicate that the interval continues into infinity. The completed number line:

[[File:Inequalitynumberlineblah3.jpg]]

=Tips=
If x ≥ y then 1/x ≤ 1/y

=Videos teaching Inequality=

===Solving Linear Inequalities===

{{#ev:youtube | VgDe_D8ojxw | 400}}

===Other Videos===

Video 1. http://www.khanacademy.org/video/inequalities?playlist=Algebra%20I%20Worked%20Examples
- This video touches upon the concept of inequality and has a basic word problem solved.

Video 2. http://www.khanacademy.org/video/interpreting-inequalities?playlist=Algebra%20I%20Worked%20Examples
- This video is about interpreting inequalities in word problems.

Video 3. http://www.khanacademy.org/video/solving-inequalities?playlist=Algebra%20I%20Worked%20Examples
- This video is about solving basic problems regarding inequalities.

Video 4. http://www.khanacademy.org/video/inequalities-using-addition-and-subtraction?playlist=ck12.org%20Algebra%201%20Examples
- This video solves random question about inequalities with addition and subtraction.

Video 5. http://www.khanacademy.org/video/inequalities-using-multiplication-and-division?playlist=ck12.org%20Algebra%201%20Examples
- This video solves random question about inequalities with multiplication and division

Video 6. http://www.khanacademy.org/video/quadratic-inequalities?playlist=Algebra
- This video explains Quadratic Inequalities.

=Useful Links=

* http://www.purplemath.com/modules/ineqsolv.htm
* http://www.mathsisfun.com/algebra/inequality-solving.html
* http://webmath.com/solverineq.html - Allows you to check your answers after solving an Inequality
* http://www.youtube.com/user/khanacademy - Youtube channel with helpful videos on many areas of Math including Inequalities

=Group 10=

[[Course:MATH110/003/Groups/Group 10/Basic skills project]]

Course:MATH110/Archive/2010-2011/003/Groups/Group 09/Basic Skills Project

2010-12-03T09:51:20Z

EllenTsang: /* Group 10 */

For the Basic Skills Project, Group 9 plans on focusing on Inequalities.

We will give several worked out examples to cover all cases of questions concerning this topic.

Also, we will include tips & tricks for how to solve more difficult problems and possible references related to the topic.

<pre>Let's work with Group 10 for the group project :D
Thanks Micha the only person from Group 10 who replied. -- Ellen</pre>

=What is an inequality?=

It basically means when:

* An equation includes < or > or ≤ or ≥.
** E.g. x + 1 ≤ 3

The symbols and meanings of the inequalities are as follows:

> means greater than

< means less than

≥ means greater than or equal to

≤ means less than or equal to

=When to change the sign in an inequality problem?=

This is where most of the students face problems while doing problems regarding inequality. This occurs when there is ambiguity whether to change the sign of the inequality.

To solve the confusions, we ONLY change the sign of the inequality to its corresponding opposite when we '''multiply or divide with a negative number'''

Equations can be thought of as balanced scales, where the total weight
on the left balances what is on the right. Inequalities are like
unbalanced scales, where all you know is which side is "down"
(heavier). So for example

3x > 6

can be thought of as 3 unknown weights labeled "X" on the left,
heavier than a 6-gram weight on the right.

Negative numbers complicate it a bit. A negative constant can be
thought of as a helium balloon (barely) able to lift a certain weight.
A negative number times a variable might mean there is an antigravity
machine under it so it pulls up as hard as it would normally push
down!

So the equation

-4y < -36

would be 4 Y's on the left with an antigravity machine, and a "-36"
gram balloon on the right. The right side is "heavier," which in this
case means not that it is pushing down more, but that it is pulling up
less!

To solve it, let us first replace the -36 gram balloon with a 36-gram
weight and an antigravity machine:

-(4y) < -(36)

Now turn off the antigravity machines:

4y > 36

Why did I reverse the direction of the "<"? That is the key to this
whole thing: antigravity machines are like turning the whole world
upside down, so the side that was down is now up:

-4y -36
^ ^
\ | |
\ |
\ |
\ |
\
\

becomes

4y 36
| |
| v /
| /
| /
v /
/
/

because 4y, which pulled up harder before, now pushes down harder.

Now we can work with positive numbers, and divide both weights by 4 to
get

y > 9

So any number BIGGER than 9 will work. For example, for y = 10,

-4y = -40 < -36

Do you see how the larger number, 10 > 9, becomes the smaller number
(-40 < -36) when it is multiplied by a negative number? That is the
key. The rule is that when you multiply an inequality by a negative
number, you have to reverse the direction. Or if you prefer, you can
do this:

-4y < -36 Add 4y
0 < 4y - 36 Add 36
36 < 4y Divide by 4
9 < y Reverse the whole inequality
y > 9

By avoiding multiplication by a negative number, I avoided the need to
reverse signs until the end.

Now that we have seen with our imagination what is going on, let us try
to prove the rule that if

a > b

then

-a < -b.

Start with the original inequality and subtract a from both sides:

0 > b - a

Now subtract b from both sides:

-b > -a

But that is the same as

-a < -b

which we were looking for. It is really pretty simple - so simple it
does not grab your attention the way helium balloons and antigravity
machines do! That is why I like to start the way I did.

'''The solution above was taken from Dr. Math. For all the visual learners, i tried to find a video.. but failed to do so...'''

Note- http://mathforum.org/dr/math/ is a very good website to find help for any math problems. You should check it out!

=Basic Examples=
====Example 1====

Solve -2x > 2

we start by dividing both sides by -2 to solve the inequality

x < -1

====Example 2====
Solving linear inequalities is almost exactly like solving linear equations.

* Solve x + 3 < 0.

If they'd given "x + 3 = 0", we would know how to solve: we would have subtracted 3 from both sides. The same applies here.

x < -3

Then the solution is:

x < –3

====Example 3====

* Solve x – 4 > 0.

If they'd given "x – 4 = 0", then we would can solve by adding four to each side. The same applies here.
x >= 4

Then the solution is: x > 4

====Example 4====

* Solve 2x < 9.

If they had given "2x = 9", we would have divided the 2 from each side.

x <= 9/2

Then the solution is: x < 9/2

====Example 5====

* Solve (2x – 3)/4 < 2.
First, multiply through by 4. Since the "4" is positive, we don't have to flip the inequality sign:

(2x – 3)/4 < 2
(4) × (2x – 3)/4 < (4)(2)
2x – 3 < 8
2x < 11
x < 11/2 = 5.5

====Example 6====

* Solve 10 < 3x + 4 < 19.

This is what is called a "compound inequality". It works just like regular inequalities, except that it has three "sides". So, for instance, when we go to subtract the 4, I will have to subtract it from all three "sides".

=10 < 3x + 4 < 19
=6 < 3x < 15
=2 < x < 5

====Example 7====

* Solve 5x + 7 < 3(x + 1).

First we multiply through on the right-hand side, and then solve as usual:

5x + 7 < 3(x + 1)

5x + 7 < 3x + 3

2x + 7 < 3

2x < –4

x < –2

====Example 8====

* Solve <math>5(x-3)/2 < 2(3x+4)</math>

First, simplify the equation then proceed to isolate x.

* (5x-15)/2 < 6x+8
5x-15 > 2(6x+8)
5x > 12x+16+15
5x > 12x+31
5x-12x > 31
-7x > 31 ;; Dividing and multiplying by negative numbers switches the sign
<math>x<-31/7</math>

====Example 9====

* Solve <math>3 < |3x+4| </math>

Given an inequality with absolute values, say |x| > a then we assume it is –a > x > a.

For better and more in-depth information, please visit here: http://www.nipissingu.ca/calculus/tutorials/absolutevalue.html

* <math>3 < 3x+4 < -3 </math>
<math>3 -4 < 3x+4 -4 < -3 -4</math> ;; We minus 4 on both sides
<math>-1 < 3x < -7 </math> ;; Divide 3 on both sides
<math>-1/3 < x < -7/3 </math>

So the solution for <math>3 < |3x+4| </math> is <math>-1/3 < x < -7/3 </math>.

=Quadratic Inequalities=

To solve a quadratic inequality, follow these steps:

1. Solve the inequality as though it were an equation. The real solutions to the equation become boundary points for the solution to the inequality.

2. Make the boundary points solid circles if the original inequality includes equality; otherwise, make the boundary points open circles.

3. Select points from each of the regions created by the boundary points. Replace these “test points” in the original inequality.

4. If a test point satisfies the original inequality, then the region that contains that test point is part of the solutions.

5. Represent the solution in graphic form and in solution test form.

====Example 1====

Solve <math>x^2-2x-15>0</math>

Treat it like a normal quadratic equation and find the zeroes
* <math>x^2-2x-15=0</math>
<math>(x-5)(x+3)=0</math>
x-5=0 or x+3=0
x=5 or -3

The zeroes divide the number line into three regions

[[File:Crappy_number_line_thing_1.png]]

For region x>5, check for a number greater than 5 (eg.6).
*<math>(6)^2-2(6)-15>0</math>
9>0
All x-values greater than 5 will work.

For region x<-3, check for a number less than -3 (eg.-4).
*<math>(-4)^2-2(-4)-15>0</math>
9>0
All x-values less than -3 will work.

*Show answer using interval notation
<math>(-infinity,-3)U(5,infinity)</math>

Then, x>5 and x<-3.

====Example 2====

Solve <math>3x^2>-x+4</math>

Make one side equal to zero
*<math>3x^2+x-4>0</math>

Find the zeroes using the quadratic formula and factor

*(3x+4)(x-1)>0
3x+4=0 or x-1=0
x=<math>-(4/3)</math> or 1

For region x><math>-(4/3)</math>, check for a number less than <math>-(4/3)</math> (eg.-2).
*<math>3(-2)^2+(-2)-4>0</math>
6>0
All x-values less than <math>-4/3</math> will work.

For region x>1, check for a number greater than 1 (eg.4).
*<math>3(4)^2+(4)-4>0</math>
48>0
All x-values greater than 1 will work.

*Show values that produce an answer greater than 0 using interval notation
<math>(-infinity, -4/3)U(1,+infinity)</math>

====Example 3====

Solve for, <math>9x+9+3x^2>-x^2+7</math>

Move terms to one side so one side is zero
*<math>x+9+3x^2+x^2-7>0</math>
<math>4x^2+9x+2>0</math>

Then find the zeroes
*(x+2)(4x+1)>0
x+2=0 or 4x+1=0
x=<math>-2</math> or <math>-(1/4)</math>

To verify, check the to see if the numbers for region x<-2 are greater than zero by plugging in some number.

*<math>4x^2+9x+2>0</math>
<math>4(-3)^2+9(-3)+2>0</math>
<math>11>0</math>

Likewise, for the region x><math>-1/4</math>

*<math>4x^2+9x+2>0</math>
<math>4(1)^2+9(1)+2>0</math>
<math>15>0</math>

And lastly, some number between -2 and <math>-1/4</math>.

*<math>4x^2+9x+2>0</math>
<math>4(-1)^2+9(-1)+2>0</math>
<math>-3>0</math>
As you can see, -3 is not greater than zero.

Then in the end, we are assume that the solution of this question is, x<-2 and x><math>-1/4</math>.

====Example 4====

Solve for, <math>-9x^2+7+22x>11x+2-15x^2</math>

Move terms to one side so one side is zero
*<math>-9x^2+7+22x-11x-2+15x^2>0</math>
<math>6x+11x+5>0</math>

Then find the zeroes
*(6x+5)(x+1)>0
6x+5=0 or x+1=0
x=<math>-5/6</math> or -1

To verify, check the to see if the numbers for region x<<math>-5/6</math> are greater than zero by plugging in some number.

*<math>6x+11x+5>0</math>
<math>6(-4)+11(-4)+5>0</math>
<math>30>0</math>

Likewise, for the region x>-1

*<math>6x+11x+5>0</math>
<math>6(6)+11(6)+5>0</math>
<math>200>0</math>

And lastly, some number between <math>-5/6</math> and -1.

*<math>6x+11x+5>0</math>
<math>6(-.9)+11(-.9)+5>0</math>
<math>-2.86>0</math>
As you can see, it is not greater than zero.

