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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
2-4 Agenda Holt Algebra 1 Bellringer – Group Quiz Solving Absolute-Value Equations Book Problems Solving in words activity Worksheet
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
 
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 • • •
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 • • •
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
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
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
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
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
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
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
Remember! Absolute value must be nonnegative because it represents a distance.
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
Word Problems Troy’s car can go 24 miles on one gallon of gasoline. However, his gas mileage can vary from this value by 2 miles per gallon depending on where he drives. Write an absolute-value equation that you can use to find the minimum and maximum gas mileage. │ x – 24 │= 2 Solve the equation to find the minimum and maximum gas mileage. x – 24 = 2  and  x – 24 = -2 + 24  +24   + 24  + 24 x  =  26  and  x = 22
Book problems Solving Equations with Words activity. Update your flipbooks!  Most of you have not done this lately.  You are getting a grade for this.  Worksheet

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Algebra 1. 9.7 Lesson. Absolute Value

  • 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
  • 2. 2-4 Agenda Holt Algebra 1 Bellringer – Group Quiz Solving Absolute-Value Equations Book Problems Solving in words activity Worksheet
  • 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
  • 4.  
  • 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
  • 16. Word Problems Troy’s car can go 24 miles on one gallon of gasoline. However, his gas mileage can vary from this value by 2 miles per gallon depending on where he drives. Write an absolute-value equation that you can use to find the minimum and maximum gas mileage. │ x – 24 │= 2 Solve the equation to find the minimum and maximum gas mileage. x – 24 = 2 and x – 24 = -2 + 24 +24 + 24 + 24 x = 26 and x = 22
  • 17. Book problems Solving Equations with Words activity. Update your flipbooks! Most of you have not done this lately. You are getting a grade for this. Worksheet