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

2010-12-03T07:55:44Z

EdithFan: /* Example 9 */

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 from Group 10 for replying. -- 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=
References:

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

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

====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.

=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).

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.

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:

<pre>********** Ali, were you supposed to put a number line here or something? **********</pre>

=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

=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

2010-12-03T02:45:46Z

EdithFan: /* 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 from Group 10 for replying. -- 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

=Linear Inequalities=

The only difference between solving linear inequalities and solving linear equations is that '>' or '<' replaces the '=' sign.

Also, if you multiply or divide by a negative number, you have to change the sign around.

====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.

* <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=
References:

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

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

====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.

=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).

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.

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:

<pre>********** Ali, were you supposed to put a number line here or something? **********</pre>

=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

=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

2010-12-03T02:39:47Z

EdithFan: /* Example 4 */

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 from Group 10 for replying. -- 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

=Linear Inequalities=

The only difference between solving linear inequalities and solving linear equations is that '>' or '<' replaces the '=' sign.

Also, if you multiply or divide by a negative number, you have to change the sign around.

====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.

* Solve (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>

=Quadratic Inequalities=
References:

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

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

====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.

=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).

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.

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:

<pre>********** Ali, were you supposed to put a number line here or something? **********</pre>

=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

=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

2010-12-03T02:36:34Z

EdithFan: /* Example 8 */

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 from Group 10 for replying. -- 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

=Linear Inequalities=

The only difference between solving linear inequalities and solving linear equations is that '>' or '<' replaces the '=' sign.

Also, if you multiply or divide by a negative number, you have to change the sign around.

====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.

* Solve (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>

=Quadratic Inequalities=
References:

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

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

====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> x>-1.

=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).

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.

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:

<pre>********** Ali, were you supposed to put a number line here or something? **********</pre>

=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

=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

2010-12-03T02:32:38Z

EdithFan: /* 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 from Group 10 for replying. -- 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

=Linear Inequalities=

The only difference between solving linear inequalities and solving linear equations is that '>' or '<' replaces the '=' sign.

Also, if you multiply or divide by a negative number, you have to change the sign around.

====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.

* Solve -(5x-15)/2 < 6x+8 ;; Dividing and multiplying by negative numbers switches the sign
5x-15 > -2(6x+8)
5x > -12x-16+15
5x > -12x-1
5x+12x > -1
17x > -1
<math>x>-1/17</math>

=Quadratic Inequalities=
References:

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

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

====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> x>-1.

=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).

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.

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:

<pre>********** Ali, were you supposed to put a number line here or something? **********</pre>

=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

=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

2010-12-03T02:19:21Z

EdithFan: /* Quadratic 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 from Group 10 for replying. -- 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

=Linear Inequalities=

The only difference between solving linear inequalities and solving linear equations is that '>' or '<' replaces the '=' sign.

Also, if you multiply or divide by a negative number, you have to change the sign around.

====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

=Quadratic Inequalities=
References:

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

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

====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> x>-1.

=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).

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.

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:

<pre>********** Ali, were you supposed to put a number line here or something? **********</pre>

=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

=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

2010-12-03T01:39:08Z

EdithFan: /* Example 1 */

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

2010-12-03T01:37:19Z

EdithFan: /* Example 7 */

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

2010-11-12T04:51:02Z

EdithFan: /* Email addresses */

{{Infobox MATH110 Groups
| group number = 9
| member 1 = Kazi Ahmed
| member 2 = Edith Fan
| member 3 = [[User:MariaFayloga|Maria Fayloga]]
| member 4 = [[User:AliHussein|Ali Hussein]]
| member 5 = [[User:EllenTsang|Ellen Tsang]]
| member 6 =
}}

===Email addresses===
* Edith Fan | xsaret@gmail.com
* Maria Fayloga | maria.fayloga@gmail.com
* Ellen Tsang | ellentsang.nl@hotmail.com
* Ali Hussein | aliiabdi@hotmail.com
* Kazi Ahmed | kaziahammed123@hotmail.com

<pre>Guys please remember to write the Qs before your answers or it will be confusing when we review later on.

Our homework is categorized into pages BELOW! -- Ellen Tsang</pre>

'''Basic Skills Project'''

For the basic skills project, we will do inequalities and give several examples to cover all cases of questions concerning this topic.

==Subpages==

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

Course:MATH110/Archive/2010-2011/003/Groups/Group 09/Homework 4

2010-10-20T02:25:55Z

EdithFan:

Hi guys, please choose whichever Q you want to do.

Q1: Ellen

Q2:

Q3: MF (Do I have to show how I got the answers?) - yeap.

Q4:

Q5: Edith

==Problem 1==
'''Five persons named their pets after each other. From the following clues, can you decide which pet belongs to Suzan's mother?
* Tosh owns a cat,
* Bianca owns a frog that she loves,
* Jaela owns a parrot which keeps calling her "darling, darling",
* Jun owns a snake, don't mess with him,
* Suzan is the name of the frog,
* The cat is named Jun,
* The name by which they call the turtle is the name of the woman whose pet is Tosh,
* Finally, Suzan's mother's pet is Bianca.'''

Tosh owns a cat - Jun

Bianca owns a frog - Suzan

Jaela owns a parrot - Bianca or Tosh (Bianca if Jaela is Suzan's mother)

Jun owns a snake - Bianca or Tosh (Bianca if Jaela is Suzan's mother)

Suzan owns a turtle - Jaela or Jun (Name of the woman whose pet is Tosh)

==Problem 2==
'''Bohao, Stewart, Dylan, Tim and Chan are the five players of a basketball team. Two are left handed and three right handed, Two are over 2m tall and three are under 2m, Bohao and Dylan are of the same handedness, whereas Tim and Chan use different hands. Stewart and Chan are of the same height range, while Dylan and Tim are in different height ranges. If you know that the one playing centre is over 2m tall and is left handed, can you guess his name?'''

==Problem 3==
'''Adam, Bobo, Charles, Ed, Hassan, Jason, Mathieu, Pascal and Sung have formed a baseball team. The following facts are true:
* Adam does not like the catcher,
* Ed's sister is engaged to the second baseman,
* The centre fielder is taller than the right fielder,
* Hassan and the third baseman live in the same building,
* Pascal and Charles each won $20 from the pitcher at a poker game,
* Ed and the outfielders play cards during their free time,
* The pitcher's wife is the third baseman's sister,
* All the battery and infield except Charles, Hassan and Adam are shorter than Sung,
* Pascal, Adam and the shortstop lost $100 each at the race track,
* The second baseman beat Pascal, Hassan, Bobo and the catcher at billiards,
* Sung is in the process of getting a divorce,
* The catcher and the third baseman each have two legitimate children,
* Ed, Pascal Jason, the right fielder and the centre fielder are bachelors, the others are all married
* The shortstop, the third baseman and Bobo all attended the fight,
* Mathieu is the shortest player of the team,
Determine the positions of each player on the baseball team.'''