Then in the end, we are assume that the solution of this question is, x<<math>-5/6</math> and x>-1.

====References====

http://www.purplemath.com/modules/ineqquad.htm

http://www.analyzemath.com/Inequalities_Polynomial/quadratic_inequalities.html

=Graphing Inequalities=

Number Line

1. Simplify the inequality you are going to graph.

<pre>E.g.

-2x2 + 5x < -6(x + 1)

-2x2 + 5x < -6x – 6</pre>

2. Move all terms to one side so the other is zero. ''(It will be easiest if the highest power variable is positive.)''

<pre>E.g.

0 < 2x2 -6x - 5x - 6

0 < 2x2 -11x – 6</pre>

3. Pretend that the inequality sign is an equal sign and find all values of the variable.

<pre>E.g.

0 = 2x2 -11x - 6

0 = (2x + 1)(x - 6)

2x + 1 = 0, x - 6 = 0

2x = -1, x = 6

x = -1/2</pre>

4. Draw a number line including the variable solutions (in order).

[[File:Inequalitiesnumberline1.jpg]]

5. Draw a circle on the points. If the inequality symbol means less than or more than (> or <), draw an empty circle over the variable solution(s). If it means less/more than and equal to (≤ or ≥) fill in that circle.

*In this case our equation was greater than zero, so use open circles.

[[File:Inequalitynumberlineshizz2.jpg]]

6. Take a number from each of the resulting intervals and plug it back into the equality. If you get a true statement once solved, shade this region of the number line.

In the interval from (-∞,-1/2) we will take -1 and plug it into the original inequality.

<pre>0 < 2x2 -11x - 6

0 < 2(-1)2 -11(-1) - 6

0 < 2(1) + 11 - 6

0 < 7

Zero is less than 7 is correct, so shade (-∞, -1/2) on the number line.</pre>

7. Next, on the interval from (-1/2, 6) we will use zero.

<pre>0 < 2(0)2 -11(0) - 6

0 < 0 + 0 - 6

0 < -6

Zero is not less than negative six, so do not shade (-1/2,6).</pre>

Lastly, we will take 10 from the interval (6,∞).

<pre>0 < 2(10)2 - 11(10) + 6

0 < 2(100) - 110 + 6

0 < 200 - 110 + 6

0 < 96

Zero is less than 96 is correct, so shade (6,∞) as well.</pre>

Use arrows on the end of shading to indicate that the interval continues into infinity. The completed number line:

[[File:Inequalitynumberlineblah3.jpg]]

=Tips=
If x ≥ y then 1/x ≤ 1/y

=Videos teaching Inequality=

===Solving Linear Inequalities===

{{#ev:youtube | VgDe_D8ojxw | 400}}
<ref>Algebra: Solving Inequalities - http://www.youtube.com/watch?v=VgDe_D8ojxw
</ref>

===Other Videos===

Video 1. http://www.khanacademy.org/video/inequalities?playlist=Algebra%20I%20Worked%20Examples
- This video touches upon the concept of inequality and has a basic word problem solved.

Video 2. http://www.khanacademy.org/video/interpreting-inequalities?playlist=Algebra%20I%20Worked%20Examples
- This video is about interpreting inequalities in word problems.

Video 3. http://www.khanacademy.org/video/solving-inequalities?playlist=Algebra%20I%20Worked%20Examples
- This video is about solving basic problems regarding inequalities.

Video 4. http://www.khanacademy.org/video/inequalities-using-addition-and-subtraction?playlist=ck12.org%20Algebra%201%20Examples
- This video solves random question about inequalities with addition and subtraction.

Video 5. http://www.khanacademy.org/video/inequalities-using-multiplication-and-division?playlist=ck12.org%20Algebra%201%20Examples
- This video solves random question about inequalities with multiplication and division

Video 6. http://www.khanacademy.org/video/quadratic-inequalities?playlist=Algebra
- This video explains Quadratic Inequalities.

=Useful Links=

* http://www.purplemath.com/modules/ineqsolv.htm
* http://www.mathsisfun.com/algebra/inequality-solving.html
* http://webmath.com/solverineq.html - Allows you to check your answers after solving an Inequality
* http://www.youtube.com/user/khanacademy - Youtube channel with helpful videos on many areas of Math including Inequalities

=Group 10=

[[Course:MATH110/003/Groups/Group 10/Basic skills project]]

Course:MATH110/Archive/2010-2011/003/Groups/Group 10/Basic skills project

2010-12-03T09:50:24Z

EllenTsang:

Here is what we have so far from our Group 10 Page. We will be adding more. Can you help with formatting like you have done for your section?

Solving Quadratic Inequalities

To solve a quadratic inequality, follow these steps:

1. Solve the inequality as though it were an equation. The real solutions to the equation become boundary points for the solution to the inequality.

2. Make the boundary points solid circles if the original inequality includes equality; otherwise, make the boundary points open circles.

3. Select points from each of the regions created by the boundary points. Replace these “test points” in the original inequality.

4. If a test point satisfies the original inequality, then the region that contains that test point is part of the solutions.

5. Represent the solution in graphic form and in solution test form.

Example 1: Solve (x-3)(x+2)>0

By the zero product property, x-3=0 or x+2=0, x=3 and x=-2

Make the boundary points.

Here, the boundary points are open circles because the original inequality does not include equality.

Select points from different regions created.

Three regions are created:

X=-3

X=0

X=4

See if the test points satisfy the original inequality

(x-3)(x+2)>0

(-3-3)(-3+2)>0

6>0 therefore, it works

(x-3)(x+2)>0

(0-3)(0+2)>0

-6>0 no, it does not work

(x-3)(x+2)>0

(4-3)(-3+2)>0

6>0 therefore, it works

One of the first steps to solving inequalities is the symbol and meanings of the inequalities.

> means greater than

< means less than

≥ means greater than or equal to

≤ means less than or equal to

Because you want to get x alone, you can rather:

• Add or subtract a number from both sides

• Multiply or divide both sides by a positive number

• Simplify a side

However, doing the following things will change the direction of the inequality:

• Multiplying or dividing both sides by a negative number

• Swapping left and right hand sides

Have a look at: http://www.mathsisfun.com/algebra/inequality-solving.html for more information.

If you’re having trouble in the book there is a good section starting on page 1061 which is a review of algebra and sets of real numbers. It gives number lines and shows inequalities to match.

If you want to check your answer on how to solve an inequality try: http://webmath.com/solverineq.html

'''What is an inequality?'''

It is exactly what it sounds, it's definition is that two numbers are not equal to each other. Such as "y" is not equal to "x".

1. It may be that "x" is greater than "y" which can be written as

x>y.

2. x is greater than or equal to y

x <math>\geq</math> y

3. x is less than y

x < y

4. x is less than or equal to y

x <math>\leq</math> y

5. x is not equal to y

x <math>\neq</math> y

'''How to represent the solutions of inequalities?'''

They can be represented in an interval notation ('''such as the notation used for defining a domain'''), as it specifies what group of number that fits under this rule.

'''How to solve linear inequalities?'''

Solve it like a linear equation.

'''Goal''': to isolate the variable so that you can determine the interval of "x"

'''Addition/Subtraction'''

The equation can be seen as a normal equation, instead of the greater than, less than, not equal to sign, just let the = sign replace them for now.
Then solve the equation algebraically, afterwards substitute the equations' original inequality sign.

'''EX'''

x + 8 > 10 <math>\longrightarrow</math> substitute the = sign

x + 8 = 10 <math>\longrightarrow</math> isolate the variable

x = 2 <math>\longrightarrow</math> replace the equal sign with the original inequality sign

x > 2

In this example "x" can be any number grater than two, and thus can be written like this (2,<math>\infty</math>)

'''Multiplication/Division'''

is similar to solving addition/subtraction equations

if 2x <math>\geq</math> 5 <math>\longrightarrow</math> then to isolate x, divide both sides by 2

x <math>\geq</math> <math>\{5 \over 2}\</math>

'''How to solve quadratic inequalities?'''

Let us look at an example of a quadratic inequality:

-x2 + 9 < 0

Our first step is to associate this with and equation. Therefore:

y= -x2 + 9

The next step is to find out where the equation cuts the graph on the x-axis. This means equating it to 0

-x2 + 9 = 0

Now we can solve:

-x2 + 9 = 0

x2 + 9 = 0

(x+3) (x-3) = 0

x=-3 x=3 Now we know that our quadratic equation crosses the x-axis at these values

We now get 3 different intervals at the points where the x-axis is cut

1.) x < -3

2.) -3 < x < 3

3.) x > 3

The next step is to find out on which intervals is the graph below the x-axis. In order to do this we must look at our original equation:

y = -x2 + 9 since it is a negative quadratic we can say that the graph would be facing down. Thus, in order to solve our original inequality of

-x2 + 9 < 0 we must look at where our y values are less than 0

From this we can see that the solution is: x < -3 or x > 3

'''Online References/extension'''

[http://www.purplemath.com/modules/ineqgrph.htm] Written step-by-step explanation

[http://www.youtube.com/watch?v=0X-bMeIN53I] Video Explanation

http://www.youtube.com/watch?v=VgDe_D8ojxw

Course:MATH110/Archive/2010-2011/003/Groups/Group 09/Basic Skills Project

2010-12-03T09:46:31Z

EllenTsang: /* Useful Links */

For the Basic Skills Project, Group 9 plans on focusing on Inequalities.

We will give several worked out examples to cover all cases of questions concerning this topic.

Also, we will include tips & tricks for how to solve more difficult problems and possible references related to the topic.

<pre>Let's work with Group 10 for the group project :D
Thanks Micha the only person from Group 10 who replied. -- Ellen</pre>

=What is an inequality?=

It basically means when:

* An equation includes < or > or ≤ or ≥.
** E.g. x + 1 ≤ 3

The symbols and meanings of the inequalities are as follows:

> means greater than

< means less than

≥ means greater than or equal to

≤ means less than or equal to

=When to change the sign in an inequality problem?=

This is where most of the students face problems while doing problems regarding inequality. This occurs when there is ambiguity whether to change the sign of the inequality.

To solve the confusions, we ONLY change the sign of the inequality to its corresponding opposite when we '''multiply or divide with a negative number'''

Equations can be thought of as balanced scales, where the total weight
on the left balances what is on the right. Inequalities are like
unbalanced scales, where all you know is which side is "down"
(heavier). So for example

3x > 6

can be thought of as 3 unknown weights labeled "X" on the left,
heavier than a 6-gram weight on the right.

Negative numbers complicate it a bit. A negative constant can be
thought of as a helium balloon (barely) able to lift a certain weight.
A negative number times a variable might mean there is an antigravity
machine under it so it pulls up as hard as it would normally push
down!

So the equation

-4y < -36

would be 4 Y's on the left with an antigravity machine, and a "-36"
gram balloon on the right. The right side is "heavier," which in this
case means not that it is pushing down more, but that it is pulling up
less!

To solve it, let us first replace the -36 gram balloon with a 36-gram
weight and an antigravity machine:

-(4y) < -(36)

Now turn off the antigravity machines:

4y > 36

Why did I reverse the direction of the "<"? That is the key to this
whole thing: antigravity machines are like turning the whole world
upside down, so the side that was down is now up:

-4y -36
^ ^
\ | |
\ |
\ |
\ |
\
\

becomes

4y 36
| |
| v /
| /
| /
v /
/
/

because 4y, which pulled up harder before, now pushes down harder.

Now we can work with positive numbers, and divide both weights by 4 to
get

y > 9

So any number BIGGER than 9 will work. For example, for y = 10,

-4y = -40 < -36

Do you see how the larger number, 10 > 9, becomes the smaller number
(-40 < -36) when it is multiplied by a negative number? That is the
key. The rule is that when you multiply an inequality by a negative
number, you have to reverse the direction. Or if you prefer, you can
do this:

-4y < -36 Add 4y
0 < 4y - 36 Add 36
36 < 4y Divide by 4
9 < y Reverse the whole inequality
y > 9

By avoiding multiplication by a negative number, I avoided the need to
reverse signs until the end.

