Transcript Solving Absolute-Value Equations

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
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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
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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
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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
73
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
x2= 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
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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
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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
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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
08
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
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Solving Absolute-Value Equations
Remember!
Absolute value must be nonnegative because it
represents distance.
Holt Algebra 1
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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
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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
x4 = 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.