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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. - PowerPoint PPT Presentation
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Holt McDougal Algebra 1
1-7 Solving Absolute-Value Equations1-7 Solving Absolute-Value Equations
Holt Algebra 1
Lesson QuizLesson Quiz
Lesson PresentationLesson Presentation
Warm UpWarm Up
Holt McDougal Algebra 1
Holt McDougal Algebra 1
1-7 Solving Absolute-Value Equations
Solve equations in one variable that contain absolute-value expressions.
Objectives
Holt McDougal Algebra 1
1-7 Solving Absolute-Value Equations
Holt McDougal Algebra 1
1-7 Solving Absolute-Value Equations
Additional Example 1A: Solving Absolute-Value Equations
Solve the equation.
|x| = 12|x| = 12
Case 1 x = 12
Case 2 x = –12
The solutions are {12, –12}.
Think: What numbers are 12 units from 0?
Rewrite the equation as two cases.
12 units 12 units
10 8 6 4 0 2 4 6 8 1012 2 12•••
Holt McDougal Algebra 1
1-7 Solving Absolute-Value Equations
Solve the equation.
Check It Out! Example 1a
|x| – 3 = 4|x| – 3 = 4
+ 3 +3|x| = 7
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.
Holt McDougal Algebra 1
1-7 Solving Absolute-Value Equations
Solve the equation.
Check It Out! Example 1b
8 =|x 2.5| Think: What numbers are
8 units from 0?
Case 18 = x 2.5
+2.5 +2.5
10.5 = x
+2.5 +2.55.5 = x
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.
Holt McDougal Algebra 1
1-7 Solving Absolute-Value EquationsAdditional Example 1B: Solving Absolute-Value Equations
3|x + 7| = 24
|x + 7| = 8
The solutions are {1, –15}.
Case 1 x + 7 = 8
Case 2 x + 7 = –8
– 7 –7 – 7 – 7x = 1 x = –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.
Holt McDougal Algebra 1
1-7 Solving Absolute-Value EquationsAdditional Example 2A: Special Cases of Absolute-Value
Equations
Solve the equation. 8 = |x + 2| 8
8 = |x + 2| 8+8 + 8
0 = |x + 2|
0 = x + 22 22 = x
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}.
Holt McDougal Algebra 1
1-7 Solving Absolute-Value EquationsAdditional Example 2B: Special Cases of Absolute-Value
Equations
Solve the equation.
3 + |x + 4| = 0
3 + |x + 4| = 03 3
|x + 4| = 3
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.
Holt McDougal Algebra 1
1-7 Solving Absolute-Value Equations
Remember!
Absolute value must be nonnegative because it represents a distance.
Holt McDougal Algebra 1
1-7 Solving Absolute-Value EquationsCheck It Out! Example 2a
Solve the equation.
2 |2x 5| = 7
2 |2x 5| = 7 2 2
|2x 5| = 5
Since 2 is added to –|2x – 5|, subtract 2 from both sides to undo the addition.
Absolute value cannot be negative.
|2x 5| = 5
This equation has no solution.
Since |2x – 5| is multiplied by negative 1, divide both sides by negative 1.
Holt McDougal Algebra 1
1-7 Solving Absolute-Value Equations
Check It Out! Example 2b
Solve the equation.
6 + |x 4| = 6
6 + |x 4| = 6+6 +6
|x 4| = 0
x 4 = 0+ 4 +4
x = 4
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.
Holt McDougal Algebra 1
1-7 Solving Absolute-Value EquationsLesson 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