Now that we have seen with our imagination what is going on, let us try
to prove the rule that if

a > b

then

-a < -b.

Start with the original inequality and subtract a from both sides:

0 > b - a

Now subtract b from both sides:

-b > -a

But that is the same as

-a < -b

which we were looking for. It is really pretty simple - so simple it
does not grab your attention the way helium balloons and antigravity
machines do! That is why I like to start the way I did.

'''The solution above was taken from Dr. Math. For all the visual learners, i tried to find a video.. but failed to do so...'''

Note- http://mathforum.org/dr/math/ is a very good website to find help for any math problems. You should check it out!

=Basic Examples=
====Example 1====

Solve -2x > 2

we start by dividing both sides by -2 to solve the inequality

x < -1

====Example 2====
Solving linear inequalities is almost exactly like solving linear equations.

* Solve x + 3 < 0.

If they'd given "x + 3 = 0", we would know how to solve: we would have subtracted 3 from both sides. The same applies here.

x < -3

Then the solution is:

x < –3

====Example 3====

* Solve x – 4 > 0.

If they'd given "x – 4 = 0", then we would can solve by adding four to each side. The same applies here.
x >= 4

Then the solution is: x > 4

====Example 4====

* Solve 2x < 9.

If they had given "2x = 9", we would have divided the 2 from each side.

x <= 9/2

Then the solution is: x < 9/2

====Example 5====

* Solve (2x – 3)/4 < 2.
First, multiply through by 4. Since the "4" is positive, we don't have to flip the inequality sign:

(2x – 3)/4 < 2
(4) × (2x – 3)/4 < (4)(2)
2x – 3 < 8
2x < 11
x < 11/2 = 5.5

====Example 6====

* Solve 10 < 3x + 4 < 19.

This is what is called a "compound inequality". It works just like regular inequalities, except that it has three "sides". So, for instance, when we go to subtract the 4, I will have to subtract it from all three "sides".

=10 < 3x + 4 < 19
=6 < 3x < 15
=2 < x < 5

====Example 7====

* Solve 5x + 7 < 3(x + 1).

First we multiply through on the right-hand side, and then solve as usual:

5x + 7 < 3(x + 1)

5x + 7 < 3x + 3

2x + 7 < 3

2x < –4

x < –2

====Example 8====

* Solve <math>5(x-3)/2 < 2(3x+4)</math>

First, simplify the equation then proceed to isolate x.

* (5x-15)/2 < 6x+8
5x-15 > 2(6x+8)
5x > 12x+16+15
5x > 12x+31
5x-12x > 31
-7x > 31 ;; Dividing and multiplying by negative numbers switches the sign
<math>x<-31/7</math>

====Example 9====

* Solve <math>3 < |3x+4| </math>

Given an inequality with absolute values, say |x| > a then we assume it is –a > x > a.

For better and more in-depth information, please visit here: http://www.nipissingu.ca/calculus/tutorials/absolutevalue.html

* <math>3 < 3x+4 < -3 </math>
<math>3 -4 < 3x+4 -4 < -3 -4</math> ;; We minus 4 on both sides
<math>-1 < 3x < -7 </math> ;; Divide 3 on both sides
<math>-1/3 < x < -7/3 </math>

So the solution for <math>3 < |3x+4| </math> is <math>-1/3 < x < -7/3 </math>.

=Quadratic Inequalities=

To solve a quadratic inequality, follow these steps:

1. Solve the inequality as though it were an equation. The real solutions to the equation become boundary points for the solution to the inequality.

2. Make the boundary points solid circles if the original inequality includes equality; otherwise, make the boundary points open circles.

3. Select points from each of the regions created by the boundary points. Replace these “test points” in the original inequality.

4. If a test point satisfies the original inequality, then the region that contains that test point is part of the solutions.

5. Represent the solution in graphic form and in solution test form.

====Example 1====

Solve <math>x^2-2x-15>0</math>

Treat it like a normal quadratic equation and find the zeroes
* <math>x^2-2x-15=0</math>
<math>(x-5)(x+3)=0</math>
x-5=0 or x+3=0
x=5 or -3

The zeroes divide the number line into three regions

[[File:Crappy_number_line_thing_1.png]]

For region x>5, check for a number greater than 5 (eg.6).
*<math>(6)^2-2(6)-15>0</math>
9>0
All x-values greater than 5 will work.

For region x<-3, check for a number less than -3 (eg.-4).
*<math>(-4)^2-2(-4)-15>0</math>
9>0
All x-values less than -3 will work.

*Show answer using interval notation
<math>(-infinity,-3)U(5,infinity)</math>

Then, x>5 and x<-3.

====Example 2====

Solve <math>3x^2>-x+4</math>

Make one side equal to zero
*<math>3x^2+x-4>0</math>

Find the zeroes using the quadratic formula and factor

*(3x+4)(x-1)>0
3x+4=0 or x-1=0
x=<math>-(4/3)</math> or 1

For region x><math>-(4/3)</math>, check for a number less than <math>-(4/3)</math> (eg.-2).
*<math>3(-2)^2+(-2)-4>0</math>
6>0
All x-values less than <math>-4/3</math> will work.

For region x>1, check for a number greater than 1 (eg.4).
*<math>3(4)^2+(4)-4>0</math>
48>0
All x-values greater than 1 will work.

*Show values that produce an answer greater than 0 using interval notation
<math>(-infinity, -4/3)U(1,+infinity)</math>

====Example 3====

Solve for, <math>9x+9+3x^2>-x^2+7</math>

Move terms to one side so one side is zero
*<math>x+9+3x^2+x^2-7>0</math>
<math>4x^2+9x+2>0</math>

Then find the zeroes
*(x+2)(4x+1)>0
x+2=0 or 4x+1=0
x=<math>-2</math> or <math>-(1/4)</math>

To verify, check the to see if the numbers for region x<-2 are greater than zero by plugging in some number.

*<math>4x^2+9x+2>0</math>
<math>4(-3)^2+9(-3)+2>0</math>
<math>11>0</math>

Likewise, for the region x><math>-1/4</math>

*<math>4x^2+9x+2>0</math>
<math>4(1)^2+9(1)+2>0</math>
<math>15>0</math>

And lastly, some number between -2 and <math>-1/4</math>.

*<math>4x^2+9x+2>0</math>
<math>4(-1)^2+9(-1)+2>0</math>
<math>-3>0</math>
As you can see, -3 is not greater than zero.

Then in the end, we are assume that the solution of this question is, x<-2 and x><math>-1/4</math>.

====Example 4====

Solve for, <math>-9x^2+7+22x>11x+2-15x^2</math>

Move terms to one side so one side is zero
*<math>-9x^2+7+22x-11x-2+15x^2>0</math>
<math>6x+11x+5>0</math>

Then find the zeroes
*(6x+5)(x+1)>0
6x+5=0 or x+1=0
x=<math>-5/6</math> or -1

To verify, check the to see if the numbers for region x<<math>-5/6</math> are greater than zero by plugging in some number.

*<math>6x+11x+5>0</math>
<math>6(-4)+11(-4)+5>0</math>
<math>30>0</math>

Likewise, for the region x>-1

*<math>6x+11x+5>0</math>
<math>6(6)+11(6)+5>0</math>
<math>200>0</math>

And lastly, some number between <math>-5/6</math> and -1.

*<math>6x+11x+5>0</math>
<math>6(-.9)+11(-.9)+5>0</math>
<math>-2.86>0</math>
As you can see, it is not greater than zero.

Then in the end, we are assume that the solution of this question is, x<<math>-5/6</math> and x>-1.

====References====

http://www.purplemath.com/modules/ineqquad.htm

http://www.analyzemath.com/Inequalities_Polynomial/quadratic_inequalities.html

=Graphing Inequalities=

Number Line

1. Simplify the inequality you are going to graph.

<pre>E.g.

-2x2 + 5x < -6(x + 1)

-2x2 + 5x < -6x – 6</pre>

2. Move all terms to one side so the other is zero. ''(It will be easiest if the highest power variable is positive.)''

<pre>E.g.

0 < 2x2 -6x - 5x - 6

0 < 2x2 -11x – 6</pre>

3. Pretend that the inequality sign is an equal sign and find all values of the variable.

<pre>E.g.

0 = 2x2 -11x - 6

0 = (2x + 1)(x - 6)

2x + 1 = 0, x - 6 = 0

2x = -1, x = 6

x = -1/2</pre>

4. Draw a number line including the variable solutions (in order).

[[File:Inequalitiesnumberline1.jpg]]

5. Draw a circle on the points. If the inequality symbol means less than or more than (> or <), draw an empty circle over the variable solution(s). If it means less/more than and equal to (≤ or ≥) fill in that circle.

*In this case our equation was greater than zero, so use open circles.

[[File:Inequalitynumberlineshizz2.jpg]]

6. Take a number from each of the resulting intervals and plug it back into the equality. If you get a true statement once solved, shade this region of the number line.

In the interval from (-∞,-1/2) we will take -1 and plug it into the original inequality.

<pre>0 < 2x2 -11x - 6

0 < 2(-1)2 -11(-1) - 6

0 < 2(1) + 11 - 6

0 < 7

Zero is less than 7 is correct, so shade (-∞, -1/2) on the number line.</pre>

7. Next, on the interval from (-1/2, 6) we will use zero.

<pre>0 < 2(0)2 -11(0) - 6

0 < 0 + 0 - 6

0 < -6

Zero is not less than negative six, so do not shade (-1/2,6).</pre>

Lastly, we will take 10 from the interval (6,∞).

<pre>0 < 2(10)2 - 11(10) + 6

0 < 2(100) - 110 + 6

0 < 200 - 110 + 6

0 < 96

Zero is less than 96 is correct, so shade (6,∞) as well.</pre>

Use arrows on the end of shading to indicate that the interval continues into infinity. The completed number line:

[[File:Inequalitynumberlineblah3.jpg]]

=Tips=
If x ≥ y then 1/x ≤ 1/y

=Videos teaching Inequality=

===Solving Linear Inequalities===

{{#ev:youtube | VgDe_D8ojxw | 400}}
<ref>Algebra: Solving Inequalities - http://www.youtube.com/watch?v=VgDe_D8ojxw
</ref>

===Other Videos===

Video 1. http://www.khanacademy.org/video/inequalities?playlist=Algebra%20I%20Worked%20Examples
- This video touches upon the concept of inequality and has a basic word problem solved.

Video 2. http://www.khanacademy.org/video/interpreting-inequalities?playlist=Algebra%20I%20Worked%20Examples
- This video is about interpreting inequalities in word problems.

Video 3. http://www.khanacademy.org/video/solving-inequalities?playlist=Algebra%20I%20Worked%20Examples
- This video is about solving basic problems regarding inequalities.

Video 4. http://www.khanacademy.org/video/inequalities-using-addition-and-subtraction?playlist=ck12.org%20Algebra%201%20Examples
- This video solves random question about inequalities with addition and subtraction.

Video 5. http://www.khanacademy.org/video/inequalities-using-multiplication-and-division?playlist=ck12.org%20Algebra%201%20Examples
- This video solves random question about inequalities with multiplication and division

Video 6. http://www.khanacademy.org/video/quadratic-inequalities?playlist=Algebra
- This video explains Quadratic Inequalities.

=Useful Links=

* http://www.purplemath.com/modules/ineqsolv.htm
* http://www.mathsisfun.com/algebra/inequality-solving.html
* http://webmath.com/solverineq.html - Allows you to check your answers after solving an Inequality
* http://www.youtube.com/user/khanacademy - Youtube channel with helpful videos on many areas of Math including Inequalities

=Group 10=

Here is what we have so far from our Group 10 Page. We will be adding more. Can you help with formatting like you have done for your section?

Solving Quadratic Inequalities

To solve a quadratic inequality, follow these steps:

1. Solve the inequality as though it were an equation. The real solutions to the equation become boundary points for the solution to the inequality.

2. Make the boundary points solid circles if the original inequality includes equality; otherwise, make the boundary points open circles.

3. Select points from each of the regions created by the boundary points. Replace these “test points” in the original inequality.

