1-7 Solving Absolute-Value Equations Warm Up Lesson Presentation Holt Algebra 1 Warm Up Lesson Presentation Lesson Quiz Holt McDougal Algebra 1

Objectives Solve equations in one variable that contain absolute-value expressions.

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

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

Check It Out! Example 1b Solve the equation. 8 =|x  2.5| Think: What numbers are 8 units from 0? 8 =|x  2.5| Rewrite the equations as two cases. Case 1 8 = x  2.5 Case 2  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 The solutions are {10.5, –5.5}.

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

Additional Example 2A: Special Cases of Absolute-Value Equations Solve the equation. 8 = |x + 2|  8 Since 8 is subtracted from |x + 2|, add 8 to both sides to undo the subtraction. 8 = |x + 2|  8 +8 + 8 0 = |x + 2| There is only one case. Since 2 is added to x, subtract 2 from both sides to undo the addition. 0 = x + 2 2 2 2 = x The solution is {2}.

Additional Example 2B: Special Cases of Absolute-Value Equations Solve the equation. 3 + |x + 4| = 0 Since 3 is added to |x + 4|, subtract 3 from both sides to undo the addition. 3 + |x + 4| = 0 3 3 |x + 4| = 3 Absolute value cannot be negative. This equation has no solution.

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

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

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

Lesson Quiz Solve each equation. 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 6. Inline skates typically have wheels with a diameter of 74 mm. The wheels are manufactured so that the diameters vary from this value by at most 0.1 mm. Write and solve an absolute-value equation to find the minimum and maximum diameters of the wheels. |x – 74| = 0.1; 73.9 mm; 74.1 mm