How do you solve |2x−3|−|x+4|=8?

1 Answer
Alan P.
Apr 2, 2015

There are 2 values of x for which the absolute values become significant
x=32
and x=−4
So we need to consider 3 ranges:
x<−4

−4<x<32
and
32<x

If x<−4 **
the arguments of both absolute values are negative, so without absolute values the equations could be re-written as
(3−2x)−(−(x+4))=8
→x+7=8
which would imply x=1 but** x<−4 so this solution can be ignored as being extraneous.

** If −4<x<32
then only the |2x−3| would contain a negative argument (which is reversed by the absolute value and the equations could be re-written as
(3−2x)−(x+4)=8
→−3x−1=8
which implies x=−3 (which is acceptably within our range.

If x>4
then neither absolute value function has any effect and the equation could equivalently be written as
(2x−3)−(x+4)=8
→x−7=8
which implies x=15

The only two valid solutions are
x=−3
and
x=15
graph{abs(2x-3)-abs(x+4)-8 [-28.86, 28.85, -14.43, 14.43]}

Related questions
See all questions in Absolute Value Equations
Impact of this question
2278 views around the world
You can reuse this answer
Creative Commons License