4. If a test point satisfies the original inequality, then the region that contains that test point is part of the solutions.

5. Represent the solution in graphic form and in solution test form.

Example 1: Solve (x-3)(x+2)>0

By the zero product property, x-3=0 or x+2=0, x=3 and x=-2

Make the boundary points.

Here, the boundary points are open circles because the original inequality does not include equality.

Select points from different regions created.

Three regions are created:

X=-3

X=0

X=4

See if the test points satisfy the original inequality

(x-3)(x+2)>0

(-3-3)(-3+2)>0

6>0 therefore, it works

(x-3)(x+2)>0

(0-3)(0+2)>0

-6>0 no, it does not work

(x-3)(x+2)>0

(4-3)(-3+2)>0

6>0 therefore, it works

Because you want to get x alone, you can rather:

• Add or subtract a number from both sides

• Multiply or divide both sides by a positive number

• Simplify a side

However, doing the following things will change the direction of the inequality:

• Multiplying or dividing both sides by a negative number

• Swapping left and right hand sides

Have a look at: http://www.mathsisfun.com/algebra/inequality-solving.html for more information.

If you’re having trouble in the book there is a good section starting on page 1061 which is a review of algebra and sets of real numbers. It gives number lines and shows inequalities to match.

If you want to check your answer on how to solve an inequality try: http://webmath.com/solverineq.html

'''What is an inequality?'''

It is exactly what it sounds, it's definition is that two numbers are not equal to each other. Such as "y" is not equal to "x".

1. It may be that "x" is greater than "y" which can be written as

x>y.

2. x is greater than or equal to y

x <math>\geq</math> y

3. x is less than y

x < y

4. x is less than or equal to y

x <math>\leq</math> y

5. x is not equal to y

x <math>\neq</math> y

'''How to represent the solutions of inequalities?'''

They can be represented in an interval notation ('''such as the notation used for defining a domain'''), as it specifies what group of number that fits under this rule.

'''How to solve linear inequalities?'''

Solve it like a linear equation.

'''Goal''': to isolate the variable so that you can determine the interval of "x"

'''Addition/Subtraction'''

The equation can be seen as a normal equation, instead of the greater than, less than, not equal to sign, just let the = sign replace them for now.
Then solve the equation algebraically, afterwards substitute the equations' original inequality sign.

'''EX'''

x + 8 > 10 <math>\longrightarrow</math> substitute the = sign

x + 8 = 10 <math>\longrightarrow</math> isolate the variable

x = 2 <math>\longrightarrow</math> replace the equal sign with the original inequality sign

x > 2

In this example "x" can be any number grater than two, and thus can be written like this (2,<math>\infty</math>)

'''Multiplication/Division'''

is similar to solving addition/subtraction equations

if 2x <math>\geq</math> 5 <math>\longrightarrow</math> then to isolate x, divide both sides by 2

x <math>\geq</math> <math>\{5 \over 2}\</math>

'''How to solve quadratic inequalities?'''

Let us look at an example of a quadratic inequality:

-x2 + 9 < 0

Our first step is to associate this with and equation. Therefore:

y= -x2 + 9

The next step is to find out where the equation cuts the graph on the x-axis. This means equating it to 0

-x2 + 9 = 0

Now we can solve:

-x2 + 9 = 0

x2 + 9 = 0

(x+3) (x-3) = 0

x=-3 x=3 Now we know that our quadratic equation crosses the x-axis at these values

We now get 3 different intervals at the points where the x-axis is cut

1.) x < -3

2.) -3 < x < 3

3.) x > 3

The next step is to find out on which intervals is the graph below the x-axis. In order to do this we must look at our original equation:

y = -x2 + 9 since it is a negative quadratic we can say that the graph would be facing down. Thus, in order to solve our original inequality of

-x2 + 9 < 0 we must look at where our y values are less than 0

From this we can see that the solution is: x < -3 or x > 3

'''Online References/extension'''

[http://www.purplemath.com/modules/ineqgrph.htm] Written step-by-step explanation

[http://www.youtube.com/watch?v=0X-bMeIN53I] Video Explanation

http://www.youtube.com/watch?v=VgDe_D8ojxw

Course:MATH110/Archive/2010-2011/003/Groups/Group 09/Basic Skills Project

2010-12-03T09:44:18Z

EllenTsang: /* Videos teaching Inequality */

For the Basic Skills Project, Group 9 plans on focusing on Inequalities.

We will give several worked out examples to cover all cases of questions concerning this topic.

Also, we will include tips & tricks for how to solve more difficult problems and possible references related to the topic.

<pre>Let's work with Group 10 for the group project :D
Thanks Micha the only person from Group 10 who replied. -- Ellen</pre>

=What is an inequality?=

It basically means when:

* An equation includes < or > or ≤ or ≥.
** E.g. x + 1 ≤ 3

The symbols and meanings of the inequalities are as follows:

> means greater than

< means less than

≥ means greater than or equal to

≤ means less than or equal to

=When to change the sign in an inequality problem?=

This is where most of the students face problems while doing problems regarding inequality. This occurs when there is ambiguity whether to change the sign of the inequality.

To solve the confusions, we ONLY change the sign of the inequality to its corresponding opposite when we '''multiply or divide with a negative number'''

Equations can be thought of as balanced scales, where the total weight
on the left balances what is on the right. Inequalities are like
unbalanced scales, where all you know is which side is "down"
(heavier). So for example

3x > 6

can be thought of as 3 unknown weights labeled "X" on the left,
heavier than a 6-gram weight on the right.

Negative numbers complicate it a bit. A negative constant can be
thought of as a helium balloon (barely) able to lift a certain weight.
A negative number times a variable might mean there is an antigravity
machine under it so it pulls up as hard as it would normally push
down!

So the equation

-4y < -36

would be 4 Y's on the left with an antigravity machine, and a "-36"
gram balloon on the right. The right side is "heavier," which in this
case means not that it is pushing down more, but that it is pulling up
less!

To solve it, let us first replace the -36 gram balloon with a 36-gram
weight and an antigravity machine:

-(4y) < -(36)

Now turn off the antigravity machines:

4y > 36

Why did I reverse the direction of the "<"? That is the key to this
whole thing: antigravity machines are like turning the whole world
upside down, so the side that was down is now up:

-4y -36
^ ^
\ | |
\ |
\ |
\ |
\
\

becomes

4y 36
| |
| v /
| /
| /
v /
/
/

because 4y, which pulled up harder before, now pushes down harder.

Now we can work with positive numbers, and divide both weights by 4 to
get

y > 9

So any number BIGGER than 9 will work. For example, for y = 10,

-4y = -40 < -36

Do you see how the larger number, 10 > 9, becomes the smaller number
(-40 < -36) when it is multiplied by a negative number? That is the
key. The rule is that when you multiply an inequality by a negative
number, you have to reverse the direction. Or if you prefer, you can
do this:

-4y < -36 Add 4y
0 < 4y - 36 Add 36
36 < 4y Divide by 4
9 < y Reverse the whole inequality
y > 9

By avoiding multiplication by a negative number, I avoided the need to
reverse signs until the end.

Now that we have seen with our imagination what is going on, let us try
to prove the rule that if

a > b

then

-a < -b.

Start with the original inequality and subtract a from both sides:

0 > b - a

Now subtract b from both sides:

-b > -a

But that is the same as

-a < -b

which we were looking for. It is really pretty simple - so simple it
does not grab your attention the way helium balloons and antigravity
machines do! That is why I like to start the way I did.

'''The solution above was taken from Dr. Math. For all the visual learners, i tried to find a video.. but failed to do so...'''

Note- http://mathforum.org/dr/math/ is a very good website to find help for any math problems. You should check it out!

=Basic Examples=
====Example 1====

Solve -2x > 2

we start by dividing both sides by -2 to solve the inequality

x < -1

====Example 2====
Solving linear inequalities is almost exactly like solving linear equations.

* Solve x + 3 < 0.

If they'd given "x + 3 = 0", we would know how to solve: we would have subtracted 3 from both sides. The same applies here.

x < -3

Then the solution is:

x < –3

====Example 3====

* Solve x – 4 > 0.

If they'd given "x – 4 = 0", then we would can solve by adding four to each side. The same applies here.
x >= 4

Then the solution is: x > 4

====Example 4====

* Solve 2x < 9.

If they had given "2x = 9", we would have divided the 2 from each side.

x <= 9/2

Then the solution is: x < 9/2

====Example 5====

* Solve (2x – 3)/4 < 2.
First, multiply through by 4. Since the "4" is positive, we don't have to flip the inequality sign:

(2x – 3)/4 < 2
(4) × (2x – 3)/4 < (4)(2)
2x – 3 < 8
2x < 11
x < 11/2 = 5.5

====Example 6====

* Solve 10 < 3x + 4 < 19.

This is what is called a "compound inequality". It works just like regular inequalities, except that it has three "sides". So, for instance, when we go to subtract the 4, I will have to subtract it from all three "sides".

=10 < 3x + 4 < 19
=6 < 3x < 15
=2 < x < 5

====Example 7====

* Solve 5x + 7 < 3(x + 1).

First we multiply through on the right-hand side, and then solve as usual:

5x + 7 < 3(x + 1)

5x + 7 < 3x + 3

2x + 7 < 3

2x < –4

x < –2

====Example 8====

* Solve <math>5(x-3)/2 < 2(3x+4)</math>

First, simplify the equation then proceed to isolate x.

* (5x-15)/2 < 6x+8
5x-15 > 2(6x+8)
5x > 12x+16+15
5x > 12x+31
5x-12x > 31
-7x > 31 ;; Dividing and multiplying by negative numbers switches the sign
<math>x<-31/7</math>

====Example 9====

* Solve <math>3 < |3x+4| </math>

Given an inequality with absolute values, say |x| > a then we assume it is –a > x > a.

For better and more in-depth information, please visit here: http://www.nipissingu.ca/calculus/tutorials/absolutevalue.html

* <math>3 < 3x+4 < -3 </math>
<math>3 -4 < 3x+4 -4 < -3 -4</math> ;; We minus 4 on both sides
<math>-1 < 3x < -7 </math> ;; Divide 3 on both sides
<math>-1/3 < x < -7/3 </math>

So the solution for <math>3 < |3x+4| </math> is <math>-1/3 < x < -7/3 </math>.

=Quadratic Inequalities=

To solve a quadratic inequality, follow these steps:

1. Solve the inequality as though it were an equation. The real solutions to the equation become boundary points for the solution to the inequality.

2. Make the boundary points solid circles if the original inequality includes equality; otherwise, make the boundary points open circles.

3. Select points from each of the regions created by the boundary points. Replace these “test points” in the original inequality.

4. If a test point satisfies the original inequality, then the region that contains that test point is part of the solutions.

5. Represent the solution in graphic form and in solution test form.

====Example 1====

Solve <math>x^2-2x-15>0</math>

Treat it like a normal quadratic equation and find the zeroes
* <math>x^2-2x-15=0</math>
<math>(x-5)(x+3)=0</math>
x-5=0 or x+3=0
x=5 or -3

The zeroes divide the number line into three regions

[[File:Crappy_number_line_thing_1.png]]

For region x>5, check for a number greater than 5 (eg.6).
*<math>(6)^2-2(6)-15>0</math>
9>0
All x-values greater than 5 will work.

For region x<-3, check for a number less than -3 (eg.-4).
*<math>(-4)^2-2(-4)-15>0</math>
9>0
All x-values less than -3 will work.

*Show answer using interval notation
<math>(-infinity,-3)U(5,infinity)</math>

Then, x>5 and x<-3.

====Example 2====

Solve <math>3x^2>-x+4</math>

Make one side equal to zero
*<math>3x^2+x-4>0</math>

Find the zeroes using the quadratic formula and factor

*(3x+4)(x-1)>0
3x+4=0 or x-1=0
x=<math>-(4/3)</math> or 1

For region x><math>-(4/3)</math>, check for a number less than <math>-(4/3)</math> (eg.-2).
*<math>3(-2)^2+(-2)-4>0</math>
6>0
All x-values less than <math>-4/3</math> will work.

