1. Welcome: Solve each equation and inequality: |3x + 5| = -5
No solution; (2) No solution; (3) All real numbers

2. HW Key: Solving Absolute Value Equations & Inequalities WKS
{4, 12}
(1, 7)
(-, -10] U [-4, )
[-0.5, 4.5]
{-10, 4}
(-, 0] U [6, )
[-4.25, .25]
(-2, 6)

3. Solving Equations and Inequalities Involving Two Absolute Value Expressions

4. Objectives and HW:
The students will be able to solve equations and inequalities involving two absolute value expressions.
HW: p. 125: all –> Do (a) – (c) for each

5. Solving |ax + b| = |cx + d|
To solve the equation |ax + b| = |cx + d| analytically, solve the compound equation: ax + b = cx + d or ax + b = -(cx + d)

6. Example 1: Solve analytically: |2x + 3| = |4x – 7|
Answer: {5, 2/3}

7. Example 2: Solve graphically: |2x + 3| = |4x – 7|
Let y1 = |2x + 3| and y2 = |4x – 7|.
Since y1 = y2, then y3 = y1 – y2
y3 = |2x + 3| – |4x – 7|
Put y3 in the calculator and graph.
Then find the zeros (2nd Trace, Zero) of y3, i.e. find the x-intercepts.
Note: Absolute value is found under Math, Num, 1:abs(.
Answer: {5, 2/3}

8. Example 3: Solve graphically: |2x + 3| > |4x – 7|
Using the same graph in example 2, simply analyze f(x) > 0 where f(x) = |2x + 3| – |4x – 7|
Answer: (2/3, 5); Note: Vertex is at about (1.75, 6.5).

9. Example 4: Solve graphically: |2x + 3| < |4x – 7|
Using the same graph in example 2, simply analyze f(x) < 0 where f(x) = |2x + 3| – |4x – 7|
Answer: (-, 2/3) U (5, ); Note: Vertex is at about (1.75, 6.5).

10. Students Try!!! 1) Solve analytically: |x + 6| = |3x – 5|.
Solve graphically:
a) |x + 6| > |3x – 5|
b) |x + 6| < |3x – 5|
Answers: (1) {-.25, 5.5}; (2a) (-.25, 5.5); (2b) (-, -.25) U (5.5, )