How do you solve abs(2x-3) - abs(x+4) = 8?

1 Answer
Alan P.
Apr 2, 2015

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

-4 < x < 3/2
and
3/2 < 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
rarr x+7 = 8
which would imply x=1 but** x<-4 so this solution can be ignored as being extraneous.

** If -4 < x <3/2
then only the abs(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
rarr -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
rarr 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]}

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