For region x>1, check for a number greater than 1 (eg.4).
*<math>3(4)^2+(4)-4>0</math>
48>0
All x-values greater than 1 will work.

*Show values that produce an answer greater than 0 using interval notation
<math>(-infinity, -4/3)U(1,+infinity)</math>

====Example 3====

Solve for, <math>9x+9+3x^2>-x^2+7</math>

Move terms to one side so one side is zero
*<math>x+9+3x^2+x^2-7>0</math>
<math>4x^2+9x+2>0</math>

Then find the zeroes
*(x+2)(4x+1)>0
x+2=0 or 4x+1=0
x=<math>-2</math> or <math>-(1/4)</math>

To verify, check the to see if the numbers for region x<-2 are greater than zero by plugging in some number.

*<math>4x^2+9x+2>0</math>
<math>4(-3)^2+9(-3)+2>0</math>
<math>11>0</math>

Likewise, for the region x><math>-1/4</math>

*<math>4x^2+9x+2>0</math>
<math>4(1)^2+9(1)+2>0</math>
<math>15>0</math>

And lastly, some number between -2 and <math>-1/4</math>.

*<math>4x^2+9x+2>0</math>
<math>4(-1)^2+9(-1)+2>0</math>
<math>-3>0</math>
As you can see, -3 is not greater than zero.

Then in the end, we are assume that the solution of this question is, x<-2 and x><math>-1/4</math>.

====Example 4====

Solve for, <math>-9x^2+7+22x>11x+2-15x^2</math>

Move terms to one side so one side is zero
*<math>-9x^2+7+22x-11x-2+15x^2>0</math>
<math>6x+11x+5>0</math>

Then find the zeroes
*(6x+5)(x+1)>0
6x+5=0 or x+1=0
x=<math>-5/6</math> or -1

To verify, check the to see if the numbers for region x<<math>-5/6</math> are greater than zero by plugging in some number.

*<math>6x+11x+5>0</math>
<math>6(-4)+11(-4)+5>0</math>
<math>30>0</math>

Likewise, for the region x>-1

*<math>6x+11x+5>0</math>
<math>6(6)+11(6)+5>0</math>
<math>200>0</math>

And lastly, some number between <math>-5/6</math> and -1.

*<math>6x+11x+5>0</math>
<math>6(-.9)+11(-.9)+5>0</math>
<math>-2.86>0</math>
As you can see, it is not greater than zero.

Then in the end, we are assume that the solution of this question is, x<<math>-5/6</math> and x>-1.

====References====

http://www.purplemath.com/modules/ineqquad.htm

http://www.analyzemath.com/Inequalities_Polynomial/quadratic_inequalities.html

=Graphing Inequalities=

Number Line

1. Simplify the inequality you are going to graph.

<pre>E.g.

-2x2 + 5x < -6(x + 1)

-2x2 + 5x < -6x – 6</pre>

2. Move all terms to one side so the other is zero. ''(It will be easiest if the highest power variable is positive.)''

<pre>E.g.

0 < 2x2 -6x - 5x - 6

0 < 2x2 -11x – 6</pre>

3. Pretend that the inequality sign is an equal sign and find all values of the variable.

<pre>E.g.

0 = 2x2 -11x - 6

0 = (2x + 1)(x - 6)

2x + 1 = 0, x - 6 = 0

2x = -1, x = 6

x = -1/2</pre>

4. Draw a number line including the variable solutions (in order).

[[File:Inequalitiesnumberline1.jpg]]

5. Draw a circle on the points. If the inequality symbol means less than or more than (> or <), draw an empty circle over the variable solution(s). If it means less/more than and equal to (≤ or ≥) fill in that circle.

*In this case our equation was greater than zero, so use open circles.

[[File:Inequalitynumberlineshizz2.jpg]]

6. Take a number from each of the resulting intervals and plug it back into the equality. If you get a true statement once solved, shade this region of the number line.

In the interval from (-∞,-1/2) we will take -1 and plug it into the original inequality.

<pre>0 < 2x2 -11x - 6

0 < 2(-1)2 -11(-1) - 6

0 < 2(1) + 11 - 6

0 < 7

Zero is less than 7 is correct, so shade (-∞, -1/2) on the number line.</pre>

7. Next, on the interval from (-1/2, 6) we will use zero.

<pre>0 < 2(0)2 -11(0) - 6

0 < 0 + 0 - 6

0 < -6

Zero is not less than negative six, so do not shade (-1/2,6).</pre>

Lastly, we will take 10 from the interval (6,∞).

<pre>0 < 2(10)2 - 11(10) + 6

0 < 2(100) - 110 + 6

0 < 200 - 110 + 6

0 < 96

Zero is less than 96 is correct, so shade (6,∞) as well.</pre>

Use arrows on the end of shading to indicate that the interval continues into infinity. The completed number line:

[[File:Inequalitynumberlineblah3.jpg]]

=Tips=
If x ≥ y then 1/x ≤ 1/y

=Videos teaching Inequality=

===Solving Linear Inequalities===

{{#ev:youtube | VgDe_D8ojxw | 400}}
<ref>Algebra: Solving Inequalities - http://www.youtube.com/watch?v=VgDe_D8ojxw
</ref>

===Other Videos===

Video 1. http://www.khanacademy.org/video/inequalities?playlist=Algebra%20I%20Worked%20Examples
- This video touches upon the concept of inequality and has a basic word problem solved.

Video 2. http://www.khanacademy.org/video/interpreting-inequalities?playlist=Algebra%20I%20Worked%20Examples
- This video is about interpreting inequalities in word problems.

Video 3. http://www.khanacademy.org/video/solving-inequalities?playlist=Algebra%20I%20Worked%20Examples
- This video is about solving basic problems regarding inequalities.

Video 4. http://www.khanacademy.org/video/inequalities-using-addition-and-subtraction?playlist=ck12.org%20Algebra%201%20Examples
- This video solves random question about inequalities with addition and subtraction.

Video 5. http://www.khanacademy.org/video/inequalities-using-multiplication-and-division?playlist=ck12.org%20Algebra%201%20Examples
- This video solves random question about inequalities with multiplication and division

Video 6. http://www.khanacademy.org/video/quadratic-inequalities?playlist=Algebra
- This video explains Quadratic Inequalities.

=Useful Links=

* http://www.purplemath.com/modules/ineqsolv.htm
* http://www.mathsisfun.com/algebra/inequality-solving.html
* http://webmath.com/solverineq.html (If you want to check your answers after solving an inequality)

=Group 10=

Here is what we have so far from our Group 10 Page. We will be adding more. Can you help with formatting like you have done for your section?

Solving Quadratic Inequalities

To solve a quadratic inequality, follow these steps:

1. Solve the inequality as though it were an equation. The real solutions to the equation become boundary points for the solution to the inequality.

2. Make the boundary points solid circles if the original inequality includes equality; otherwise, make the boundary points open circles.

3. Select points from each of the regions created by the boundary points. Replace these “test points” in the original inequality.

4. If a test point satisfies the original inequality, then the region that contains that test point is part of the solutions.

5. Represent the solution in graphic form and in solution test form.

Example 1: Solve (x-3)(x+2)>0

By the zero product property, x-3=0 or x+2=0, x=3 and x=-2

Make the boundary points.

Here, the boundary points are open circles because the original inequality does not include equality.

Select points from different regions created.

Three regions are created:

X=-3

X=0

X=4

See if the test points satisfy the original inequality

(x-3)(x+2)>0

(-3-3)(-3+2)>0

6>0 therefore, it works

(x-3)(x+2)>0

(0-3)(0+2)>0

-6>0 no, it does not work

(x-3)(x+2)>0

(4-3)(-3+2)>0

6>0 therefore, it works

Because you want to get x alone, you can rather:

• Add or subtract a number from both sides

• Multiply or divide both sides by a positive number

• Simplify a side

However, doing the following things will change the direction of the inequality:

• Multiplying or dividing both sides by a negative number

• Swapping left and right hand sides

Have a look at: http://www.mathsisfun.com/algebra/inequality-solving.html for more information.

If you’re having trouble in the book there is a good section starting on page 1061 which is a review of algebra and sets of real numbers. It gives number lines and shows inequalities to match.

If you want to check your answer on how to solve an inequality try: http://webmath.com/solverineq.html

'''What is an inequality?'''

It is exactly what it sounds, it's definition is that two numbers are not equal to each other. Such as "y" is not equal to "x".

1. It may be that "x" is greater than "y" which can be written as

x>y.

2. x is greater than or equal to y

x <math>\geq</math> y

3. x is less than y

x < y

4. x is less than or equal to y

x <math>\leq</math> y

5. x is not equal to y

x <math>\neq</math> y

'''How to represent the solutions of inequalities?'''

They can be represented in an interval notation ('''such as the notation used for defining a domain'''), as it specifies what group of number that fits under this rule.

'''How to solve linear inequalities?'''

Solve it like a linear equation.

'''Goal''': to isolate the variable so that you can determine the interval of "x"

'''Addition/Subtraction'''

The equation can be seen as a normal equation, instead of the greater than, less than, not equal to sign, just let the = sign replace them for now.
Then solve the equation algebraically, afterwards substitute the equations' original inequality sign.

'''EX'''

x + 8 > 10 <math>\longrightarrow</math> substitute the = sign

x + 8 = 10 <math>\longrightarrow</math> isolate the variable

x = 2 <math>\longrightarrow</math> replace the equal sign with the original inequality sign

x > 2

In this example "x" can be any number grater than two, and thus can be written like this (2,<math>\infty</math>)

'''Multiplication/Division'''

is similar to solving addition/subtraction equations

if 2x <math>\geq</math> 5 <math>\longrightarrow</math> then to isolate x, divide both sides by 2

x <math>\geq</math> <math>\{5 \over 2}\</math>

'''How to solve quadratic inequalities?'''

Let us look at an example of a quadratic inequality:

-x2 + 9 < 0

Our first step is to associate this with and equation. Therefore:

y= -x2 + 9

The next step is to find out where the equation cuts the graph on the x-axis. This means equating it to 0

-x2 + 9 = 0

Now we can solve:

-x2 + 9 = 0

x2 + 9 = 0

(x+3) (x-3) = 0

x=-3 x=3 Now we know that our quadratic equation crosses the x-axis at these values

We now get 3 different intervals at the points where the x-axis is cut

1.) x < -3

2.) -3 < x < 3

3.) x > 3

The next step is to find out on which intervals is the graph below the x-axis. In order to do this we must look at our original equation:

y = -x2 + 9 since it is a negative quadratic we can say that the graph would be facing down. Thus, in order to solve our original inequality of

-x2 + 9 < 0 we must look at where our y values are less than 0

From this we can see that the solution is: x < -3 or x > 3

'''Online References/extension'''

[http://www.purplemath.com/modules/ineqgrph.htm] Written step-by-step explanation

[http://www.youtube.com/watch?v=0X-bMeIN53I] Video Explanation

http://www.youtube.com/watch?v=VgDe_D8ojxw

Course:MATH110/Archive/2010-2011/003/Groups/Group 09/Basic Skills Project

2010-12-03T09:35:05Z

EllenTsang: /* Videos teaching Inequality */

For the Basic Skills Project, Group 9 plans on focusing on Inequalities.

We will give several worked out examples to cover all cases of questions concerning this topic.

Also, we will include tips & tricks for how to solve more difficult problems and possible references related to the topic.

<pre>Let's work with Group 10 for the group project :D
Thanks Micha the only person from Group 10 who replied. -- Ellen</pre>

=What is an inequality?=

It basically means when:

* An equation includes < or > or ≤ or ≥.
** E.g. x + 1 ≤ 3

The symbols and meanings of the inequalities are as follows:

> means greater than

< means less than

≥ means greater than or equal to

≤ means less than or equal to

=When to change the sign in an inequality problem?=

This is where most of the students face problems while doing problems regarding inequality. This occurs when there is ambiguity whether to change the sign of the inequality.