''Note: On a baseball team there are three outfielders (right, centre and left), four infielders (first baseman, second baseman, third baseman and shortstop) and the battery (pitcher and catcher).''

* Adam is the third baseman
* Bobo is the centre fielder
* Charles is the catcher
* Ed is the first baseman
* Hassan is the pitcher
* Jason is the second baseman
* Mathieu is the right fielder
* Pascal is the left fielder
* Sung is the shortstop

==Problem 4==
'''Six players - Petra, Carla, Janet, Sandra, Li and Fernanda - are competing in a chess tournament over a period of five days. Each player plays each of the others once. Three matches are played simultaneously during each of the five days. The first day, Carla beats Petra after 36 moves. The second day, Carla was again victorious when Janet failed to complete 40 moves within the required time limit. The third day had the most exciting match of all when Janet declared that she would checkmate Li in 8 moves and succeeded in doing so. On the fourth day, Petra defeated Sandra. Who played against Fernanda on the fifth day?'''

==Problem 5==
'''Homer finally had a week off from his job at the nuclear power plant and intended to spend all nine days of his vacation (Saturday through the following Sunday) sleeping late. But his plans were foiled by some of the people who work in his neighbourhood.'''

*On Saturday, his first morning off, Homer was wakened by the doorbell; it was a salesman of magazine subscriptions.
*On Sunday, the barking of the neighbour's dog abruptly ended Homer's sleep.
*On Monday, he was again wakened by the persistent salesman but was able to fall asleep again, only to be disturbed by the construction workers next door.
*In fact, the salesman, the neighbour's dog and the construction workers combined to wake Homer at least once each day of his vacation, with only one exception.
*The salesman woke him again on Wednesday; the construction workers on the second Saturday; the dog on Wednesday and on the final Sunday.
*No one of the three noisemakers was quiet for three consecutive days; but yet, no pair of them made noise on more than one day during Homer's vacation. On which day of his holiday was Homer actually able to sleep late?

S = Salesman

D = Neighbour's dog

C = Construction Workers

Confirmed:

Sat: S

Sun: D

Mon: S+C

Tues: ?

Wed: S+D

Thurs: ?

Fri: ?

Sat: C

Sun: D

Important!

== //No one of the three noisemakers was quiet for three consecutive days

== //No pair of them made noise on more than one day during Homer's vacation

--> Then it means that there could be a pair of construction workers + dog on a certain day

// The construction worker has to bother Homer on Thursday .

// From above information, the salesman has to come by on Friday (as the last confirmed date he came by is Wednesday) as he can't come by on Thursday.

// The last time the dog barks is on Sunday so the dog has to bark on a day not past Thursday. But the above information states on pair makes noise for more than a day so the dog has to bark on Thursday. It will work as the construction workers and dog didn't combine their forces for any time in the confirmed information.

Chart:

Sat: S

Sun: D

Mon: S+C

Tues: ? ==> REST!!

Wed: S+D

Thurs: ? ==> C+D

Fri: ? ==> S

Sat: C

Sun: D

Then with all the combined information, the holiday day where Homer gets to sleep in has to be Tuesday.

Course:MATH110/Archive/2010-2011/003/Groups/Group 09/Homework 4

2010-10-20T02:25:36Z

EdithFan:

Hi guys, please choose whichever Q you want to do.

Q1: Ellen

Q2:

Q3: MF (Do I have to show how I got the answers?) - yeap.

Q4:

Q5: Edith

==Problem 1==
'''Five persons named their pets after each other. From the following clues, can you decide which pet belongs to Suzan's mother?
* Tosh owns a cat,
* Bianca owns a frog that she loves,
* Jaela owns a parrot which keeps calling her "darling, darling",
* Jun owns a snake, don't mess with him,
* Suzan is the name of the frog,
* The cat is named Jun,
* The name by which they call the turtle is the name of the woman whose pet is Tosh,
* Finally, Suzan's mother's pet is Bianca.'''

Tosh owns a cat - Jun

Bianca owns a frog - Suzan

Jaela owns a parrot - Bianca or Tosh (Bianca if Jaela is Suzan's mother)

Jun owns a snake - Bianca or Tosh (Bianca if Jaela is Suzan's mother)

Suzan owns a turtle - Jaela or Jun (Name of the woman whose pet is Tosh)

==Problem 2==
'''Bohao, Stewart, Dylan, Tim and Chan are the five players of a basketball team. Two are left handed and three right handed, Two are over 2m tall and three are under 2m, Bohao and Dylan are of the same handedness, whereas Tim and Chan use different hands. Stewart and Chan are of the same height range, while Dylan and Tim are in different height ranges. If you know that the one playing centre is over 2m tall and is left handed, can you guess his name?'''

==Problem 3==
'''Adam, Bobo, Charles, Ed, Hassan, Jason, Mathieu, Pascal and Sung have formed a baseball team. The following facts are true:
* Adam does not like the catcher,
* Ed's sister is engaged to the second baseman,
* The centre fielder is taller than the right fielder,
* Hassan and the third baseman live in the same building,
* Pascal and Charles each won $20 from the pitcher at a poker game,
* Ed and the outfielders play cards during their free time,
* The pitcher's wife is the third baseman's sister,
* All the battery and infield except Charles, Hassan and Adam are shorter than Sung,
* Pascal, Adam and the shortstop lost $100 each at the race track,
* The second baseman beat Pascal, Hassan, Bobo and the catcher at billiards,
* Sung is in the process of getting a divorce,
* The catcher and the third baseman each have two legitimate children,
* Ed, Pascal Jason, the right fielder and the centre fielder are bachelors, the others are all married
* The shortstop, the third baseman and Bobo all attended the fight,
* Mathieu is the shortest player of the team,
Determine the positions of each player on the baseball team.'''

''Note: On a baseball team there are three outfielders (right, centre and left), four infielders (first baseman, second baseman, third baseman and shortstop) and the battery (pitcher and catcher).''

