Download Solve the equation.

2-6
Solving Absolute-Value Equations
Warm Up
Simplify.
1. x – 10 = 4
14
2. s + 5 = –2
–7
3. 32 = –8y
–4
4.
10
5. –14 = x – 5 –9
6. 2t + 5 = 45 20
Holt McDougal Algebra 1
2-6
Solving Absolute-Value Equations
Objectives
3.6 Solve equations in one variable that
contain absolute-value expressions.
Holt McDougal Algebra 1
2-6
Solving Absolute-Value Equations
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 units
6 5 4 3 2 1
Holt McDougal Algebra 1
5 units
0
1
2
3
4
5
6
2-6
Solving Absolute-Value Equations
Holt McDougal Algebra 1
2-6
Solving Absolute-Value Equations
Isolate absolute value first.
Then you must consider two cases…
Holt McDougal Algebra 1
2-6
Solving Absolute-Value Equations
Example 1
Solve the equation.
|x| = 12
|x| = 12
Think: What numbers are 12 units
from 0?
•
12 units
12 10 8 6 4 2
Case 1
x = 12
Case 2
x = –12
•
0
12 units
2
6
•
8 10 12
Rewrite the equation as two
cases.
The solutions are {12, –12}.
Holt McDougal Algebra 1
4
2-6
Solving Absolute-Value Equations
Example 2
Solve the equation.
3|x + 7| = 24
|x + 7| = 8
Since |x + 7| is multiplied by 3,
divide both sides by 3 to undo
the multiplication.
Think: What numbers are 8
units from 0?
Case 2
Case 1
Rewrite the equations as two
x + 7 = 8 x + 7 = –8
cases. Since 7 is added to x
– 7 –7
–7 –7
subtract 7 from both sides
x
=1 x
= –15
of each equation.
The solutions are {1, –15}.
Holt McDougal Algebra 1
2-6
Solving Absolute-Value Equations
Example 4
Solve the equation.
|x| – 3
|x| – 3
+3
|x|
Case 1
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
Rewrite the equation as two
x = –7
cases.
The solutions are {7, –7}.
Holt McDougal Algebra 1
2-6
Solving Absolute-Value Equations
Not all absolute-value equations have two
solutions.
• If the absolute-value expression equals
zero, there is one solution.
• If an equation states that an absolutevalue is negative, there are no
solutions.
Holt McDougal Algebra 1
2-6
Solving Absolute-Value Equations
Example 5
Solve the equation.
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.
The solution is {2}.
Holt McDougal Algebra 1
2-6
Solving Absolute-Value Equations
Example 6
Solve the equation.
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 value cannot be
negative.
This equation has no solution.
Holt McDougal Algebra 1
2-6
Solving Absolute-Value Equations
Remember!
Absolute value must be nonnegative because it
represents a distance.
Holt McDougal Algebra 1
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Solving Absolute-Value Equations
Homework
Page 171
#s 2-20, 72-74
Holt McDougal Algebra 1
2-6
Solving Absolute-Value Equations
Exit Card
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
Holt McDougal Algebra 1
2-6
Solving Absolute-Value Equations
Solve each equation.
1. 15 = |x| –15, 15
Exit Card
2. 2|x – 7| = 14
0, 14
3. |x + 1|– 9 = –9 –1 4. |5 + x| – 3 = –2 –6, –4
5. 7 + |x – 8| = 6 no solution
Holt McDougal Algebra 1