2. Type Example

3. Solution Example Find the solution set: a) |x| = 6; b) |x| = 0; c) |x| = –2 a) We interpret |x| = 6 to mean that the number x is 6 units from zero on a number line. Thus the solution set is {–6, 6}. 6 66 b) We interpret |x| = 0 to mean that x is 0 units from zero on a number line. The only number that satisfies this is zero itself. Thus the solution set is {0}. c) Since distance is always nonnegative, |x| = –2 has no solution. Thus, the solution set is

5. ExampleFind the solution set: a) |2x +1| = 5; b) |3 – 4x| = –10 The solution set is {–3, 2}. x = –3 or x = 2 2x = –6 or 2x = 4 Substituting a) We use the absolute-value principle, knowing that 2x + 1 must be either 5 or –5: |2x +1| = 5 |X| = p 2x +1 = –5 or 2x +1 = 5

6. Find the zeros using “CALC”. The solution set is {–3, 2}.

7. b) The absolute-value principle reminds us that absolute value is always nonnegative. The equation |3 – 4x| = –10 has no solution. The solution set is ExampleFind the solution set: a) |2x +1| = 5; b) |3 – 4x| = –10 No zeros! The solution set is

8. Solution ExampleGiven that f (x) = 3|x+5| – 4, find all x for which f (x) = 11 Since we are looking for f (x) = 11, we substitute: Replacing f (x) with 3|x + 5|  4 x +5 = –5 or x +5 = 5 x = –10 or x = 0 The solution set is {–10, 0}. 3|x+5| – 4 = 11 f (x) = 11 3|x+5| = 15 |x+5| = 5

9. Sometimes an equation has two absolute-value expressions. Consider |a| = |b|. This means that a and b are the same distance from zero. If a and b are the same distance from zero, then either they are the same number or they are opposites.

10. ExampleTo solve |3x – 5| = |8 + 4x| we would consider the two cases. and solve each equation. 3x – 5 = 8 + 4x This assumes these numbers are the same. This assumes these numbers are opposites. 3x – 5 = –(8 + 4x) or Solution

11. ExampleSolve |x| < 3. Then graph. The solutions of |x| < 3 are all numbers whose distance from zero is less than 3. By substituting we see that numbers like –2, –1, –1/2, 0, 1/3, 1, and 2 are all solutions. The solution set is {x| –3 < x < 3}. In interval notation, the solution is (–3, 3). The graph is as follows:  3 3

12. Solution ExampleSolve |x| < 3. Then graph. The solutions of |x| < 3 are all numbers whose distance from zero is less than 3. By substituting we see that numbers like –2, –1, –1/2, 0, 1/3, 1, and 2 are all solutions. The solution set is {x| –3 < x < 3}. In interval notation, the solution is (–3, 3). The graph is as follows:  3 3

13. Solution The solutions of are all numbers whose distance from zero great than or equal to 3 units. The solution set is In interval notation, the solution is The graph is as follows: Example  3 3

15. ExampleSolve.