* Adam is the third baseman
* Bobo is the centre fielder
* Charles is the catcher
* Ed is the first baseman
* Hassan is the pitcher
* Jason is the second baseman
* Mathieu is the right fielder
* Pascal is the left fielder
* Sung is the shortstop

==Problem 4==
'''Six players - Petra, Carla, Janet, Sandra, Li and Fernanda - are competing in a chess tournament over a period of five days. Each player plays each of the others once. Three matches are played simultaneously during each of the five days. The first day, Carla beats Petra after 36 moves. The second day, Carla was again victorious when Janet failed to complete 40 moves within the required time limit. The third day had the most exciting match of all when Janet declared that she would checkmate Li in 8 moves and succeeded in doing so. On the fourth day, Petra defeated Sandra. Who played against Fernanda on the fifth day?'''

==Problem 5==
'''Homer finally had a week off from his job at the nuclear power plant and intended to spend all nine days of his vacation (Saturday through the following Sunday) sleeping late. But his plans were foiled by some of the people who work in his neighbourhood.'''

*On Saturday, his first morning off, Homer was wakened by the doorbell; it was a salesman of magazine subscriptions.
*On Sunday, the barking of the neighbour's dog abruptly ended Homer's sleep.
*On Monday, he was again wakened by the persistent salesman but was able to fall asleep again, only to be disturbed by the construction workers next door.
*In fact, the salesman, the neighbour's dog and the construction workers combined to wake Homer at least once each day of his vacation, with only one exception.
*The salesman woke him again on Wednesday; the construction workers on the second Saturday; the dog on Wednesday and on the final Sunday.
*No one of the three noisemakers was quiet for three consecutive days; but yet, no pair of them made noise on more than one day during Homer's vacation. On which day of his holiday was Homer actually able to sleep late?

S = Salesman
D = Neighbour's dog
C = Construction Workers

Confirmed:

Sat: S

Sun: D

Mon: S+C

Tues: ?

Wed: S+D

Thurs: ?

Fri: ?

Sat: C

Sun: D

Important!

== //No one of the three noisemakers was quiet for three consecutive days

== //No pair of them made noise on more than one day during Homer's vacation

--> Then it means that there could be a pair of construction workers + dog on a certain day

// The construction worker has to bother Homer on Thursday .

// From above information, the salesman has to come by on Friday (as the last confirmed date he came by is Wednesday) as he can't come by on Thursday.

// The last time the dog barks is on Sunday so the dog has to bark on a day not past Thursday. But the above information states on pair makes noise for more than a day so the dog has to bark on Thursday. It will work as the construction workers and dog didn't combine their forces for any time in the confirmed information.

Chart:

Sat: S

Sun: D

Mon: S+C

Tues: ? ==> REST!!

Wed: S+D

Thurs: ? ==> C+D

Fri: ? ==> S

Sat: C

Sun: D

Then with all the combined information, the holiday day where Homer gets to sleep in has to be Tuesday.

Course:MATH110/Archive/2010-2011/003/Groups/Group 09/Homework 4

2010-10-20T02:05:00Z

EdithFan:

Hi guys, please choose whichever Q you want to do.

Q1: Ellen

Q2:

Q3: MF (Do I have to show how I got the answers?)

Q4:

Q5: Edith

==Problem 1==
'''Five persons named their pets after each other. From the following clues, can you decide which pet belongs to Suzan's mother?
* Tosh owns a cat,
* Bianca owns a frog that she loves,
* Jaela owns a parrot which keeps calling her "darling, darling",
* Jun owns a snake, don't mess with him,
* Suzan is the name of the frog,
* The cat is named Jun,
* The name by which they call the turtle is the name of the woman whose pet is Tosh,
* Finally, Suzan's mother's pet is Bianca.'''

Tosh owns a cat - Jun

Bianca owns a frog - Suzan

Jaela owns a parrot - Bianca or Tosh (Bianca if Jaela is Suzan's mother)

Jun owns a snake - Bianca or Tosh (Bianca if Jaela is Suzan's mother)

Suzan owns a turtle - Jaela or Jun (Name of the woman whose pet is Tosh)

==Problem 2==
'''Bohao, Stewart, Dylan, Tim and Chan are the five players of a basketball team. Two are left handed and three right handed, Two are over 2m tall and three are under 2m, Bohao and Dylan are of the same handedness, whereas Tim and Chan use different hands. Stewart and Chan are of the same height range, while Dylan and Tim are in different height ranges. If you know that the one playing centre is over 2m tall and is left handed, can you guess his name?'''

==Problem 3==
'''Adam, Bobo, Charles, Ed, Hassan, Jason, Mathieu, Pascal and Sung have formed a baseball team. The following facts are true:
* Adam does not like the catcher,
* Ed's sister is engaged to the second baseman,
* The centre fielder is taller than the right fielder,
* Hassan and the third baseman live in the same building,
* Pascal and Charles each won $20 from the pitcher at a poker game,
* Ed and the outfielders play cards during their free time,
* The pitcher's wife is the third baseman's sister,
* All the battery and infield except Charles, Hassan and Adam are shorter than Sung,
* Pascal, Adam and the shortstop lost $100 each at the race track,
* The second baseman beat Pascal, Hassan, Bobo and the catcher at billiards,
* Sung is in the process of getting a divorce,
* The catcher and the third baseman each have two legitimate children,
* Ed, Pascal Jason, the right fielder and the centre fielder are bachelors, the others are all married
* The shortstop, the third baseman and Bobo all attended the fight,
* Mathieu is the shortest player of the team,
Determine the positions of each player on the baseball team.'''

''Note: On a baseball team there are three outfielders (right, centre and left), four infielders (first baseman, second baseman, third baseman and shortstop) and the battery (pitcher and catcher).''

* Adam is the third baseman
* Bobo is the centre fielder
* Charles is the catcher
* Ed is the first baseman
* Hassan is the pitcher
* Jason is the second baseman
* Mathieu is the right fielder
* Pascal is the left fielder
* Sung is the shortstop

==Problem 4==
'''Six players - Petra, Carla, Janet, Sandra, Li and Fernanda - are competing in a chess tournament over a period of five days. Each player plays each of the others once. Three matches are played simultaneously during each of the five days. The first day, Carla beats Petra after 36 moves. The second day, Carla was again victorious when Janet failed to complete 40 moves within the required time limit. The third day had the most exciting match of all when Janet declared that she would checkmate Li in 8 moves and succeeded in doing so. On the fourth day, Petra defeated Sandra. Who played against Fernanda on the fifth day?'''

==Problem 5==
'''Homer finally had a week off from his job at the nuclear power plant and intended to spend all nine days of his vacation (Saturday through the following Sunday) sleeping late. But his plans were foiled by some of the people who work in his neighbourhood.'''

