Transcript Absolute Value Equations and Inequalities

10TH EDITION
COLLEGE ALGEBRA
LIAL
HORNSBY
SCHNEIDER
1.8 - 1
1.8
Absolute Value Equations and
Inequalities
Absolute Value Equations
Absolute Value Inequalities
Special Cases
Absolute Value Models for Distance and
Tolerance
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Distance
is 3.
Distance
is greater
than 3.
Distance
is 3.
Distance
Distance is
is greater
less than 3.
than 3.
Distance is
less than 3.
–3
0
3
By definition, the equation x = 3 can be
solved by finding real numbers at a distance
of three units from 0. Two numbers satisfy
this equation, 3 and – 3.
So the solution set is 3,3.
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Properties of Absolute Value
1. For b  0, a  b if and only if a  b or a  b.
2. a  b if and only if a  b or a  b.
For any positive number b:
3. a  b if and only if  b  a  b.
4. a  b if and only if a  b or a  b.
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Example 1
SOLVING ABSOLUTE VALUE
EQUATIONS
Solve
a. 5  3 x  12
Solution
For the given expression 5 – 3x to have
absolute value 12, it must represent either 12
or –12 . This requires applying Property 1,
with a = 5 – 3x and b = 12.
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Example 1
SOLVING ABSOLUTE VALUE
EQUATIONS
Solve
a. 5  3 x  12
Solution
5  3 x  12
5  3x  12
or
5  3x  12
Property 1
3x  7
or
3x  17
Subtract 5.
7
x
3
or
17
x
3
Divide by
–3.
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Example 1
SOLVING ABSOLUTE VALUE
EQUATIONS
Solve
a. 5  3 x  12
Solution
17
7
or
x
x
Divide by –3.
3
3
Check the solutions by substituting them in
the original absolute value equation. The
solution set is
7 17
 ,
.
3 3


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Example 1
SOLVING ABSOLUTE VALUE
EQUATIONS
Solve
b. 4 x  3  x  6
Solution
4x  3  x  6
4x  3  x  6 or 4 x  3  ( x  6) Property 2
3x  9
or 4x  3   x  6
or
5x  3
3
3
The solutionset is  ,3 . x  
5
5
x 3
 
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Example 2
SOLVING ABSOLUTE VALUE
INEQUALITIES
Solve
a. 2 x  1  7
Solution
Use Property 3, replacing a with 2x + 1 and b
with 7.
2x  1  7
7  2x  1  7
8  2x  6
4  x  3
Property 3
Subtract 1 from each
part.
Divide each part by 2.
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Example 2
SOLVING ABSOLUTE VALUE
INEQUALITIES
Solve
a. 2 x  1  7
Solution
4  x  3
Divide each part by 2.
The final inequality gives the solution set (–4, 3).
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Example 2
SOLVING ABSOLUTE VALUE
INEQUALITIES
Solve
b. 2 x  1  7
Solution
2x  1  7
2x  1  7 or
2x   8 or
x  4
or
2x  1  7
2x  6
x 3
Property 4
Subtract 1 from
each side.
Divide each
part by 2.
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Example 2
SOLVING ABSOLUTE VALUE
INEQUALITIES
Solve
b. 2 x  1  7
Solution
x  4
or
x 3
Divide each
part by 2.
The solution set is ( , 4)  (3, ).
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Example 3
SOLVING AN ABSOLUTE VALUE INEQUALITY
REQUIRING A TRANSFORMATION
Solve 2  7 x  1  4.
Solution
2  7x  1  4
Add 1 to each
side.
2  7x  5
2  7x  5 or
7x  7
x 1
or
or
2  7x  5
7x  3
3
x
7
Property 4
Subtract 2.
Divide by –7;
reverse the
direction of each
inequality.
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Example 3
SOLVING AN ABSOLUTE VALUE INEQUALITY
REQUIRING A TRANSFORMATION
Solve 2  7 x  1  4.
Solution
x 1
or
3
x
7
Divide by –7;
reverse the
direction of each
inequality.
3

The solution set is   ,    1,   .
7

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Example 4
SOLVING SPECIAL CASES OF
ABSOLUTE VALUE EQUATIONS AND
INEQULAITIES
Solve
a. 2  5 x  4
Solution Since the absolute value of a
number is always nonnegative, the inequality
is always true. The solution set includes all
real numbers.
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Example 4
SOLVING SPECIAL CASES OF
ABSOLUTE VALUE EQUATIONS AND
INEQULAITIES
Solve
b. 4 x  7  3
Solution There is no number whose
absolute value is less than –3 (or less than
any negative number). The solution set is .
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Example 4
SOLVING SPECIAL CASES OF
ABSOLUTE VALUE EQUATIONS AND
INEQULAITIES
Solve
c. 5 x  15  0
Solution The absolute value of a number
will be 0 only if that number is 0. Therefore,
5 x  15  0 is equivalent to 5x  15  0
which has solution set {–3}. Check by
substituting into the original equation.
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Example 5
USING ABSOLUTE INEQUALITIES
TO DESCRIBE DISTANCES
Write each statement using an absolute
value inequality.
a. k is no less than 5 units from 8.
Solution
Since the distance from k to 8, written
k – 8 or 8 – k , is no less than 5, the
distance is greater than or equal to 5. This
can be written as
k  8  5, or equivalently 8  k  5.
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Example 5
USING ABSOLUTE INEQUALITIES
TO DESCRIBE DISTANCES
Write each statement using an absolute
value inequality.
b. n is within .001 unit of 6.
Solution
This statement indicates that the distance
between n and 6 is less than .001, written
n  6  .001 or equivalently 6  n  .001.
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Example 6
USING ABSOLUTE VALUE TO
MODEL TOLERANCE
Suppose y = 2x + 1 and we want y to be within
.01 unit of 4. For what values of x will this be
true?
Solution
y  4  .01
2 x  1  4  .01
Write an absolute
value inequality.
Substitute 2x + 1 for y.
2 x  3  .01
.01  2x  3  .01 Property 3
2.99  2x  3.01
Add three to each part.
1.495  x  1.505
Divide each part by 2.
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