1.8
Absolute Value Equations and
Inequalities
Absolute Value Equations
Absolute Value Inequalities
1.8 - 1
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}.
1.8 - 2
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.
1.8 - 3
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.
1.8 - 4
Example 1
SOLVING ABSOLUTE VALUE
EQUATIONS
Solve
a. 5 − 3 x =
12
Solution
5 − 3x =
12
5 − 3x =
12 or
−3 x =
7
7
x= −
3
or
or
5 − 3x =
−12
Property 1
−3 x =
−17
Subtract 5.
17
x=
3
Divide by
–3.
1.8 - 5
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
{
}
1.8 - 6
Example 1
SOLVING ABSOLUTE VALUE
EQUATIONS
Solve
b. 4 x − 3 = x + 6
Solution
4x − 3 = x + 6
4 x − 3 = x + 6 or 4 x − 3 =−( x + 6) Property 2
3x = 9
or 4 x − 3 =− x − 6
or
5 x = −3
3
3
The solutionset is − ,3 . x = −
5
5
x =3
{ }
1.8 - 7
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 < 2 x + 1 < 7
−8 < 2 x < 6
−4 < x < 3
Property 3
Subtract 1 from each
part.
Divide each part by 2.
1.8 - 8
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).
1.8 - 9
Example 2
SOLVING ABSOLUTE VALUE
INEQUALITIES
Solve
b. 2 x + 1 > 7
Solution
2x + 1 > 7
2 x + 1 < −7 or
2 x < − 8 or
x < −4
or
2x + 1 > 7
2x > 6
x >3
Property 4
Subtract 1 from
each side.
Divide each
part by 2.
1.8 - 10
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, ∞ ).
1.8 - 11
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 − 7 x < −5 or
−7 x < −7
x >1
or
or
2 − 7x > 5
Property 4
−7 x > 3
3
x<−
7
Subtract 2.
Divide by –7;
reverse the
direction of each
inequality.
1.8 - 12
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

1.8 - 13
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.
1.8 - 14
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 φ.
1.8 - 15
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 5 x + 15 =
0
which has solution set {–3}. Check by
substituting into the original equation.
1.8 - 16