Answers and Explanations 143
Answers
1–100
49.
no real solutions
To solve an absolute value equation of the form a b
= , where b > 0, you must solve the
two equations a = b and a = –b. However, in this case, you have 4
5 18 13
x − + = , which
gives you 4
5
5
x − = − . Because the absolute value of a number can’t be negative, this
equation has no solutions.
50.
x = –3, 9
To solve an absolute value equation of the form a b
= , where b > 0, you must solve the
two equations a = b and a = –b. So for the equation x
x
2
6
27
−
= , you must solve x
2
– 6x
= 27 and x
2
– 6x = –27.
To solve x
2
– 6x = 27, set the equation equal to zero and factor using trial and error:
x
x
x
x
2
6
27 0
9
3 0
−
− =
−
+ =
(
)(
)
Setting each factor equal to zero gives you x – 9 = 0, which has the solution x = 9, and
x + 3 = 0, which has the solution x = –3.
Then set x
2
– 6x = –27 equal to zero, giving you x
2
– 6x + 27 = 0. This equation doesn’t
factor nicely, so use the quadratic equation, with a = 1, b = –6, and c = 27:
x
b
b
ac
a
= − ±
−
=
− − ± −
( ) −
= ± −
2
2
4
2
6
6
4 1 27
2 1
6
72
2
( )
( )( )
( )
The number beneath the radical is negative, so this part of the absolute value has no
real solutions.
Therefore, the only real solutions to x
x
2
6
27
−
=
are x = –3 and x = 9.
51.
x = –3, 2
To solve an absolute value equation of the form a b
= , you must solve the two
equations a = b and a = –b. So for the equation 15
5 35 5
x
x
− =
−
, you have to
solve 15x – 5 = 35 – 5x and 15x – 5 = –(35 – 5x).
For the first equation, you have
15
5 35 5
20
40
2
x
x
x
x
− = −
=
=
And for the second equation, you have
15
5
35 5
15
5
35 5
10
30
3
x
x
x
x
x
x
− = −
−
− = − +
= −
= −
(
)
Therefore, the solutions are x = –3 and x = 2.