To solve the confusions, we ONLY change the sign of the inequality to its corresponding opposite when we '''multiply or divide with a negative number'''

Equations can be thought of as balanced scales, where the total weight
on the left balances what is on the right. Inequalities are like
unbalanced scales, where all you know is which side is "down"
(heavier). So for example

3x > 6

can be thought of as 3 unknown weights labeled "X" on the left,
heavier than a 6-gram weight on the right.

Negative numbers complicate it a bit. A negative constant can be
thought of as a helium balloon (barely) able to lift a certain weight.
A negative number times a variable might mean there is an antigravity
machine under it so it pulls up as hard as it would normally push
down!

So the equation

-4y < -36

would be 4 Y's on the left with an antigravity machine, and a "-36"
gram balloon on the right. The right side is "heavier," which in this
case means not that it is pushing down more, but that it is pulling up
less!

To solve it, let us first replace the -36 gram balloon with a 36-gram
weight and an antigravity machine:

-(4y) < -(36)

Now turn off the antigravity machines:

4y > 36

Why did I reverse the direction of the "<"? That is the key to this
whole thing: antigravity machines are like turning the whole world
upside down, so the side that was down is now up:

-4y -36
^ ^
\ | |
\ |
\ |
\ |
\
\

becomes

4y 36
| |
| v /
| /
| /
v /
/
/

because 4y, which pulled up harder before, now pushes down harder.

Now we can work with positive numbers, and divide both weights by 4 to
get

y > 9

So any number BIGGER than 9 will work. For example, for y = 10,

-4y = -40 < -36

Do you see how the larger number, 10 > 9, becomes the smaller number
(-40 < -36) when it is multiplied by a negative number? That is the
key. The rule is that when you multiply an inequality by a negative
number, you have to reverse the direction. Or if you prefer, you can
do this:

-4y < -36 Add 4y
0 < 4y - 36 Add 36
36 < 4y Divide by 4
9 < y Reverse the whole inequality
y > 9

By avoiding multiplication by a negative number, I avoided the need to
reverse signs until the end.

Now that we have seen with our imagination what is going on, let us try
to prove the rule that if

a > b

then

-a < -b.

Start with the original inequality and subtract a from both sides:

0 > b - a

Now subtract b from both sides:

-b > -a

But that is the same as

-a < -b

which we were looking for. It is really pretty simple - so simple it
does not grab your attention the way helium balloons and antigravity
machines do! That is why I like to start the way I did.

'''The solution above was taken from Dr. Math. For all the visual learners, i tried to find a video.. but failed to do so...'''

Note- http://mathforum.org/dr/math/ is a very good website to find help for any math problems. You should check it out!

=Basic Examples=
====Example 1====

Solve -2x > 2

we start by dividing both sides by -2 to solve the inequality

x < -1

====Example 2====
Solving linear inequalities is almost exactly like solving linear equations.

* Solve x + 3 < 0.

If they'd given "x + 3 = 0", we would know how to solve: we would have subtracted 3 from both sides. The same applies here.

x < -3

Then the solution is:

x < –3

====Example 3====

* Solve x – 4 > 0.

If they'd given "x – 4 = 0", then we would can solve by adding four to each side. The same applies here.
x >= 4

Then the solution is: x > 4

====Example 4====

* Solve 2x < 9.

If they had given "2x = 9", we would have divided the 2 from each side.

x <= 9/2

Then the solution is: x < 9/2

====Example 5====

* Solve (2x – 3)/4 < 2.
First, multiply through by 4. Since the "4" is positive, we don't have to flip the inequality sign:

(2x – 3)/4 < 2
(4) × (2x – 3)/4 < (4)(2)
2x – 3 < 8
2x < 11
x < 11/2 = 5.5

====Example 6====

* Solve 10 < 3x + 4 < 19.

This is what is called a "compound inequality". It works just like regular inequalities, except that it has three "sides". So, for instance, when we go to subtract the 4, I will have to subtract it from all three "sides".

=10 < 3x + 4 < 19
=6 < 3x < 15
=2 < x < 5

====Example 7====

* Solve 5x + 7 < 3(x + 1).

First we multiply through on the right-hand side, and then solve as usual:

5x + 7 < 3(x + 1)

5x + 7 < 3x + 3

2x + 7 < 3

2x < –4

x < –2

====Example 8====

* Solve <math>5(x-3)/2 < 2(3x+4)</math>

First, simplify the equation then proceed to isolate x.

* (5x-15)/2 < 6x+8
5x-15 > 2(6x+8)
5x > 12x+16+15
5x > 12x+31
5x-12x > 31
-7x > 31 ;; Dividing and multiplying by negative numbers switches the sign
<math>x<-31/7</math>

====Example 9====

* Solve <math>3 < |3x+4| </math>

Given an inequality with absolute values, say |x| > a then we assume it is –a > x > a.

For better and more in-depth information, please visit here: http://www.nipissingu.ca/calculus/tutorials/absolutevalue.html

* <math>3 < 3x+4 < -3 </math>
<math>3 -4 < 3x+4 -4 < -3 -4</math> ;; We minus 4 on both sides
<math>-1 < 3x < -7 </math> ;; Divide 3 on both sides
<math>-1/3 < x < -7/3 </math>

So the solution for <math>3 < |3x+4| </math> is <math>-1/3 < x < -7/3 </math>.

=Quadratic Inequalities=

To solve a quadratic inequality, follow these steps:

1. Solve the inequality as though it were an equation. The real solutions to the equation become boundary points for the solution to the inequality.

2. Make the boundary points solid circles if the original inequality includes equality; otherwise, make the boundary points open circles.

3. Select points from each of the regions created by the boundary points. Replace these “test points” in the original inequality.

4. If a test point satisfies the original inequality, then the region that contains that test point is part of the solutions.

5. Represent the solution in graphic form and in solution test form.

====Example 1====

Solve <math>x^2-2x-15>0</math>

Treat it like a normal quadratic equation and find the zeroes
* <math>x^2-2x-15=0</math>
<math>(x-5)(x+3)=0</math>
x-5=0 or x+3=0
x=5 or -3

The zeroes divide the number line into three regions

[[File:Crappy_number_line_thing_1.png]]

For region x>5, check for a number greater than 5 (eg.6).
*<math>(6)^2-2(6)-15>0</math>
9>0
All x-values greater than 5 will work.

For region x<-3, check for a number less than -3 (eg.-4).
*<math>(-4)^2-2(-4)-15>0</math>
9>0
All x-values less than -3 will work.

*Show answer using interval notation
<math>(-infinity,-3)U(5,infinity)</math>

Then, x>5 and x<-3.

====Example 2====

Solve <math>3x^2>-x+4</math>

Make one side equal to zero
*<math>3x^2+x-4>0</math>

Find the zeroes using the quadratic formula and factor

*(3x+4)(x-1)>0
3x+4=0 or x-1=0
x=<math>-(4/3)</math> or 1

For region x><math>-(4/3)</math>, check for a number less than <math>-(4/3)</math> (eg.-2).
*<math>3(-2)^2+(-2)-4>0</math>
6>0
All x-values less than <math>-4/3</math> will work.

For region x>1, check for a number greater than 1 (eg.4).
*<math>3(4)^2+(4)-4>0</math>
48>0
All x-values greater than 1 will work.

*Show values that produce an answer greater than 0 using interval notation
<math>(-infinity, -4/3)U(1,+infinity)</math>

====Example 3====

Solve for, <math>9x+9+3x^2>-x^2+7</math>

Move terms to one side so one side is zero
*<math>x+9+3x^2+x^2-7>0</math>
<math>4x^2+9x+2>0</math>

Then find the zeroes
*(x+2)(4x+1)>0
x+2=0 or 4x+1=0
x=<math>-2</math> or <math>-(1/4)</math>

To verify, check the to see if the numbers for region x<-2 are greater than zero by plugging in some number.

*<math>4x^2+9x+2>0</math>
<math>4(-3)^2+9(-3)+2>0</math>
<math>11>0</math>

Likewise, for the region x><math>-1/4</math>

*<math>4x^2+9x+2>0</math>
<math>4(1)^2+9(1)+2>0</math>
<math>15>0</math>

And lastly, some number between -2 and <math>-1/4</math>.

*<math>4x^2+9x+2>0</math>
<math>4(-1)^2+9(-1)+2>0</math>
<math>-3>0</math>
As you can see, -3 is not greater than zero.

Then in the end, we are assume that the solution of this question is, x<-2 and x><math>-1/4</math>.

====Example 4====

Solve for, <math>-9x^2+7+22x>11x+2-15x^2</math>

Move terms to one side so one side is zero
*<math>-9x^2+7+22x-11x-2+15x^2>0</math>
<math>6x+11x+5>0</math>

Then find the zeroes
*(6x+5)(x+1)>0
6x+5=0 or x+1=0
x=<math>-5/6</math> or -1

To verify, check the to see if the numbers for region x<<math>-5/6</math> are greater than zero by plugging in some number.

*<math>6x+11x+5>0</math>
<math>6(-4)+11(-4)+5>0</math>
<math>30>0</math>

Likewise, for the region x>-1

*<math>6x+11x+5>0</math>
<math>6(6)+11(6)+5>0</math>
<math>200>0</math>

And lastly, some number between <math>-5/6</math> and -1.

*<math>6x+11x+5>0</math>
<math>6(-.9)+11(-.9)+5>0</math>
<math>-2.86>0</math>
As you can see, it is not greater than zero.

Then in the end, we are assume that the solution of this question is, x<<math>-5/6</math> and x>-1.

====References====

http://www.purplemath.com/modules/ineqquad.htm

http://www.analyzemath.com/Inequalities_Polynomial/quadratic_inequalities.html

=Graphing Inequalities=

Number Line

1. Simplify the inequality you are going to graph.

<pre>E.g.

-2x2 + 5x < -6(x + 1)

-2x2 + 5x < -6x – 6</pre>

2. Move all terms to one side so the other is zero. ''(It will be easiest if the highest power variable is positive.)''

<pre>E.g.

0 < 2x2 -6x - 5x - 6

0 < 2x2 -11x – 6</pre>

3. Pretend that the inequality sign is an equal sign and find all values of the variable.

<pre>E.g.

0 = 2x2 -11x - 6

0 = (2x + 1)(x - 6)

2x + 1 = 0, x - 6 = 0

2x = -1, x = 6

x = -1/2</pre>

4. Draw a number line including the variable solutions (in order).

[[File:Inequalitiesnumberline1.jpg]]

5. Draw a circle on the points. If the inequality symbol means less than or more than (> or <), draw an empty circle over the variable solution(s). If it means less/more than and equal to (≤ or ≥) fill in that circle.

*In this case our equation was greater than zero, so use open circles.

[[File:Inequalitynumberlineshizz2.jpg]]

6. Take a number from each of the resulting intervals and plug it back into the equality. If you get a true statement once solved, shade this region of the number line.

In the interval from (-∞,-1/2) we will take -1 and plug it into the original inequality.

<pre>0 < 2x2 -11x - 6

0 < 2(-1)2 -11(-1) - 6

0 < 2(1) + 11 - 6

0 < 7

Zero is less than 7 is correct, so shade (-∞, -1/2) on the number line.</pre>

7. Next, on the interval from (-1/2, 6) we will use zero.

<pre>0 < 2(0)2 -11(0) - 6

0 < 0 + 0 - 6

0 < -6

Zero is not less than negative six, so do not shade (-1/2,6).</pre>

Lastly, we will take 10 from the interval (6,∞).

<pre>0 < 2(10)2 - 11(10) + 6

0 < 2(100) - 110 + 6

0 < 200 - 110 + 6

0 < 96

Zero is less than 96 is correct, so shade (6,∞) as well.</pre>

Use arrows on the end of shading to indicate that the interval continues into infinity. The completed number line:

[[File:Inequalitynumberlineblah3.jpg]]

=Tips=
If x ≥ y then 1/x ≤ 1/y

=Videos teaching Inequality=

Video 1. http://www.khanacademy.org/video/inequalities?playlist=Algebra%20I%20Worked%20Examples
- This video touches upon the concept of inequality and has a basic word problem solved.

Video 2. http://www.khanacademy.org/video/interpreting-inequalities?playlist=Algebra%20I%20Worked%20Examples
- This video is about interpreting inequalities in word problems.

