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.

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

What is an inequality?

It basically means when:

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

The symbols and meanings of the inequalities are as follows:

> means greater than

< means less than

≥ means greater than or equal to

≤ means less than or equal to

When to change the sign in an inequality problem?

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

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

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

   3x > 6

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

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

So the equation

   -4y < -36

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

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

   -(4y) < -(36)

Now turn off the antigravity machines:

   4y > 36

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

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

becomes

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

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

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

   y > 9

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

   -4y = -40 < -36

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

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

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

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

   a > b

then

   -a < -b.

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

   0 > b - a

Now subtract b from both sides:

   -b > -a

But that is the same as

   -a < -b

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

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

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

Basic Examples

Example 1

Solve -2x > 2

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

x < -1

Example 2

Solving linear inequalities is almost exactly like solving linear equations.

   * Solve x + 3 < 0.

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

           x < -3

     Then the solution is:

           x < –3

Example 3

* Solve x – 4 > 0.

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

     Then the solution is: x > 4

Example 4

   * Solve 2x < 9.

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

           x <= 9/2

     Then the solution is: x < 9/2

Example 5

   * Solve (2x – 3)/4  < 2.

First, multiply through by 4. Since the "4" is positive, we don't have to flip the inequality sign:

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

Example 6 - Compound Inequality

   * Solve 10 < 3x + 4 < 19.

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

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

Example 7

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

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

5x + 7 < 3(x + 1)

5x + 7 < 3x + 3

2x + 7 < 3

2x < –4

x < –2

Example 8

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

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
       $x<-31/7$

Example 9

   * Solve  $3<|3x+4|$

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

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

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

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 $x^{2}-2x-15>0$

Treat it like a normal quadratic equation and find the zeroes

   *  $x^{2}-2x-15=0$ 
      $(x-5)(x+3)=0$ 
     x-5=0 or x+3=0
     x=5 or -3

The zeroes divide the number line into three regions

For region x>5, check for a number greater than 5 (eg.6).

   * $(6)^{2}-2(6)-15>0$ 
    9>0
    All x-values greater than 5 will work.

For region x<-3, check for a number less than -3 (eg.-4).

   * $(-4)^{2}-2(-4)-15>0$ 
    9>0
    All x-values less than -3 will work.

   *Show answer using interval notation
    $(-infinity,-3)U(5,infinity)$

Then, x>5 and x<-3.

Example 2

Solve $3x^{2}>-x+4$

Make one side equal to zero

   * $3x^{2}+x-4>0$

Find the zeroes using the quadratic formula and factor

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

For region x> $-(4/3)$ , check for a number less than $-(4/3)$ (eg.-2).

   * $3(-2)^{2}+(-2)-4>0$ 
    6>0 
    All x-values less than  $-4/3$  will work.

For region x>1, check for a number greater than 1 (eg.4).

   * $3(4)^{2}+(4)-4>0$ 
    48>0
    All x-values greater than 1 will work.

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

Example 3

Solve for, $9x+9+3x^{2}>-x^{2}+7$

Move terms to one side so one side is zero

   * $x+9+3x^{2}+x^{2}-7>0$ 
     $4x^{2}+9x+2>0$

Then find the zeroes

   *(x+2)(4x+1)>0 
     x+2=0 or 4x+1=0
     x= $-2$  or  $-(1/4)$

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

   * $4x^{2}+9x+2>0$ 
     $4(-3)^{2}+9(-3)+2>0$ 
     $11>0$

Likewise, for the region x> $-1/4$

   * $4x^{2}+9x+2>0$ 
     $4(1)^{2}+9(1)+2>0$ 
     $15>0$

And lastly, some number between -2 and $-1/4$ .

   * $4x^{2}+9x+2>0$ 
     $4(-1)^{2}+9(-1)+2>0$ 
     $-3>0$ 
    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> $-1/4$ .

Example 4

Solve for, $-9x^{2}+7+22x>11x+2-15x^{2}$

Move terms to one side so one side is zero

   * $-9x^{2}+7+22x-11x-2+15x^{2}>0$ 
     $6x+11x+5>0$

Then find the zeroes

   *(6x+5)(x+1)>0 
     6x+5=0 or x+1=0
     x= $-5/6$  or -1

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

   * $6x+11x+5>0$ 
     $6(-4)+11(-4)+5>0$ 
     $30>0$

Likewise, for the region x>-1

   * $6x+11x+5>0$ 
     $6(6)+11(6)+5>0$ 
     $200>0$

And lastly, some number between $-5/6$ and -1.

   * $6x+11x+5>0$ 
     $6(-.9)+11(-.9)+5>0$ 
     $-2.86>0$ 
    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< $-5/6$ and x>-1.

References

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

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

Graphing Inequalities

Number Line

1. Simplify the inequality you are going to graph.

E.g.

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

-2x2 + 5x < -6x – 6

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

E.g.

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

0 < 2x2 -11x – 6

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

E.g.

0 = 2x2 -11x - 6

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

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

2x = -1, x = 6

x = -1/2

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.

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.

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

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).

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

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.

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

Tips

If x ≥ y then 1/x ≤ 1/y

Videos teaching Inequality

Solving Linear Inequalities

Useful Links

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

Group 10

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

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

What is an inequality?

When to change the sign in an inequality problem?

Basic Examples

Example 1

Example 2

Example 3

Example 4

Example 5

Example 6 - Compound Inequality

Example 7

Example 8

Example 9

Quadratic Inequalities

Example 1

Example 2

Example 3

Example 4

References

Graphing Inequalities

Tips

Videos teaching Inequality

Solving Linear Inequalities

Other Videos

Useful Links

Group 10