Algebra 1: Solving absolute value equations

1. Group Quiz Solve each equation. Bellringer: Finish this group quiz from yesterday. 1. 15 = | x | 2. 2| x – 7| = 14 3. | x + 1|– 9 = –9 4. |5 + x | – 3 = –2 5. 7 + | x – 8| = 6 – 15, 15 0, 14 – 1 – 6, –4 no solution

3. Recall that the absolute-value of a number is that number’s distance from zero on a number line. For example, |–5| = 5 and |5| = 5.  5  4  3  2 0 1 2 3 4 5  6  1 6 To write this statement using algebra, you would write | x | = 5. This equation asks, “What values of x have an absolute value of 5?” The solutions are 5 and –5. Notice this equation has two solutions. 5 units 5 units

5. Additional Example 1A: Solving Absolute-Value Equations Solve the equation. | x | = 12 | x | = 12 The solutions are {12, –12}. Think: What numbers are 12 units from 0? Rewrite the equation as two cases. Case 1 x = 12 Case 2 x = –12 12 units 12 units  10  8  6  4 0 2 4 6 8 10  12  2 12 • • •

6. Additional Example 1A: Solving Absolute-Value Equations Solve the equation. | x | = 9 | x | = 9 The solutions are {9, –9}. Think: What numbers are 9 units from 0? Rewrite the equation as two cases. Case 1 x = 9 Case 2 x = –9 9 units 9 units  10  8  6  4 0 2 4 6 8 10  12  2 12 • • •

7. Solve the equation. Check It Out! Example 1b 8 =| x  2.5| Think: What numbers are 8 units from 0? Case 1 8 = x  2.5 Case 2  8 = x  2.5 Rewrite the equations as two cases. The solutions are {10.5, –5.5}. 8 =| x  2.5| Since 2.5 is subtracted from x add 2.5 to both sides of each equation. +2.5 +2.5 10.5 = x +2.5 +2.5  5.5 = x

8. Solve the equation. Check It Out! Example 1b 7 =| x  4| Case 1 7 = x  4 Case 2  7 = x  4 The solutions are {11, –3}. 7 =| x  4| +4 +4 11 = x +4 +4 -3 = x

9. Solve the equation. Check It Out! Example 1a | x| – 3 = 4 Case 1 x = 7 Case 2 x = –7 The solutions are {7, –7}. Since 3 is subtracted from |x|, add 3 to both sides. Think: What numbers are 7 units from 0? Rewrite the equation as two cases. | x| – 3 = 4 + 3 +3 | x | = 7

10. Additional Example 1B: Solving Absolute-Value Equations 3| x + 7| = 24 | x + 7| = 8 The solutions are {1, –15}. Since |x + 7| is multiplied by 3, divide both sides by 3 to undo the multiplication. Think: What numbers are 8 units from 0? Rewrite the equations as two cases. Since 7 is added to x subtract 7 from both sides of each equation. Solve the equation. Case 1 x + 7 = 8 Case 2 x + 7 = –8 – 7 –7 – 7 – 7 x = 1 x = –15

11. Check It Out! Example 2b Solve the equation.  6 + | x  4| =  6 Since –6 is added to |x  4|, add 6 to both sides. There is only one case. Since 4 is subtracted from x, add 4 to both sides to undo the addition.  6 + | x  4| =  6 +6 +6 | x  4 | = 0 x  4 = 0 + 4 +4 x = 4

12. 2 nd Problem Solve the equation.  7 + | x  3| = -2 x  4 = 5  7 + | x  4| =  2 +7 +7 | x  4 | = 5 x  4 = -5 + 4 +4 x = -1 x  4 = 5 + 4 +4 x = 9

13. Additional Example 2B: Special Cases of Absolute-Value Equations Solve the equation. 3 + | x + 4| = 0 Absolute value cannot be negative. This equation has no solution. 3 + | x + 4| = 0  3  3 | x + 4| =  3

14. Remember! Absolute value must be nonnegative because it represents a distance.

15. Check It Out! Example 2a Solve the equation. 2  |2 x  5| = 7 Since 2 is added to –|2x – 5|, subtract 2 from both sides to undo the addition. Absolute value cannot be negative. |2 x  5| =  5 This equation has no solution. Since |2x – 5| is multiplied by negative 1, divide both sides by negative 1. 2  |2 x  5| = 7  2  2  |2 x  5| = 5