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Solving Absolute-Value Equations. 2-Ext. Holt Algebra 1. Lesson Presentation. Objective. Solve equations in one variable that contain absolute-value expressions. 5 units. The absolute-value of a number is that numbers distance from zero on a number line. For example, |–5| = 5.  1. 6.

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  1. Solving Absolute-Value Equations 2-Ext Holt Algebra 1 Lesson Presentation

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

  3. 5units The absolute-value of a number is that numbers distance from zero on a number line. For example, |–5| = 5. 1 6 4 3 0 1 2 3 4 5 5 2 6 Both 5 and –5 are a distance of 5 units from 0, so both 5 and –5 have an absolute value of 5. To write this 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.

  4. To solve absolute-value equations, perform inverse operations to isolate the absolute-value expression on one side of the equation. Then you must consider two cases.

  5. Case 2 x = –12 Case 1 x = 12 |x| = 12 |x| = 12 Check |12| 12 |12| 12   12 12 12 12 Example 1A: Solving Absolute-Value Equations Solve each equation. Check your answer. |x| = 12 Think: What numbers are 12 units from 0? |x| = 12 Rewrite the equation as two cases. The solutions are 12 and –12.

  6. Case 2 x + 7 = –8 Case 1 x + 7 = 8 – 7 –7 – 7 –7 x = 1 x = –15 Example 1B: Solving Absolute-Value Equations 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 and –15.

  7. Check 3|x + 7| = 24 3|x + 7| = 24 3|1+ 7| 24 3|15 + 7| 24 3|8| 24 3|8| 24 3(8) 24 3(8) 24   24 24 24 24 Example 1B Continued 3|x + 7| = 24 The solutions are 1 and –15.

  8. |x| –3 = 4 + 3 +3 |x| = 7 Case 2 –x = 7 Case 1 x = 7 –1(–x) = –1(7) x = 7 x = –7 Check It Out! Example 1a Solve each equation. Check your answer. |x| –3 = 4 Since 3 is subtracted from |x|, add 3 to both sides. Think: what numbers are 7 units from 0? Rewrite the case 2 equation by multiplying by –1 to change the minus x to a positive.. The solutions are 7 and –7.

  9. Check |x| 3 = 4 |x| 3 = 4 | 7| 3 4 |7|  3 4 7 3 4 7  3 4   4 4 4 4 Check It Out! Example 1a Continued Solve the equation. Check your answer. |x| 3 = 4 The solutions are 7 and 7.

  10. +2 +2 Case 1 x 2= 8 Case 2 x 2= 8 +2 +2 x = 10 x = 6 Check It Out! Example 1b Solve the equation. Check your answer. |x 2| = 8 Think: what numbers are 8 units from 0? |x 2| = 8 Rewrite the equations as two cases. Since 2 is subtracted from x add 2 to both sides of each equation. The solutions are 10 and 6.

  11. Check |x2| = 8 |x2| = 8 | 6 + (2)| 8 |102| 8 6 + 2 8 10 2| 8   8 8 8 8 Check It Out! Example 1b Continued Solve the equation. Check your answer. |x 2| = 8 The solutions are 10 and 6.

  12. Not all absolute-value equations have two solutions. If the absolute-value expression equals 0, there is one solution. If an equation states that an absolute-value is negative, there are no solutions.

  13. 8 = |x + 2|  8 +8 + 8 0 = |x +2| 0 = x + 2 2 2 2 = x Example 2A: Special Cases of Absolute-Value Equations Solve the equation. Check your answer. 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.

  14. Check 8 =|x + 2|  8 8 |2+ 2|  8 8 |0|  8 8 0  8  8 8 Example 2A Continued Solve the equation. Check your answer. 8 = |x +2|  8 Solution is x = 2 To check your solution, substitute 2 for x in your original equation.

  15. 3 + |x + 4| = 0 3 3 |x + 4| = 3 Example 2B: Special Cases of Absolute-Value Equations Solve the equation. Check your answer. 3 + |x + 4| = 0 Since 3 is added to |x + 4|, subtract 3 from both sides to undo the addition. Absolute values cannot be negative. This equation has no solution.

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

  17. 2  |2x 5| = 7 2 2  |2x 5| = 5 |2x  5| = 5 Check It Out! Example 2a Solve the equation. Check your answer. 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 a negative 1, divide both sides by negative 1. Absolute values cannot be negative. This equation has no solution.

  18. 6 + |x 4| = 6 +6 +6 |x 4| = 0 x 4 = 0 + 4 +4 x = 4 Check It Out! Example 2b Solve the equation. Check your answer. 6 + |x 4| = 6 Since 6 is subtracted from |x  4|, add 6 to both sides to undo the subtraction. There is only one case. Since 4 is subtracted from x, add 4 to both sides to undo the addition.

  19. 6 + |4 4| 6 6 +|0| 6 6 + 0 6 6 6 Check It Out! Example 2b Continued Solve the equation. Check your answer. 6 + |x 4| = 6 The solution is x = 4. To check your solution, substitute 4 for x in your original equation. 6 + |x 4| = 6

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