How would you solve
$\displaystyle |x+2|+|x-3|=7$
recall that for real numbers, $\displaystyle |x|^2 = x^2$
thus, by squaring, we can get rid of the absolute values.
so square both sides, we get:
$\displaystyle (x + 2)^2 + 2|x + 2||x - 3| + (x - 3)^2 = 49$
$\displaystyle \Rightarrow 2|x + 2||x - 3| = 49 - (x + 2)^2 - (x - 3)^2$
now, you can expand and simplify the right hand side, and then square both sides again. you will get a fourth degree polynomial as a result, which might not be too hard to solve. be sure to check for erroneous solutions.
alternatively, take note of what skeeter and i did in your other thread. that is, note the sign changes that the absolute values give rise to, and do all the cases separately.
that is,
case 1: both x + 2 and x - 3 are positive
case 2: both x + 2 and x - 3 are negative
case 3: x + 2 is positive but x - 3 is negative
case 4: x + 2 is negative but x - 3 is positive
choose whichever method you prefer. either will be a pain, but neither should be too difficult
another alternative, which is similar to the first. is to bring one of the absolute value expressions over the equal sign and then square. the algebra will be easier