1. Section 4.3 Solving Absolute Value Equations and Inequalities
Solving Equations of Form |X| = k
| 3x – 2 | = 5
Equations with Two Absolute Values
| 5x + 3 | = | 3x + 25 |
Solving Inequalities of Form |X| < k
| 2x – 3 | < 9
Solving Inequalities of Form |X| > k
| 3 – x | > 30

2. Absolute Value
Absolute value can allow 2 ways for an equation to be true
|5| = 5 and |-5| = 5
|y| = 8 has 2 solutions: y = 8 and y = -8
|7 – 4| = 3 and |-7 + 4| = 3
Sometimes there are not 2 solutions
|x| = -3 has no solutions
|z| = 0 has only one solution z = 0
Let X be any algebraic expression, then … |X| = p (p is a positive number) has 2 solutions
X = p and X = -p

3. x = 7 or x = -7
Impossible – no solutions
3x – 5 = 10 or 3x – 5 = -10 3x = x = -5 x = 5 or x = -5/3

4. Class Exercise
What are the solutions for:

5. Caution: Isolate the Absolute Value
What if the absolute value is negative? -|x + 2| = -3
First multiply both sides by -1, then solve:
(-1)(-|x + 2|) = (-1)(-3)
|x + 2| = 3
x + 2 = 3 or x + 2 = -3
x = or x = -5
-2|x + 6| = What do you do first?
|x + 6| = So… the solution?
3|4 – x| = 12
|4 – x| = 4
4 – x = 4 or 4 – x = -4
x = 0 or x = 8
Impossible !

6. More Isolation – Do Add or Subtract First
Try: |x/2 – 5| - 4 = -1
Add 4 to each side 3|x/2 – 5| = 3
Divide by |x/2 – 5| = 1
Now solve as usual x/2 – 5 = 1 or x/2 – 5 = x/2 = x/2 = x = 12 or x = 8
? -⅓|2 – x| + 2 = -8
? 5 - |3x – 5| = -4

7. Class Exercise
What are the solutions for:

8. Two Absolute Values?
|3| = |3| or |3| = |-3| or |-3| = |3| or |-3| = |-3|
Only 2 cases: They are equal, or opposite
If |X| = |Y| then solve X = Y or X = -Y
Let’s try it: |5x + 3| = |3x + 25|
5x + 3 = 3x or 5x + 3 = -(3x + 25)
or 5x + 3 = -3x – 25
2x = or x = -28
x = or x = -7/2

9. Class Exercise
What are the solutions for:

11. Examples | 5x – 12 | < -5 |2x – 5 | > 25
Let g(x) = | 4(3x + 2) – 2 | For what values of x is g(x) ≤ 4
Impossible – no solutions
2x – 5 < or 2x – 5 > 25
2x < or x > 30
x < or x > 15
(-∞,-10)U(15, ∞)
-4 ≤ 4(3x + 2) – 2 ≤ 4
-2 ≤ 4(3x + 2) ≤ 6
-2 ≤ 12x + 8 ≤ 6
-10 ≤ 12x ≤ -2
-5/6 ≤ x ≤ -1/6
[-5/6,-1/6]

12. Class Exercise: Solve & Graph < (and) > (or)
(-3,17/3)
――(―――)― /3
| -1 – 2x | > 5
-1 – 2x < -5 or -1 – 2x > 5
-2x < x > 6
x > x < -3
(-∞,-3)U(2,∞)
――)―――(――

13. What Next?
Present Section 4.4 Linear Inequalities with Two Variables