Solving Absolute Value Equations

1. Solving Absolute Value Equations

2. What is Absolute Value? • The absolute value of a number is the number of units it is from zero on the number line. • 5 and -5 have the same absolute value. • The symbol |x| represents the absolute value of the number x.

3. Solving Absolute Value Equations • Absolute value is denoted by the bars |3|. • Absolute value represents the distance a number is from 0. Thus, it is always positive. • |8| = 8 and |-8| = 8

4. |-1000| = 1000 • |4| = 4 • You try: • |15| = ? • |-23| = ? • Absolute Value can also be defined as : • if a >0, then |a| = a • if a < 0, then |a| = -a

5. Solving absolute value equations • First, isolate the absolute value expression. • Set up two equations to solve. • For the first equation, drop the absolute value bars and solve the equation. • For the second equation, drop the bars, negate the opposite side, and solve the equation. • Always check the solutions.

6. 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

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

8. 6|5x + 2| = 312 • Isolate the absolute value expression by dividing by 6. 6|5x + 2| = 312 |5x + 2| = 52 • Set up two equations to solve. • 5x + 2 = 52 5x + 2 = -52 • 5x = 50 5x = -54 • x = 10 or x = -10.8 • Check:6|5x + 2| = 312 6|5x + 2| = 312 • 6|5(10)+2| = 312 6|5(-10.8)+ 2| = 312 • 6|52| = 312 6|-52| = 312 • 312 = 312 312 = 312

9. 3|x + 2| -7 = 14 • Isolate the absolute value expression by adding 7 and dividing by 3. 3|x + 2| -7 = 14 3|x + 2| = 21 |x + 2| = 7 • Set up two equations to solve. • x + 2 = 7x + 2 = -7 • x = 5 or x = -9 • Check:3|x + 2| - 7 = 14 3|x + 2| -7 = 14 • 3|5 + 2| - 7 = 14 3|-9+ 2| -7 = 14 3|7| - 7 = 14 3|-7| -7 = 14 • 21 - 7 = 14 21 - 7 = 14 • 14 = 14 14 = 14

10. 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 = -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 |y + 4| - 3 = 0

11. 2x + 4 > 12 or 2x + 4 < -12 2x > 8 2x < -16 x > 4 or x < -8 x < -8 or x > 4 4 -8 0 Solve: |2x + 4| > 12

12. 9 -1 0 Solve: 2|4 - x| < 10 |4 - x| < 5 4 - x < 5 and 4 - x> -5 - x < 1 - x > -9 x > -1 and x < 9 -1 < x < 9

13. |3d - 9| + 6 = 0 First isolate the variable by subtracting 6 from both sides. |3d - 9| = -6 There is no need to go any further with this problem! • Absolute value is never negative. • Therefore, the solution is the empty set!

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

15. Solve: |8 + 5a| = 14 - a 8 + 5a = 14 - a and 8 + 5a = -(14 – a) Set up your 2 equations, but make sure to negate the entire right side of the second equation. 8 + 5a = 14 - a and 8 + 5a = -14 + a 6a = 6 4a = -22 a = 1 a = -5.5 Check: |8 + 5a| = 14 - a |8 + 5(1)| = 14 - 1 |8 + 5(-5.5) = 14 - (-5.5) |13| = 13 |-19.5| = 19.5 13 = 13 19.5 = 19.5