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1-7. Solving Absolute-Value Equations. Holt Algebra 1. Warm Up. Lesson Presentation. Lesson Quiz. Holt McDougal Algebra 1. Objectives. Solve equations in one variable that contain absolute-value expressions. Case 2 x = –12. Case 1 x = 12. 12 units. 12 units. •. •. •.  2. 12.

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

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

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

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

  5. +2.5 +2.5 +2.5 +2.5 5.5 = x 10.5= x 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. The solutions are {10.5, –5.5}.

  6. Case 2 x + 7= –8 Case 1 x + 7= 8 – 7 –7 – 7 –7 x = 1 x = –15 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 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}.

  7. 8 = |x + 2|  8 +8 + 8 0 = |x + 2| 0 = x + 2 2 2 2 = x 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. There is only one case. Since 2 is added to x, subtract 2 from both sides to undo the addition. The solution is {2}.

  8. 3 + |x + 4| = 0 3 3 |x + 4| = 3 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. Absolute value cannot be negative. This equation has no solution.

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

  10. 2  |2x 5| = 7 2 2  |2x 5| = 5 |2x  5| = 5 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. Since |2x – 5| is multiplied by negative 1, divide both sides by negative 1. Absolute value cannot be negative. This equation has no solution.

  11. 6 + |x 4| = 6 +6 +6 |x 4| = 0 x 4 = 0 + 4 +4 x = 4 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.

  12. 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

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