2. You walk directly east from your house one block. How far from your house are you? 1 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. 2-7-6-5-4-3-21573 0468 4 units away from 0

3. What we are after here are values of x such that they are 6 away from 0. 2-7-6-5-4-3-21573 0468 6 and -6 are both 6 units away from 0 The "stuff" inside the absolute value signs could = 10 (the positive direction) or the "stuff" inside the absolute value signs could =  10 (the negative direction) Let's check it:

4. 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

5. 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

6. 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

7. Solve: 3|x - 5| = 12 |x - 5| = 4 x - 5 = 4and -(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

8. Based 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". To 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 -46 333 So x is inbetween or equal to - 4/3 and 2 2-7-6-5-4-3-21573 0468 Let's graph the solution:

9. What if the inequality is greater than? This is asking, "When is our distance from 0 more than 5 units away?" 2-7-6-5-4-3-21573 0468 Everything outside these lines is more than 5 units away from 0 So if we have it is equivalent to Everything outside these lines is more than 5 units away from 0 We'll have to express this with two difference pieces OR In interval notation:

10. Let's look at absolute value with an inequality. This is asking, "For what numbers is the distance from 0 less than 5 units?" 2-7-6-5-4-3-21573 0468 Everything inbetween these lines is less than 5 units away from 0 Inequality notationInterval notation So if we have it is equivalent to This means x is greater than -a and x is less than a (or x is inbetween -a and a)

11. This 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 6 From 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 Isolate x, remembering that if you multiply or divide by a negative you must turn the sign. - 2 We are dividing by a negative so turn the signs! or 2-7-6-5-4-3-21573 0468

12. Solving Absolute Value Inequalities 1.ax+b 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

13. Ex: Solve & graph. Becomes an “and” problem -3 7 8

14. Solve & graph. Get absolute value by itself first. Becomes an “or” problem -2 3 4