Question

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

Answer 1

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]}