Video 3. http://www.khanacademy.org/video/solving-inequalities?playlist=Algebra%20I%20Worked%20Examples
- This video is about solving basic problems regarding inequalities.

Video 4. http://www.khanacademy.org/video/inequalities-using-addition-and-subtraction?playlist=ck12.org%20Algebra%201%20Examples
- This video solves random question about inequalities with addition and subtraction.

Video 5. http://www.khanacademy.org/video/inequalities-using-multiplication-and-division?playlist=ck12.org%20Algebra%201%20Examples
- This video solves random question about inequalities with multiplication and division

Video 6. http://www.khanacademy.org/video/quadratic-inequalities?playlist=Algebra
- This video explains Quadratic Inequalities.

=Useful Links=

* http://www.purplemath.com/modules/ineqsolv.htm
* http://www.mathsisfun.com/algebra/inequality-solving.html
* http://webmath.com/solverineq.html (If you want to check your answers after solving an inequality)

=Group 10=

Here is what we have so far from our Group 10 Page. We will be adding more. Can you help with formatting like you have done for your section?

Solving Quadratic Inequalities

To solve a quadratic inequality, follow these steps:

1. Solve the inequality as though it were an equation. The real solutions to the equation become boundary points for the solution to the inequality.

2. Make the boundary points solid circles if the original inequality includes equality; otherwise, make the boundary points open circles.

3. Select points from each of the regions created by the boundary points. Replace these “test points” in the original inequality.

4. If a test point satisfies the original inequality, then the region that contains that test point is part of the solutions.

5. Represent the solution in graphic form and in solution test form.

Example 1: Solve (x-3)(x+2)>0

By the zero product property, x-3=0 or x+2=0, x=3 and x=-2

Make the boundary points.

Here, the boundary points are open circles because the original inequality does not include equality.

Select points from different regions created.

Three regions are created:

X=-3

X=0

X=4

See if the test points satisfy the original inequality

(x-3)(x+2)>0

(-3-3)(-3+2)>0

6>0 therefore, it works

(x-3)(x+2)>0

(0-3)(0+2)>0

-6>0 no, it does not work

(x-3)(x+2)>0

(4-3)(-3+2)>0

6>0 therefore, it works

Because you want to get x alone, you can rather:

• Add or subtract a number from both sides

• Multiply or divide both sides by a positive number

• Simplify a side

However, doing the following things will change the direction of the inequality:

• Multiplying or dividing both sides by a negative number

• Swapping left and right hand sides

Have a look at: http://www.mathsisfun.com/algebra/inequality-solving.html for more information.

If you’re having trouble in the book there is a good section starting on page 1061 which is a review of algebra and sets of real numbers. It gives number lines and shows inequalities to match.

If you want to check your answer on how to solve an inequality try: http://webmath.com/solverineq.html

'''What is an inequality?'''

It is exactly what it sounds, it's definition is that two numbers are not equal to each other. Such as "y" is not equal to "x".

1. It may be that "x" is greater than "y" which can be written as

x>y.

2. x is greater than or equal to y

x <math>\geq</math> y

3. x is less than y

x < y

4. x is less than or equal to y

x <math>\leq</math> y

5. x is not equal to y

x <math>\neq</math> y

'''How to represent the solutions of inequalities?'''

They can be represented in an interval notation ('''such as the notation used for defining a domain'''), as it specifies what group of number that fits under this rule.

'''How to solve linear inequalities?'''

Solve it like a linear equation.

'''Goal''': to isolate the variable so that you can determine the interval of "x"

'''Addition/Subtraction'''

The equation can be seen as a normal equation, instead of the greater than, less than, not equal to sign, just let the = sign replace them for now.
Then solve the equation algebraically, afterwards substitute the equations' original inequality sign.

'''EX'''

x + 8 > 10 <math>\longrightarrow</math> substitute the = sign

x + 8 = 10 <math>\longrightarrow</math> isolate the variable

x = 2 <math>\longrightarrow</math> replace the equal sign with the original inequality sign

x > 2

In this example "x" can be any number grater than two, and thus can be written like this (2,<math>\infty</math>)

'''Multiplication/Division'''

is similar to solving addition/subtraction equations

if 2x <math>\geq</math> 5 <math>\longrightarrow</math> then to isolate x, divide both sides by 2

x <math>\geq</math> <math>\{5 \over 2}\</math>

'''How to solve quadratic inequalities?'''

Let us look at an example of a quadratic inequality:

-x2 + 9 < 0

Our first step is to associate this with and equation. Therefore:

y= -x2 + 9

The next step is to find out where the equation cuts the graph on the x-axis. This means equating it to 0

-x2 + 9 = 0

Now we can solve:

-x2 + 9 = 0

x2 + 9 = 0

(x+3) (x-3) = 0

x=-3 x=3 Now we know that our quadratic equation crosses the x-axis at these values

We now get 3 different intervals at the points where the x-axis is cut

1.) x < -3

2.) -3 < x < 3

3.) x > 3

The next step is to find out on which intervals is the graph below the x-axis. In order to do this we must look at our original equation:

y = -x2 + 9 since it is a negative quadratic we can say that the graph would be facing down. Thus, in order to solve our original inequality of

-x2 + 9 < 0 we must look at where our y values are less than 0

From this we can see that the solution is: x < -3 or x > 3

'''Online References/extension'''

[http://www.purplemath.com/modules/ineqgrph.htm] Written step-by-step explanation

[http://www.youtube.com/watch?v=0X-bMeIN53I] Video Explanation

http://www.youtube.com/watch?v=VgDe_D8ojxw

Course:MATH110/Archive/2010-2011/003/Groups/Group 09/Basic Skills Project

2010-12-03T09:28:07Z

EllenTsang:

For the Basic Skills Project, Group 9 plans on focusing on Inequalities.

We will give several worked out examples to cover all cases of questions concerning this topic.

Also, we will include tips & tricks for how to solve more difficult problems and possible references related to the topic.

<pre>Let's work with Group 10 for the group project :D
Thanks Micha the only person from Group 10 who replied. -- Ellen</pre>

=What is an inequality?=

It basically means when:

* An equation includes < or > or ≤ or ≥.
** E.g. x + 1 ≤ 3

The symbols and meanings of the inequalities are as follows:

> means greater than

< means less than

≥ means greater than or equal to

≤ means less than or equal to

=When to change the sign in an inequality problem?=

This is where most of the students face problems while doing problems regarding inequality. This occurs when there is ambiguity whether to change the sign of the inequality.

To solve the confusions, we ONLY change the sign of the inequality to its corresponding opposite when we '''multiply or divide with a negative number'''

Equations can be thought of as balanced scales, where the total weight
on the left balances what is on the right. Inequalities are like
unbalanced scales, where all you know is which side is "down"
(heavier). So for example

3x > 6

can be thought of as 3 unknown weights labeled "X" on the left,
heavier than a 6-gram weight on the right.

Negative numbers complicate it a bit. A negative constant can be
thought of as a helium balloon (barely) able to lift a certain weight.
A negative number times a variable might mean there is an antigravity
machine under it so it pulls up as hard as it would normally push
down!

So the equation

-4y < -36

would be 4 Y's on the left with an antigravity machine, and a "-36"
gram balloon on the right. The right side is "heavier," which in this
case means not that it is pushing down more, but that it is pulling up
less!

To solve it, let us first replace the -36 gram balloon with a 36-gram
weight and an antigravity machine:

-(4y) < -(36)

Now turn off the antigravity machines:

4y > 36

Why did I reverse the direction of the "<"? That is the key to this
whole thing: antigravity machines are like turning the whole world
upside down, so the side that was down is now up:

-4y -36
^ ^
\ | |
\ |
\ |
\ |
\
\

becomes

4y 36
| |
| v /
| /
| /
v /
/
/

because 4y, which pulled up harder before, now pushes down harder.

Now we can work with positive numbers, and divide both weights by 4 to
get

y > 9

So any number BIGGER than 9 will work. For example, for y = 10,

-4y = -40 < -36

Do you see how the larger number, 10 > 9, becomes the smaller number
(-40 < -36) when it is multiplied by a negative number? That is the
key. The rule is that when you multiply an inequality by a negative
number, you have to reverse the direction. Or if you prefer, you can
do this:

-4y < -36 Add 4y
0 < 4y - 36 Add 36
36 < 4y Divide by 4
9 < y Reverse the whole inequality
y > 9

By avoiding multiplication by a negative number, I avoided the need to
reverse signs until the end.

Now that we have seen with our imagination what is going on, let us try
to prove the rule that if

a > b

then

-a < -b.

Start with the original inequality and subtract a from both sides:

0 > b - a

Now subtract b from both sides:

-b > -a

But that is the same as

-a < -b

which we were looking for. It is really pretty simple - so simple it
does not grab your attention the way helium balloons and antigravity
machines do! That is why I like to start the way I did.

'''The solution above was taken from Dr. Math. For all the visual learners, i tried to find a video.. but failed to do so...'''

Note- http://mathforum.org/dr/math/ is a very good website to find help for any math problems. You should check it out!

=Basic Examples=
====Example 1====

Solve -2x > 2

we start by dividing both sides by -2 to solve the inequality

x < -1

====Example 2====
Solving linear inequalities is almost exactly like solving linear equations.

* Solve x + 3 < 0.

If they'd given "x + 3 = 0", we would know how to solve: we would have subtracted 3 from both sides. The same applies here.

x < -3

Then the solution is:

x < –3

====Example 3====

* Solve x – 4 > 0.

If they'd given "x – 4 = 0", then we would can solve by adding four to each side. The same applies here.
x >= 4

Then the solution is: x > 4

====Example 4====

* Solve 2x < 9.

If they had given "2x = 9", we would have divided the 2 from each side.

x <= 9/2

Then the solution is: x < 9/2

====Example 5====

* Solve (2x – 3)/4 < 2.
First, multiply through by 4. Since the "4" is positive, we don't have to flip the inequality sign:

(2x – 3)/4 < 2
(4) × (2x – 3)/4 < (4)(2)
2x – 3 < 8
2x < 11
x < 11/2 = 5.5

====Example 6====

* Solve 10 < 3x + 4 < 19.

This is what is called a "compound inequality". It works just like regular inequalities, except that it has three "sides". So, for instance, when we go to subtract the 4, I will have to subtract it from all three "sides".

=10 < 3x + 4 < 19
=6 < 3x < 15
=2 < x < 5

====Example 7====

* Solve 5x + 7 < 3(x + 1).

First we multiply through on the right-hand side, and then solve as usual:

5x + 7 < 3(x + 1)

5x + 7 < 3x + 3

2x + 7 < 3

2x < –4

x < –2

====Example 8====

* Solve <math>5(x-3)/2 < 2(3x+4)</math>

First, simplify the equation then proceed to isolate x.

* (5x-15)/2 < 6x+8
5x-15 > 2(6x+8)
5x > 12x+16+15
5x > 12x+31
5x-12x > 31
-7x > 31 ;; Dividing and multiplying by negative numbers switches the sign
<math>x<-31/7</math>

====Example 9====

* Solve <math>3 < |3x+4| </math>

Given an inequality with absolute values, say |x| > a then we assume it is –a > x > a.

For better and more in-depth information, please visit here: http://www.nipissingu.ca/calculus/tutorials/absolutevalue.html

* <math>3 < 3x+4 < -3 </math>
<math>3 -4 < 3x+4 -4 < -3 -4</math> ;; We minus 4 on both sides
<math>-1 < 3x < -7 </math> ;; Divide 3 on both sides
<math>-1/3 < x < -7/3 </math>

So the solution for <math>3 < |3x+4| </math> is <math>-1/3 < x < -7/3 </math>.

=Quadratic Inequalities=

To solve a quadratic inequality, follow these steps:

1. Solve the inequality as though it were an equation. The real solutions to the equation become boundary points for the solution to the inequality.

2. Make the boundary points solid circles if the original inequality includes equality; otherwise, make the boundary points open circles.

3. Select points from each of the regions created by the boundary points. Replace these “test points” in the original inequality.

4. If a test point satisfies the original inequality, then the region that contains that test point is part of the solutions.

