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

Apr 2, 2015

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

$- 4 < x < \frac{3}{2}$
and
$\frac{3}{2} < x$

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

** If $- 4 < x < \frac{3}{2}$
then only the $\left\mid 2 x - 3 \right\mid$ would contain a negative argument (which is reversed by the absolute value and the equations could be re-written as
$\left(3 - 2 x\right) - \left(x + 4\right) = 8$
$\rightarrow - 3 x - 1 = 8$
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
$\left(2 x - 3\right) - \left(x + 4\right) = 8$
$\rightarrow x - 7 = 8$
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]}