*On Saturday, his first morning off, Homer was wakened by the doorbell; it was a salesman of magazine subscriptions.
*On Sunday, the barking of the neighbour's dog abruptly ended Homer's sleep.
*On Monday, he was again wakened by the persistent salesman but was able to fall asleep again, only to be disturbed by the construction workers next door.
*In fact, the salesman, the neighbour's dog and the construction workers combined to wake Homer at least once each day of his vacation, with only one exception.
*The salesman woke him again on Wednesday; the construction workers on the second Saturday; the dog on Wednesday and on the final Sunday.
*No one of the three noisemakers was quiet for three consecutive days; but yet, no pair of them made noise on more than one day during Homer's vacation. On which day of his holiday was Homer actually able to sleep late?

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

2010-10-13T06:04:36Z

EdithFan:

Group members:
* Kazi Ahmed
* Edith Fan | xsaret@gmail.com
* Maria Fayloga | maria.fayloga@gmail.com
* Ali Hussein
* Ellen Tsang | ellentsang.nl@hotmail.com

14. One day in the maternity ward, the name tags for four girl babies became mixed up. (i) In how many different ways could two of the babies be tagged correctly and two of the babies be tagged incorrectly? (ii) In how many different ways could three of the babies be tagged correctly and one baby be tagged incorrectly?

ANS: (I)there are six different ways of tagging two of the babies correctly and two incorrectly. (II) it is impossible to get three of the babies tagged correctly and one incorrectly since there are only four in number and if three are tagged correctly, then the remaining will obviously be tagged correctly also.

15. Alex says to you, “I'll bet you any amount of money that if I shuffle this deck of cards, there will always be as many red cards in the first half of the deck as there are black cards in the second half of the deck.” Should you accept his bet?

ANS: i would neither accept nor refuse since there are 52 cards in a deck and consists of 26 red cards and 26 black cards. this is because the probility of getting red or blacks cards in the first half and second half is the same; 1/2

=Homework 3, third part - Problem-Solving Skills=

==Q1==
'''1. A bus traveled from the terminal to the airport at an average speed of 30 mi/hr and the trip took an hour and 20 min. The bus then traveled from the airport back to the terminal and again averaged 30 mi/hr. However, the return trip required 80 min. Explain.'''

The time taken for the bus to travel from the terminal to the airport is:
1 hour = 60 min
1 hour 20 min = 60 + 20 = 80 min

This is the same as the time taken for the bus to travel from the airport back to the terminal, 80 min.

The average speed for both is the same, 30 mi/hr, hence the time taken to and back from the airport is the same, 80 min.

==Q21==

Sven = middle of all runners so the total amount of runners should be odd -> middle of 7 is 4; but there is no middle of say, 8.

Dan is slower than Sven (in tenth place) so -> Sven > 10

Lars is in 16th place so Sven > Dan > 16

- assumed that Lars is not last as it was not stated explicitly in the question then the total runners > 16.

Therefore, the total number of runners has to be 17 because if the total number of runners exceed that, say, 19 then Sven would be number 20. However, that is not possible as the question states that he is slower than Sven at 10th place; Sven and Dan can not coexist in the same place number.

==Q22==

If we designate raining as R and sunny periods of time as S then we can make a simple table.

First, if it rains in the morning ---> afternoon is sunny

- rainy afternoon ---> sunny morning

Total R = 13

11 Sunny mornings

12 Sunny afternoons

Morn - R S S S S R R S S S S S R R S S R R

Noon - S S R R S S S R R S R S S S R S S S

^Sunny Mornings = 11

Sunny Noon = 12

Rainy = 13

--> Total days of vacation = 18

==Q23==

Products of age = 36

Oldest child = x

Second child = y

Third child = z

- then x * y * z = 36

Sum of all ages = a date

- then x + y + z = d ;; but d <= 31 (the largest amount of days in a month is 31)

A way to go about this question is to list every possibility that x, y and z will multiply to 36.

[[File:Tttable.png]]

The original question states that Paul didn't get it even though it the ages add up to a date. The table above shows that all of the sums (or dates) are unique so the only possibilities are ages 2, 2, 9 and ages 1, 6, 6. However, if we assume that the oldest child and the second are not the same age (or twins) then we can say that x > y => z. So it leaves that Paula's children are ages 2, 2 and 9 respectively.

==Q24==

Time taken for candle A to burn out = 6 hours

Time taken for candle B to burn out = 3 hours

And to solve the question, we can say that Candle A = 2*Candle B

If we assume time starts at 0 and 'burning of the candle' is constant then:

Candle A = 6 hours then it means it will burn 1/6 per hour

Candle B = same as above; it will burn 1/3 per hour

Then Candle A = (6-x)/6

Candle B = (3-x)/6

x = time in hours

- as stated above: Candle A = 2(Candle B)

Then if we calculate the two states of Candle B before it burns out and not at its full length:

<math>2((x-3)/3) = (6-x)/6</math>

<math>6(6-2x) = 3(6-x)</math>

<math>36-12x = 18-3x</math>

<math>18 = 9x</math>

<math>x = 2 hours</math>

So the time it takes for one of the candle to be half of the other candle is 2 hours.

==Q25==

Candle A has length L and is lit at 6 h.

Candle B has length (L + 1) and is lit at 4.5 h.

At 8.5 h, Candle A = Candle B

Candle A burns out at 10 h while Candle B burns out at 10.5 h. Then the total burning time of Candle A = 10-6 = 4h; Candle B = 10.5-4.5 = 6h.

Burning time till 8.5h for Candle A = 8.5 - 6 = 2.5h

Burning time till 8.5h for Candle B = 8.5 - 4.5 = 4h

let x = candle a length ;; aka. L

let y = candle b length

--> y = x+1

let b1 = burning constant for x

let b2 = burning constant for y

-> Candle A = Candle B at 8.5 h

<math>x-4*B1 = x+1-2.5B2</math>

--> x cancels out to

<math>4*B1 = 1-2.5B2</math>

---> What at B1 and B2 then?

