You walk directly east from your house one block. How far from your house are you?

1 block1 block

You walk directly west from your house one block. How far from your house are you?

It didn't matter which direction you walked, you were still 1 block from your house.

This is like absolute value. It is the distance from zero. It doesn't matter whether we are in the positive direction or the negative direction, we just care about how far away we are.

2 -7 -6 -5 -4 -3 -2 -1 1 5 7 3 0 4 6 8

44 4 units away from 044 4 units away from 0

6x What we are after here are values of x such that they are 6 away from 0.

2 -7 -6 -5 -4 -3 -2 -1 1 5 7 3 0 4 6 8

6 and -6 are both 6 units away from 0

6or 6 xx

104 xThe "stuff" inside the absolute value signs could = 10 (the positive direction) or the "stuff" inside the absolute value signs could = 10 (the negative direction)

104 x 104 x 14or 6 xx

Let's check it: 1046

1010

10414

1010

We can evaluate expressions that contain absolute value symbols.

• Think of the | | bars as grouping symbols. • Evaluate |9x -3| + 5 if x = -2

|9(-2) -3| + 5

|-18 -3| + 5

|-21| + 5

21+ 5=26

Equations may also contain absolute value expressions

• When solving an equation, isolate the absolute value expression first.

• Rewrite the equation as two separate equations. • Consider the equation | x | = 3. The equation has two solutions

since x can equal 3 or -3.• Solve each equation.• Always check your solutions.Example: Solve |x + 8| = 3

( x + 8) = 3 and -(x + 8) = 3 EXPLAIN ! -x-8=3 -x=11 x=-11

x = -5 x = -11 Check: |x + 8| = 3 |-5 + 8| = 3 |-11 + 8| = 3

|3| = 3 |-3| = 3 3 = 3 3 = 3

Now Try These• Solve |y + 4| - 3 = 0 |y + 4| = 3 You must first isolate the variable by adding

3 to both sides.

• Write the two separate equations. (y + 4) = 3 & -(y + 4) = 3

y = -1 -y-4=3 -y=7 y = -7• Check: |y + 4| - 3 = 0 |-1 + 4| -3 = 0 |-7 + 4| - 3 = 0

|-3| - 3 = 0 |-3| - 3 = 0 3 - 3 = 0 3 - 3 = 0

0 = 0 0 = 0

• Solve: 3|x - 5| = 12 |x - 5| = 4x - 5 = 4 and -(x – 5) = 4

x = 9 x = 1

• Check: 3|x - 5| = 12 3|9 - 5| = 12 3|1 - 5| = 12 3|4| = 12 3|-4| = 12 3(4) = 12 3(4) = 12 12 = 12 12 = 12

513 xBased on what we just observed, the "stuff" inside the absolute value signs is inbetween -5 and 5 or equal to either end since the inequality sign has "or equal to".

5135 xTo solve this we get x isolated in the middle. Whatever steps we do to get it alone, we do to each end. We keep in our minds the fact that if we multiply or divide by a negative, we must turn the signs the other way.

+1 +1+1

-4 6

634 x3 3 3

23

4 x So x is inbetween or equal to

- 4/3 and 2

2,

3

4

2 -7 -6 -5 -4 -3 -2 -1 1 5 7 3 0 4 6 8 Let's graph the solution:

What if the inequality is greater than? 5x

This is asking, "When is our distance from 0 more than 5 units away?"

2 -7 -6 -5 -4 -3 -2 -1 1 5 7 3 0 4 6 8 Everything outside these

lines is more than 5 units away from 0

5x 5x

So if we have it is equivalent to ax axax or

Everything outside these lines is more than 5 units away from 0We'll have to express

this with two difference pieces

OR

In interval notation: ),5(or )5,(

Let's look at absolute value with an inequality. 5x

This is asking, "For what numbers is the distance from 0 less than 5 units?"

2 -7 -6 -5 -4 -3 -2 -1 1 5 7 3 0 4 6 8 Everything

inbetween these lines is less than 5 units away from 0

55 x

Inequality notation

)5,5(

Interval notation

So if we have it is equivalent to ax axa

This means x is greater than -a and x is less than a(or x is inbetween -a and a)

3321 xThis means if there are other things on the left hand side of the inequality that are outside of the absolute value signs, we must get rid of them first.

We must first isolate the absolute value.

+3 +3

6

621 xFrom what we saw previously, the "stuff" inside the absolute value is either less than or equal to -6 or greater than or equal to 6

621or 621 xxIsolate x, remembering that if you multiply or divide by a negative you must turn the sign.

72 x 52 x- 2 - 2 - 2 - 2

We are dividing by a negative so turn the signs!

2

7x

2

5xor

2 -7 -6 -5 -4 -3 -2 -1 1 5 7 3 0 4 6 8

Solving Absolute Value Inequalities

1. ax+b < c, where c>0

Becomes an “and” problem

Changes to: –c<ax+b<c

2. ax+b > c, where c>0

Becomes an “or” problem

Changes to: ax+b>c or ax+b<-c

Ex: Solve & graph.

• Becomes an “and” problem

2194 x

219421 x30412 x

2

153 x

-3 7 8

Solve & graph.

• Get absolute value by itself first.

• Becomes an “or” problem

11323 x

823 x

823or 823 xx63or 103 xx

2or 3

10 xx

-2 3 4