5. Represent the solution in graphic form and in solution test form.

====Example 1====

Solve <math>x^2-2x-15>0</math>

Treat it like a normal quadratic equation and find the zeroes
* <math>x^2-2x-15=0</math>
<math>(x-5)(x+3)=0</math>
x-5=0 or x+3=0
x=5 or -3

The zeroes divide the number line into three regions

[[File:Crappy_number_line_thing_1.png]]

For region x>5, check for a number greater than 5 (eg.6).
*<math>(6)^2-2(6)-15>0</math>
9>0
All x-values greater than 5 will work.

For region x<-3, check for a number less than -3 (eg.-4).
*<math>(-4)^2-2(-4)-15>0</math>
9>0
All x-values less than -3 will work.

*Show answer using interval notation
<math>(-infinity,-3)U(5,infinity)</math>

Then, x>5 and x<-3.

====Example 2====

Solve <math>3x^2>-x+4</math>

Make one side equal to zero
*<math>3x^2+x-4>0</math>

Find the zeroes using the quadratic formula and factor

*(3x+4)(x-1)>0
3x+4=0 or x-1=0
x=<math>-(4/3)</math> or 1

For region x><math>-(4/3)</math>, check for a number less than <math>-(4/3)</math> (eg.-2).
*<math>3(-2)^2+(-2)-4>0</math>
6>0
All x-values less than <math>-4/3</math> will work.

For region x>1, check for a number greater than 1 (eg.4).
*<math>3(4)^2+(4)-4>0</math>
48>0
All x-values greater than 1 will work.

*Show values that produce an answer greater than 0 using interval notation
<math>(-infinity, -4/3)U(1,+infinity)</math>

====Example 3====

Solve for, <math>9x+9+3x^2>-x^2+7</math>

Move terms to one side so one side is zero
*<math>x+9+3x^2+x^2-7>0</math>
<math>4x^2+9x+2>0</math>

Then find the zeroes
*(x+2)(4x+1)>0
x+2=0 or 4x+1=0
x=<math>-2</math> or <math>-(1/4)</math>

To verify, check the to see if the numbers for region x<-2 are greater than zero by plugging in some number.

*<math>4x^2+9x+2>0</math>
<math>4(-3)^2+9(-3)+2>0</math>
<math>11>0</math>

Likewise, for the region x><math>-1/4</math>

*<math>4x^2+9x+2>0</math>
<math>4(1)^2+9(1)+2>0</math>
<math>15>0</math>

And lastly, some number between -2 and <math>-1/4</math>.

*<math>4x^2+9x+2>0</math>
<math>4(-1)^2+9(-1)+2>0</math>
<math>-3>0</math>
As you can see, -3 is not greater than zero.

Then in the end, we are assume that the solution of this question is, x<-2 and x><math>-1/4</math>.

====Example 4====

Solve for, <math>-9x^2+7+22x>11x+2-15x^2</math>

Move terms to one side so one side is zero
*<math>-9x^2+7+22x-11x-2+15x^2>0</math>
<math>6x+11x+5>0</math>

Then find the zeroes
*(6x+5)(x+1)>0
6x+5=0 or x+1=0
x=<math>-5/6</math> or -1

To verify, check the to see if the numbers for region x<<math>-5/6</math> are greater than zero by plugging in some number.

*<math>6x+11x+5>0</math>
<math>6(-4)+11(-4)+5>0</math>
<math>30>0</math>

Likewise, for the region x>-1

*<math>6x+11x+5>0</math>
<math>6(6)+11(6)+5>0</math>
<math>200>0</math>

And lastly, some number between <math>-5/6</math> and -1.

*<math>6x+11x+5>0</math>
<math>6(-.9)+11(-.9)+5>0</math>
<math>-2.86>0</math>
As you can see, it is not greater than zero.

Then in the end, we are assume that the solution of this question is, x<<math>-5/6</math> and x>-1.

====References====

http://www.purplemath.com/modules/ineqquad.htm

http://www.analyzemath.com/Inequalities_Polynomial/quadratic_inequalities.html

=Graphing Inequalities=

Number Line

1. Simplify the inequality you are going to graph.

<pre>E.g.

-2x2 + 5x < -6(x + 1)

-2x2 + 5x < -6x – 6</pre>

2. Move all terms to one side so the other is zero. ''(It will be easiest if the highest power variable is positive.)''

<pre>E.g.

0 < 2x2 -6x - 5x - 6

0 < 2x2 -11x – 6</pre>

3. Pretend that the inequality sign is an equal sign and find all values of the variable.

<pre>E.g.

0 = 2x2 -11x - 6

0 = (2x + 1)(x - 6)

2x + 1 = 0, x - 6 = 0

2x = -1, x = 6

x = -1/2</pre>

4. Draw a number line including the variable solutions (in order).

[[File:Inequalitiesnumberline1.jpg]]

5. Draw a circle on the points. If the inequality symbol means less than or more than (> or <), draw an empty circle over the variable solution(s). If it means less/more than and equal to (≤ or ≥) fill in that circle.

*In this case our equation was greater than zero, so use open circles.

[[File:Inequalitynumberlineshizz2.jpg]]

6. Take a number from each of the resulting intervals and plug it back into the equality. If you get a true statement once solved, shade this region of the number line.

In the interval from (-∞,-1/2) we will take -1 and plug it into the original inequality.

<pre>0 < 2x2 -11x - 6

0 < 2(-1)2 -11(-1) - 6

0 < 2(1) + 11 - 6

0 < 7

Zero is less than 7 is correct, so shade (-∞, -1/2) on the number line.</pre>

7. Next, on the interval from (-1/2, 6) we will use zero.

<pre>0 < 2(0)2 -11(0) - 6

0 < 0 + 0 - 6

0 < -6

Zero is not less than negative six, so do not shade (-1/2,6).</pre>

Lastly, we will take 10 from the interval (6,∞).

<pre>0 < 2(10)2 - 11(10) + 6

0 < 2(100) - 110 + 6

0 < 200 - 110 + 6

0 < 96

Zero is less than 96 is correct, so shade (6,∞) as well.</pre>

Use arrows on the end of shading to indicate that the interval continues into infinity. The completed number line:

[[File:Inequalitynumberlineblah3.jpg]]

=Tips=
If x ≥ y then 1/x ≤ 1/y

=Videos teaching Inequality=

Video 1. http://www.khanacademy.org/video/inequalities?playlist=Algebra%20I%20Worked%20Examples
- This video touches upon the concept of inequality and has a basic word problem solved.

Video 2. http://www.khanacademy.org/video/interpreting-inequalities?playlist=Algebra%20I%20Worked%20Examples
- This video is about interpreting inequalities in word problems.

Video 3. http://www.khanacademy.org/video/solving-inequalities?playlist=Algebra%20I%20Worked%20Examples
- This video is about solving basic problems regarding inequalities.

Video 4. http://www.khanacademy.org/video/inequalities-using-addition-and-subtraction?playlist=ck12.org%20Algebra%201%20Examples
- This video solves random question about inequalities with addition and subtraction.

Video 5. http://www.khanacademy.org/video/inequalities-using-multiplication-and-division?playlist=ck12.org%20Algebra%201%20Examples
- This video solves random question about inequalities with multiplication and division

Video 6. http://www.khanacademy.org/video/quadratic-inequalities?playlist=Algebra
- This video explains quadratic Inequalities.

=Useful Links=

* http://www.purplemath.com/modules/ineqsolv.htm
* http://www.mathsisfun.com/algebra/inequality-solving.html
* http://webmath.com/solverineq.html (If you want to check your answers after solving an inequality)

=Group 10=

Here is what we have so far from our Group 10 Page. We will be adding more. Can you help with formatting like you have done for your section?

Solving Quadratic Inequalities

To solve a quadratic inequality, follow these steps:

1. Solve the inequality as though it were an equation. The real solutions to the equation become boundary points for the solution to the inequality.

2. Make the boundary points solid circles if the original inequality includes equality; otherwise, make the boundary points open circles.

3. Select points from each of the regions created by the boundary points. Replace these “test points” in the original inequality.

4. If a test point satisfies the original inequality, then the region that contains that test point is part of the solutions.

5. Represent the solution in graphic form and in solution test form.

Example 1: Solve (x-3)(x+2)>0

By the zero product property, x-3=0 or x+2=0, x=3 and x=-2

Make the boundary points.

Here, the boundary points are open circles because the original inequality does not include equality.

Select points from different regions created.

Three regions are created:

X=-3

X=0

X=4

See if the test points satisfy the original inequality

(x-3)(x+2)>0

(-3-3)(-3+2)>0

6>0 therefore, it works

(x-3)(x+2)>0

(0-3)(0+2)>0

-6>0 no, it does not work

(x-3)(x+2)>0

(4-3)(-3+2)>0

6>0 therefore, it works

Because you want to get x alone, you can rather:

• Add or subtract a number from both sides

• Multiply or divide both sides by a positive number

• Simplify a side

However, doing the following things will change the direction of the inequality:

• Multiplying or dividing both sides by a negative number

• Swapping left and right hand sides

Have a look at: http://www.mathsisfun.com/algebra/inequality-solving.html for more information.

If you’re having trouble in the book there is a good section starting on page 1061 which is a review of algebra and sets of real numbers. It gives number lines and shows inequalities to match.

If you want to check your answer on how to solve an inequality try: http://webmath.com/solverineq.html

'''What is an inequality?'''

It is exactly what it sounds, it's definition is that two numbers are not equal to each other. Such as "y" is not equal to "x".

1. It may be that "x" is greater than "y" which can be written as

x>y.

2. x is greater than or equal to y

x <math>\geq</math> y

3. x is less than y

x < y

4. x is less than or equal to y

x <math>\leq</math> y

5. x is not equal to y

x <math>\neq</math> y

'''How to represent the solutions of inequalities?'''

They can be represented in an interval notation ('''such as the notation used for defining a domain'''), as it specifies what group of number that fits under this rule.

'''How to solve linear inequalities?'''

Solve it like a linear equation.

'''Goal''': to isolate the variable so that you can determine the interval of "x"

'''Addition/Subtraction'''

The equation can be seen as a normal equation, instead of the greater than, less than, not equal to sign, just let the = sign replace them for now.
Then solve the equation algebraically, afterwards substitute the equations' original inequality sign.

'''EX'''

x + 8 > 10 <math>\longrightarrow</math> substitute the = sign

x + 8 = 10 <math>\longrightarrow</math> isolate the variable

x = 2 <math>\longrightarrow</math> replace the equal sign with the original inequality sign

x > 2

In this example "x" can be any number grater than two, and thus can be written like this (2,<math>\infty</math>)

'''Multiplication/Division'''

is similar to solving addition/subtraction equations

if 2x <math>\geq</math> 5 <math>\longrightarrow</math> then to isolate x, divide both sides by 2

x <math>\geq</math> <math>\{5 \over 2}\</math>

'''How to solve quadratic inequalities?'''

Let us look at an example of a quadratic inequality:

-x2 + 9 < 0

Our first step is to associate this with and equation. Therefore:

y= -x2 + 9

The next step is to find out where the equation cuts the graph on the x-axis. This means equating it to 0

-x2 + 9 = 0

Now we can solve:

-x2 + 9 = 0

x2 + 9 = 0

(x+3) (x-3) = 0

x=-3 x=3 Now we know that our quadratic equation crosses the x-axis at these values

We now get 3 different intervals at the points where the x-axis is cut

1.) x < -3

2.) -3 < x < 3

3.) x > 3

The next step is to find out on which intervals is the graph below the x-axis. In order to do this we must look at our original equation:

y = -x2 + 9 since it is a negative quadratic we can say that the graph would be facing down. Thus, in order to solve our original inequality of

-x2 + 9 < 0 we must look at where our y values are less than 0

From this we can see that the solution is: x < -3 or x > 3

'''Online References/extension'''

[http://www.purplemath.com/modules/ineqgrph.htm] Written step-by-step explanation

[http://www.youtube.com/watch?v=0X-bMeIN53I] Video Explanation

http://www.youtube.com/watch?v=VgDe_D8ojxw