<math>y=6B2</math> ;; at end time

<math>x=4B1</math> ;; at end time

<math> x = y</math>

<math>6B2 = 4B1 + 1</math> ;; the one comes from x+1

<math>B2 = (4B1+1)/6</math>

<math>B1 = (6B1-1)/4</math>
Using the above equations and substituting back into the original

---> <math>4*B2 = 1-2.5B1</math>

<math>4*B2 = 1-2.5((6B2-1)/4)</math>

<math>4*B2 = 1-((15B2-2.5)/4)</math>

<math>4(4B2-1) = 15B2-2.5</math>

<math>16B2-4 = 15B2-2.5</math>

<math>B2 = -2.5+4</math>

<math>B2=1.5</math> <-- plug in back to y=6B2

<math>4*((4B1+1)/6) = 1-2.5B1</math>

<math>B1=2</math> <-- plug in back to x = 4B1

<math> y = 6B2</math>

<math> y = 6(1.5) = 9</math>

<math> x = 4B1</math>

<math> x = 4(2) = 8</math>

Length of Candle B (longer candle) is 9 units and the length of the Candle A (shorter candle) is 8 units.

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

2010-10-13T06:03:53Z

EdithFan:

Group members:
* Kazi Ahmed
* Edith Fan | xsaret@gmail.com
* Maria Fayloga | maria.fayloga@gmail.com
* Ali Hussein
* Ellen Tsang | ellentsang.nl@hotmail.com

14. One day in the maternity ward, the name tags for four girl babies became mixed up. (i) In how many different ways could two of the babies be tagged correctly and two of the babies be tagged incorrectly? (ii) In how many different ways could three of the babies be tagged correctly and one baby be tagged incorrectly?

ANS: (I)there are six different ways of tagging two of the babies correctly and two incorrectly. (II) it is impossible to get three of the babies tagged correctly and one incorrectly since there are only four in number and if three are tagged correctly, then the remaining will obviously be tagged correctly also.

15. Alex says to you, “I'll bet you any amount of money that if I shuffle this deck of cards, there will always be as many red cards in the first half of the deck as there are black cards in the second half of the deck.” Should you accept his bet?

ANS: i would neither accept nor refuse since there are 52 cards in a deck and consists of 26 red cards and 26 black cards. this is because the probility of getting red or blacks cards in the first half and second half is the same; 1/2

=Homework 3, third part - Problem-Solving Skills=

==Q1==
'''1. A bus traveled from the terminal to the airport at an average speed of 30 mi/hr and the trip took an hour and 20 min. The bus then traveled from the airport back to the terminal and again averaged 30 mi/hr. However, the return trip required 80 min. Explain.'''

The time taken for the bus to travel from the terminal to the airport is:
1 hour = 60 min
1 hour 20 min = 60 + 20 = 80 min

This is the same as the time taken for the bus to travel from the airport back to the terminal, 80 min.

The average speed for both is the same, 30 mi/hr, hence the time taken to and back from the airport is the same, 80 min.

==Q21==

Sven = middle of all runners so the total amount of runners should be odd -> middle of 7 is 4; but there is no middle of say, 8.

Dan is slower than Sven (in tenth place) so -> Sven > 10

Lars is in 16th place so Sven > Dan > 16

- assumed that Lars is not last as it was not stated explicitly in the question then the total runners > 16.

Therefore, the total number of runners has to be 17 because if the total number of runners exceed that, say, 19 then Sven would be number 20. However, that is not possible as the question states that he is slower than Sven at 10th place; Sven and Dan can not coexist in the same place number.

==Q22==

If we designate raining as R and sunny periods of time as S then we can make a simple table.

First, if it rains in the morning ---> afternoon is sunny

- rainy afternoon ---> sunny morning

Total R = 13

11 Sunny mornings

12 Sunny afternoons

Morn - R S S S S R R S S S S S R R S S R R

Noon - S S R R S S S R R S R S S S R S S S

^Sunny Mornings = 11

Sunny Noon = 12

Rainy = 13

--> Total days of vacation = 18

==Q23==

Products of age = 36

Oldest child = x

Second child = y

Third child = z

- then x * y * z = 36

Sum of all ages = a date

- then x + y + z = d ;; but d <= 31 (the largest amount of days in a month is 31)

A way to go about this question is to list every possibility that x, y and z will multiply to 36.

[[File:Tttable.png]]

The original question states that Paul didn't get it even though it the ages add up to a date. The table above shows that all of the sums (or dates) are unique so the only possibilities are ages 2, 2, 9 and ages 1, 6, 6. However, if we assume that the oldest child and the second are not the same age (or twins) then we can say that x > y => z. So it leaves that Paula's children are ages 2, 2 and 9 respectively.

==Q24==

Time taken for candle A to burn out = 6 hours

Time taken for candle B to burn out = 3 hours

And to solve the question, we can say that Candle A = 2*Candle B

If we assume time starts at 0 and 'burning of the candle' is constant then:

Candle A = 6 hours then it means it will burn 1/6 per hour

Candle B = same as above; it will burn 1/3 per hour

Then Candle A = (6-x)/6

Candle B = (3-x)/6

x = time in hours

- as stated above: Candle A = 2(Candle B)

Then if we calculate the two states of Candle B before it burns out and not at its full length:

<math>2((x-3)/3) = (6-x)/6</math>

<math>6(6-2x) = 3(6-x)</math>

<math>36-12x = 18-3x</math>

<math>18 = 9x</math>

<math>x = 2 hours</math>

So the time it takes for one of the candle to be half of the other candle is 2 hours.

==Q25==

Candle A has length L and is lit at 6 h.

Candle B has length (L + 1) and is lit at 4.5 h.

At 8.5 h, Candle A = Candle B

Candle A burns out at 10 h while Candle B burns out at 10.5 h. Then the total burning time of Candle A = 10-6 = 4h; Candle B = 10.5-4.5 = 6h.

Burning time till 8.5h for Candle A = 8.5 - 6 = 2.5h

Burning time till 8.5h for Candle B = 8.5 - 4.5 = 4h

let x = candle a length ;; aka. L

let y = candle b length

--> y = x+1

let b1 = burning constant for x

let b2 = burning constant for y

-> Candle A = Candle B at 8.5 h

<math>x-4*B1 = x+1-2.5B2</math>

--> x cancels out to

<math>4*B1 = 1-2.5B2</math>

---> What at B1 and B2 then?

<math>y=6B2</math> ;; at end time

<math>x=4B1</math> ;; at end time

<math> x = y</math>

<math>6B2 = 4B1 + 1</math> ;; the one comes from x+1

<math>B2 = (4B1+1)/6</math>

<math>B1 = (6B1-1)/4
Using the above equations and substituting back into the original

---> <math>4*B2 = 1-2.5B1</math>

<math>4*B2 = 1-2.5((6B2-1)/4)</math>

<math>4*B2 = 1-((15B2-2.5)/4)</math>

<math>4(4B2-1) = 15B2-2.5</math>

<math>16B2-4 = 15B2-2.5</math>

<math>B2 = -2.5+4</math>

<math>B2=1.5</math> <-- plug in back to y=6B2

<math>4*((4B1+1)/6) = 1-2.5B1</math>

<math>B1=2</math> <-- plug in back to x = 4B1

<math> y = 6B2</math>

<math> y = 6(1.5) = 9</math>

<math> x = 4B1</math>

<math> x = 4(2) = 8</math>

Length of Candle B (longer candle) is 9 units and the length of the Candle A (shorter candle) is 8 units.

