Can someone see why there is only get one solution when solving following equation in this way:
The equation $|x+1|+|2x-3|=|x-5| $
$$|x+1|+|2x-3|=|x-5| $$ $$\pm (x+1) \pm(2x-3)=\pm(x-5)$$ $$\pm x \pm 1 \pm 2x \mp 3 = \pm x \mp 5$$ $$\pm x \pm 2x \mp x \pm 1 \pm 5 \mp 3=0 $$ $$\pm 2x \pm 6 \mp 3 = 0$$ $$\pm 2x \pm 3=0$$ $$\pm 2x=\mp 3$$ $$x=\frac{\mp 3}{\pm 2} = -\frac{3}{2}$$
There should be another solution as well, $\frac{7}{4}$ by constructing two graphs and finding the intercepts.
You should try to find solutions over the intervals $]-\infty,-1]$, $[-1,3/2]$, $[3/2,5]$ and $[5,\infty[$ instead, because over each of those intervals you know the sign of the absolute value and you will have a linear equation to solve.
For instance, over the interval $[-1,3/2]$, $|x+1| = x+1$, $|2x-3| = -(2x-3)$ and $|x-5| = -(x-5)$. This gives you the equation $$ x+1 -(2x-3) = -(x-5) \quad \Longrightarrow \quad x+1 = 2x-3 - (x-5) = x+2 $$ which has no solution, hence there is no solution over the interval $[-1,3/2]$. Work out all four cases similarly. You need to worry that the solution that you find is actually in the interval you are working on though.
Hope that helps,