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2-Ext Solving Solving Absolute-Value Absolute-Value Equations 2-Ext Equations Lesson Presentation Holt Algebra 1 2-Ext Solving Absolute-Value Equations Objective Solve equations in one variable that contain absolute-value expressions. Holt Algebra 1 2-Ext Solving Absolute-Value Equations The absolute-value of a number is that numbers distance from zero on a number line. For example, |–5| = 5. 5 units 6 5 4 3 2 1 0 1 2 3 4 5 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. Holt Algebra 1 2-Ext Solving Absolute-Value Equations Holt Algebra 1 2-Ext Solving Absolute-Value Equations To solve absolute-value equations, perform inverse operations to isolate the absolutevalue expression on one side of the equation. Then you must consider two cases. Holt Algebra 1 2-Ext Solving Absolute-Value Equations Example 1A: Solving Absolute-Value Equations Solve each equation. Check your answer. |x| = 12 |x| = 12 Case 1 x = 12 Think: What numbers are 12 units from 0? Case 2 x = –12 Rewrite the equation as two cases. The solutions are 12 and –12. Check Holt Algebra 1 |x| = 12 |12| 12 12 12 |x| = 12 |12| 12 12 12 2-Ext Solving Absolute-Value Equations 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 |x + 7| = 8 units from 0? Rewrite the equations as two Case 2 Case 1 cases. Since 7 is added to x x + 7 = 8 x + 7 = –8 subtract 7 from both sides – 7 –7 –7 –7 of each equation. x =1 x = –15 The solutions are 1 and –15. Holt Algebra 1 2-Ext Solving Absolute-Value Equations Example 1B Continued 3|x + 7| = 24 The solutions are 1 and –15. Check 3|x + 7| = 24 3|1 + 7| 24 3|15 + 7| 24 3|8| 24 3|8| 24 3(8) 24 3(8) 24 Holt Algebra 1 3|x + 7| = 24 24 24 24 24 2-Ext Solving Absolute-Value Equations Check It Out! Example 1a Solve each equation. Check your answer. |x| – 3 |x| – 3 +3 |x| Case 1 x =7 x=7 =4 =4 +3 =7 Since 3 is subtracted from |x|, add 3 to both sides. Think: what numbers are 7 units from 0? Case 2 –x = 7 Rewrite the case 2 equation by multiplying by –1 to change –1(–x) = –1(7) the minus x to a positive.. x = –7 The solutions are 7 and –7. Holt Algebra 1 2-Ext Solving Absolute-Value Equations Check It Out! Example 1a Continued Solve the equation. Check your answer. |x| 3 = 4 The solutions are 7 and 7. Check |x| 3 = 4 |7| 3 4 | 7| 3 4 7 3 4 73 4 4 Holt Algebra 1 |x| 3 = 4 4 4 4 2-Ext Solving Absolute-Value Equations Check It Out! Example 1b Solve the equation. Check your answer. |x 2| = 8 |x 2| = 8 Case 1 x2= 8 +2 +2 x = 10 Think: what numbers are 8 units from 0? Case 2 Rewrite the equations as two cases. Since 2 is subtracted x 2 = 8 +2 +2 from x add 2 to both sides of each equation. x = 6 The solutions are 10 and 6. Holt Algebra 1 2-Ext Solving Absolute-Value Equations Check It Out! Example 1b Continued Solve the equation. Check your answer. |x 2| = 8 The solutions are 10 and 6. Check |x 2| = 8 |10 2| 8 | 6 + (2)| 8 10 2| 8 6+2 8 8 Holt Algebra 1 |x 2| = 8 8 8 8 2-Ext Solving Absolute-Value Equations 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. Holt Algebra 1 2-Ext Solving Absolute-Value Equations Example 2A: Special Cases of Absolute-Value Equations Solve the equation. Check your answer. 8 = |x + 2| 8 8 = |x + 2| 8 +8 +8 0 = |x +2| Since 8 is subtracted from |x + 2|, add 8 to both sides to undo the subtraction. 0= x+2 2 2 2 = x There is only one case. Since 2 is added to x, subtract 2 from both sides to undo the addition. Holt Algebra 1 2-Ext Solving Absolute-Value Equations Example 2A Continued Solve the equation. Check your answer. 8 = |x +2| 8 Solution is x = 2 Check 8 =|x + 2| 8 8 8 |0| 8 8 08 8 Holt Algebra 1 |2 + 2| 8 8 To check your solution, substitute 2 for x in your original equation. 2-Ext Solving Absolute-Value Equations Example 2B: Special Cases of Absolute-Value Equations Solve the equation. Check your answer. 3 + |x + 4| = 0 3 + |x + 4| = 0 3 3 |x + 4| = 3 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. Holt Algebra 1 2-Ext Solving Absolute-Value Equations Remember! Absolute value must be nonnegative because it represents distance. Holt Algebra 1 2-Ext Solving Absolute-Value Equations Check It Out! Example 2a Solve the equation. Check your answer. 2 |2x 5| = 7 2 |2x 5| = 7 2 2 |2x 5| = 5 |2x 5| = 5 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. Holt Algebra 1 2-Ext Solving Absolute-Value Equations Check It Out! Example 2b Solve the equation. Check your answer. 6 + |x 4| = 6 6 + |x 4| = 6 +6 +6 |x 4| = 0 x4 = 0 + 4 +4 x Holt Algebra 1 =4 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. 2-Ext Solving Absolute-Value Equations Check It Out! Example 2b Continued Solve the equation. Check your answer. 6 + |x 4| = 6 The solution is x = 4. 6 + |x 4| = 6 6 + |4 4| 6 6 +|0| 6 6 + 0 6 6 6 Holt Algebra 1 To check your solution, substitute 4 for x in your original equation.