Course talk:MATH110/003/Groups/Group 09

2010-10-13T05:00:47Z

EdithFan:

=Homework 3, third part - Problem-Solving Skills=

==Q1==
'''1. A bus traveled from the terminal to the airport at an average speed of 30 mi/hr and the trip took an hour and 20 min. The bus then traveled from the airport back to the terminal and again averaged 30 mi/hr. However, the return trip required 80 min. Explain.'''

Isn't it just that the bus travelled a different (shorter) route? Haha.

i think for this question the time taken the bus is just the same! for because an hour and 20 min is quall to 80 min

==Q21==

Sven = middle of all runners so the total amount of runners should be odd -> middle of 7 is 4; but there is no middle of say, 8.

Dan is slower than Sven (in tenth place) so -> Sven > 10

Lars is in 16th place so Sven > Dan > 16

- assumed that Lars is not last as it was not stated explicitly in the question then the total runners > 16.

Therefore, the total number of runners has to be 17 because if the total number of runners exceed that, say, 19 then Sven would be number 20. However, that is not possible as the question states that he is slower than Sven at 10th place; Sven and Dan can not coexist in the same place number.

==Q22==

If we designate raining as R and sunny periods of time as S then we can make a simple table.

First, if it rains in the morning ---> afternoon is sunny

- rainy afternoon ---> sunny morning

Total R = 13

11 Sunny mornings

12 Sunny afternoons

Morn - R S S S S R R S S S S S R R S S R R

Noon - S S R R S S S R R S R S S S R S S S

^Sunny Mornings = 11

Sunny Noon = 12

Rainy = 13

--> Total days of vacation = 18

==Q23==

Products of age = 36

Oldest child = x

Second child = y

Third child = z

- then x * y * z = 36

Sum of all ages = a date

- then x + y + z = d ;; but d <= 31 (the largest amount of days in a month is 31)

A way to go about this question is to list every possibility that x, y and z will multiply to 36.

[[File:Tttable.png]]

The original question states that Paul didn't get it even though it the ages add up to a date. The table above shows that all of the sums (or dates) are unique so the only possibilities are ages 2, 2, 9 and ages 1, 6, 6. However, if we assume that the oldest child and the second are not the same age (or twins) then we can say that x > y => z. So it leaves that Paula's children are ages 2, 2 and 9 respectively.

==Q24==

Time taken for candle A to burn out = 6 hours

Time taken for candle B to burn out = 3 hours

And to solve the question, we can say that Candle A = 2*Candle B

If we assume time starts at 0 and 'burning of the candle' is constant then:

Candle A = 6 hours then it means it will burn 1/6 per hour

Candle B = same as above; it will burn 1/3 per hour

Then Candle A = (6-x)/6

Candle B = (3-x)/6

x = time in hours

- as stated above: Candle A = 2(Candle B)

Then if we calculate the two states of Candle B before it burns out and not at its full length:

<math>2((x-3)/3) = (6-x)/6</math>

<math>6(6-2x) = 3(6-x)</math>

<math>36-12x = 18-3x</math>

<math>18 = 9x</math>

<math>x = 2 hours</math>

So the time it takes for one of the candle to be half of the other candle is 2 hours.

==Q25==

Candle A has length L and is lit at 6 h.

Candle B has length (L + 1) and is lit at 4.5 h.

At 8.5 h, Candle A = Candle B

Candle A burns out at 10 h while Candle B burns out at 10.5 h. Then the total burning time of Candle A = 10-6 = 4h; Candle B = 10.5-4.5 = 2h.

Burning time till 8.5h for Candle A = 8.5 - 6 = 2.5h

Burning time till 8.5h for Candle B = 8.5 - 4.5 = 4h

-> Candle A = Candle B

<math>L-2.5a = (L+1)-4b </math>

--> Then what is a and b which are the values of burning of the candle per hour

<math> 4a = L</math> ;; total burning time from above

<math> a = L/4</math>

<math> 2b = L+1</math>

<math> b = (L+1)/2</math>

Using the above equations and substituting back into the original we get:

<math>L-2.5(L/4)=(L+1)-4(L+1)/2 </math>

<math>L=8/11</math>

Length of Candle B (longer candle) is 19/11 cm and the length of the Candle A (shorter candle) is 8/11 cm.

Course talk:MATH110/003/Groups/Group 09

2010-10-13T04:58:15Z

EdithFan:

=Homework 3, third part - Problem-Solving Skills=

==Q1==
'''1. A bus traveled from the terminal to the airport at an average speed of 30 mi/hr and the trip took an hour and 20 min. The bus then traveled from the airport back to the terminal and again averaged 30 mi/hr. However, the return trip required 80 min. Explain.'''

Isn't it just that the bus travelled a different (shorter) route? Haha.

i think for this question the time taken the bus is just the same! for because an hour and 20 min is quall to 80 min

==Q21==

Sven = middle of all runners so the total amount of runners should be odd -> middle of 7 is 4; but there is no middle of say, 8.

Dan is slower than Sven (in tenth place) so -> Sven > 10

Lars is in 16th place so Sven > Dan > 16

- assumed that Lars is not last as it was not stated explicitly in the question then the total runners > 16.

Therefore, the total number of runners has to be 17 because if the total number of runners exceed that, say, 19 then Sven would be number 20. However, that is not possible as the question states that he is slower than Sven at 10th place; Sven and Dan can not coexist in the same place number.

==Q22==

If we designate raining as R and sunny periods of time as S then we can make a simple table.

First, if it rains in the morning ---> afternoon is sunny

- rainy afternoon ---> sunny morning

Total R = 13

11 Sunny mornings

12 Sunny afternoons

Morn - R S S S S R R S S S S S R R S S R R

Noon - S S R R S S S R R S R S S S R S S S

^Sunny Mornings = 11

Sunny Noon = 12

Rainy = 13

--> Total days of vacation = 18

==Q23==

Products of age = 36

Oldest child = x

Second child = y

Third child = z

- then x * y * z = 36

Sum of all ages = a date

- then x + y + z = d ;; but d <= 31 (the largest amount of days in a month is 31)

A way to go about this question is to list every possibility that x, y and z will multiply to 36.

[[File:Tttable.png]]

The original question states that Paul didn't get it even though it the ages add up to a date. The table above shows that all of the sums (or dates) are unique so the only possibilities are ages 2, 2, 9 and ages 1, 6, 6. However, if we assume that the oldest child and the second are not the same age (or twins) then we can say that x > y => z. So it leaves that Paula's children are ages 2, 2 and 9 respectively.

==Q24==

Time taken for candle A to burn out = 6 hours

Time taken for candle B to burn out = 3 hours

And to solve the question, we can say that Candle A = 2*Candle B

If we assume time starts at 0 and 'burning of the candle' is constant then:

Candle A = 6 hours then it means it will burn 1/6 per hour

Candle B = same as above; it will burn 1/3 per hour

Then Candle A = (6-x)/6

Candle B = (3-x)/6

x = time in hours

- as stated above: Candle A = 2(Candle B)

Then if we calculate the two states of Candle B before it burns out and not at its full length:

<math>2((x-3)/3) = (6-x)/6</math>

<math>6(6-2x) = 3(6-x)</math>

<math>36-12x = 18-3x</math>

<math>18 = 9x</math>

<math>x = 2 hours</math>

So the time it takes for one of the candle to be half of the other candle is 2 hours.

==Q25==

Candle A has length L and is lit at 6 h.

Candle B has length (L + 1) and is lit at 4.5 h.

At 8.5 h, Candle A = Candle B

Candle A burns out at 10 h while Candle B burns out at 10.5 h. Then the total burning time of Candle A = 10-6 = 4h; Candle B = 10.5-4.5 = 2h.

Burning time till 8.5h for Candle A = 8.5 - 6 = 2.5h

Burning time till 8.5h for Candle B = 8.5 - 4.5 = 4h

-> Candle A = Candle B

<math>L-2.5a = (L+1)-4b </math>

--> Then what is a and b which are the values of burning of the candle per hour
<math> 4a = L</math> ;; total burning time from above

<math> a = L/4</math>

<math> 2b = L+1</math>

<math> b = (L+1)/2</math>

Using the above equations and substituting back into the original we get:

<math>L-2.5(L/4)=(L+1)-4(L+1)/2 </math>

<math>L=8/11</math>

Length of Candle B (longer candle) is 19/11 cm and the length of the Candle A (shorter candle) is 8/11 cm.

Course talk:MATH110/003/Groups/Group 09

2010-10-13T02:29:33Z

EdithFan:

=Homework 3, third part - Problem-Solving Skills=

==Q1==
'''1. A bus traveled from the terminal to the airport at an average speed of 30 mi/hr and the trip took an hour and 20 min. The bus then traveled from the airport back to the terminal and again averaged 30 mi/hr. However, the return trip required 80 min. Explain.'''

Isn't it just that the bus travelled a different (shorter) route? Haha.

==Q21==

Sven placed exactly in the middle among all runners in a race. Dan was slower than Sven, in 10th place, and Lars was in 16th place. How many runners were in the race?

Sven = x/2

y = total number of runners

Dan is slower than Sven so -> Sven > Dan

Lars is in 16th place so Sven > Dan > Lars

==Q22==

If we designate raining as R and sunny periods of time as S then we can make a simple table.

First, if it rains in the morning ---> afternoon is sunny

- rainy afternoon ---> sunny morning

Total R = 13

11 Sunny mornings

12 Sunny afternoons

Morn - R S S S S R R S S S S S R R S S R R

Noon - S S R R S S S R R S R S S S R S S S

^Sunny Mornings = 11

Sunny Noon = 12

Rainy = 13

--> Total days of vacation = 18

==Q23==

Products of age = 36

Oldest child = x

Second child = y

Third child = z

- then x * y * z = 36

Sum of all ages = a date

- then x + y + z = d ;; but d <= 31 (the largest amount of days in a month is 31)

A way to go about this question is to list every possibility that x, y and z will multiply to 36.

[[File:Tttable.png]]

The original question states that Paul didn't get it even though it the ages add up to a date. The table above shows that all of the sums (or dates) are unique so the only possibilities are ages 2, 2, 9 and ages 1, 6, 6. However, if we assume that the oldest child and the second are not the same age (or twins) then we can say that x > y => z. So it leaves that Paula's children are ages 2, 2 and 9 respectively.

==Q24==

Time taken for candle A to burn out = 6 hours

Time taken for candle B to burn out = 3 hours

And to solve the question, we can say that Candle A = 2*Candle B

If we assume time starts at 0 and 'burning of the candle' is constant then:

Candle A = 6 hours then it means it will burn 1/6 per hour

Candle B = same as above; it will burn 1/3 per hour

Then Candle A = (6-x)/6

Candle B = (3-x)/6

x = time in hours

- as stated above: Candle A = 2(Candle B)

Then if we calculate the two states of Candle B before it burns out and not at its full length:

<math>2((x-3)/3) = (6-x)/6</math>

<math>6(6-2x) = 3(6-x)</math>

<math>36-12x = 18-3x</math>

<math>18 = 9x</math>

<math>x = 2 hours</math>

So the time it takes for one of the candle to be half of the other candle is 2 hours.

==Q25==

Candle A has length L and is lit at 6 h.

Candle B has length (L + 1) and is lit at 4.5 h.

At 8.5 h, Candle A = Candle B

Candle A burns out at 10 h while Candle B burns out at 10.5 h. Then the total burning time of Candle A = 10-6 = 4h; Candle B = 10.5-4.5 = 2h.

Burning time till 8.5h for Candle A = 8.5 - 6 = 2.5h

Burning time till 8.5h for Candle B = 8.5 - 4.5 = 4h

-> Candle A = Candle B

<math>L-2.5a = (L+1)-4b </math>

--> Then what is a and b which are the values of burning of the candle per hour
<math> 4a = L</math> ;; total burning time from above

<math> a = L/4</math>

<math> 2b = L+1</math>

<math> b = (L+1)/2</math>

Using the above equations and substituting back into the original we get:

<math>L-2.5(L/4)=(L+1)-4(L+1)/2 </math>

<math>L=8/11</math>

Length of Candle B (longer candle) is 19/11 cm and the length of the Candle A (shorter candle) is 8/11 cm.

Course talk:MATH110/003/Groups/Group 09

2010-10-13T02:28:03Z

EdithFan:

=Homework 3, third part - Problem-Solving Skills=

==Q1==
'''1. A bus traveled from the terminal to the airport at an average speed of 30 mi/hr and the trip took an hour and 20 min. The bus then traveled from the airport back to the terminal and again averaged 30 mi/hr. However, the return trip required 80 min. Explain.'''

Isn't it just that the bus travelled a different (shorter) route? Haha.

----

Question 21:

Sven placed exactly in the middle among all runners in a race. Dan was slower than Sven, in 10th place, and Lars was in 16th place. How many runners were in the race?

Sven = x/2

y = total number of runners

Dan is slower than Sven so -> Sven > Dan

Lars is in 16th place so Sven > Dan > Lars

----

Question 22:

If we designate raining as R and sunny periods of time as S then we can make a simple table.

First, if it rains in the morning ---> afternoon is sunny

- rainy afternoon ---> sunny morning

Total R = 13

11 Sunny mornings

12 Sunny afternoons

Morn - R S S S S R R S S S S S R R S S R R

Noon - S S R R S S S R R S R S S S R S S S

^Sunny Mornings = 11

Sunny Noon = 12

Rainy = 13

--> Total days of vacation = 18

----

Question 23:

Products of age = 36

Oldest child = x

Second child = y

Third child = z

- then x * y * z = 36

Sum of all ages = a date

- then x + y + z = d ;; but d <= 31 (the largest amount of days in a month is 31)

A way to go about this question is to list every possibility that x, y and z will multiply to 36.

[[File:Tttable.png]]

The original question states that Paul didn't get it even though it the ages add up to a date. The table above shows that all of the sums (or dates) are unique so the only possibilities are ages 2, 2, 9 and ages 1, 6, 6. However, if we assume that the oldest child and the second are not the same age (or twins) then we can say that x > y => z. So it leaves that Paula's children are ages 2, 2 and 9 respectively.

----

Question 24:

Time taken for candle A to burn out = 6 hours

Time taken for candle B to burn out = 3 hours

And to solve the question, we can say that Candle A = 2*Candle B

If we assume time starts at 0 and 'burning of the candle' is constant then:

Candle A = 6 hours then it means it will burn 1/6 per hour

Candle B = same as above; it will burn 1/3 per hour

Then Candle A = (6-x)/6

Candle B = (3-x)/6

x = time in hours

- as stated above: Candle A = 2(Candle B)

Then if we calculate the two states of Candle B before it burns out and not at its full length:

<math>2((x-3)/3) = (6-x)/6</math>

<math>6(6-2x) = 3(6-x)</math>

<math>36-12x = 18-3x</math>

<math>18 = 9x</math>

<math>x = 2 hours</math>

So the time it takes for one of the candle to be half of the other candle is 2 hours.

----

Question 25:

Candle A has length L and is lit at 6 h.

Candle B has length (L + 1) and is lit at 4.5 h.

At 8.5 h, Candle A = Candle B

Candle A burns out at 10 h while Candle B burns out at 10.5 h. Then the total burning time of Candle A = 10-6 = 4h; Candle B = 10.5-4.5 = 2h.

Burning time till 8.5h for Candle A = 8.5 - 6 = 2.5h

Burning time till 8.5h for Candle B = 8.5 - 4.5 = 4h

-> Candle A = Candle B

<math>L-2.5a = (L+1)-4b </math>

--> Then what is a and b which are the values of burning of the candle per hour
<math> 4a = L</math> ;; total burning time from above

<math> a = L/4</math>

<math> 2b = L+1</math>

<math> b = (L+1)/2</math>

Using the above equations and substituting back into the original we get:

<math>L-2.5(L/4)=(L+1)-4(L+1)/2 </math>

<math>L=8/11</math>

Length of Candle B (longer candle) is 19/11 cm and the length of the Candle A (shorter candle) is 8/11 cm.

File:Tttable.png

2010-10-13T00:47:39Z

EdithFan:

User:EdithFan

2010-09-22T18:32:59Z

EdithFan:

Hello.

----
==Pythagorean Theorem==

The origin/theory of the Pythagorean Theorem traces back to even Ancient Egypt but it was named after Pythagoras, a Mathematician, for proving said theorem<ref>http://en.wikipedia.org/wiki/Pythagorean_theorem</ref>. If one is given a right angle triangle and asked to solve for one of the sides, when given two of the lengths, a simple equation can be used.

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

The sides of the triangle should be labelled as ‘a’ and ‘b’ (or any other letter for that matter) and the hypotenuse should be the sides squared and added together. It could also be used to determine the type of triangle given to you by seeing how <math>a^2</math> and <math>b^2</math> add up in relationship to <math>c^2</math>. If the sum of the two squared numbers is greater than <math>c^2</math>, the triangle is acute; if it is the opposite result then it is obtuse<ref>hhttp://www.jimloy.com/geometry/pythag.htm</ref>. Some of the applications of this theorem in real life could range from finding the length of a shadow of a tree or the distance traveled if you went west 4 km and south 10 km. There are also different ways to rewrite the equation, depending on which value you are trying to solve.

The different ways goes as:

<math>
b= (\sqrt( c^2 - a^2))
</math>

<math>
a= (\sqrt(c^2 - b^2))
</math>

<math>
c= (\sqrt(a^2 + b^2))
</math>

<references/>

User:EdithFan

2010-09-20T05:12:00Z

EdithFan: Created page with 'Hello. ---- The origin/theory of the Pythagorean Theorem traces back to even Ancient Egypt but it was named after Pythagoras, a Mathematician, for proving said theorem. If on…'

Hello.

----

The origin/theory of the Pythagorean Theorem traces back to even Ancient Egypt but it was named after Pythagoras, a Mathematician, for proving said theorem. If one is given a right angle triangle and asked to solve for one of the sides, when given two of the lengths, a simple equation can be used.

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

The sides of the triangle should be labelled as ‘a’ and ‘b’ (or any other letter for that matter) and the hypotenuse should be the sides squared and added together. It could also be used to determine the type of triangle given to you by seeing how <math>a^2</math> and <math>b^2</math> add up in relationship to <math>c^2</math>. If the sum of the two squared numbers is greater than <math>c^2</math>, the triangle is acute; if it is the opposite result then it is obtuse. Some of the applications of this theorem in real life could range from finding the length of a shadow of a tree or the distance traveled if you went west 4 km and south 10 km. There are also different ways to rewrite the equation, depending on which value you are trying to solve.

The different ways goes as:

<math>
b= √( c^2 - a^2)
</math>

<math>
a= √( c^2 - b^2)
</math>

<math>
c= √( a^2 + b^2)
</math>

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

http://www.jimloy.com/geometry/pythag.htm