# General Strategy For Solving Absolute Value Equations Involving The Addition Of Multiple Absolute Value Functions

I'm having trouble solving absolute value equations involving multiple absolute value functions added together. For example, take the problem

$$|x+3|-|x+1|+ x+2 =0$$

If all the outputs of the absolute values are non-negative at the same time:

$$x+3-(x+1)+ x+2 =0$$

$$x+3-x-1+ x+2 =0$$

$$2+x+2 =0$$

$$x=-4$$

And yet, when you plug in -4 into $$|x+3|-|x+1|+ x+2 =0$$, it doesn't work.

What confuses me even more is why x=-4 doesn't work when plugged into the original equation, when, if all the outputs of the absolute values are non-negative at the same time, then $$x+3−(x+1)+x+2=|x+3|−|x+1|+x+2$$.

I'm not just looking for a solution to this problem. I want to know what methods I can use to solve absolute value equations. Because the methods I use to solve absolute value problems involving only 1 absolute value function (or absolute value problems involving the multiplication and division of multiple absolute values) don't work here, as seen above.

• $|x+3|-|x+1|$ is non decreasing so $|x+3|-|x+1|+x+2$ is strictly increasing hence $|x+3|-|x+1|+x+2=0$ has only one solution which is $x = -2$ – Cesareo Sep 22 '18 at 13:03
You have to distinguish the cases: $$x\geq -3$$ and $$x\geq -1$$
$$x\geq -3$$ and $$x<-1$$ $$x<-3$$ and $$x<-1$$ the case $$x<-3$$ and $$x>-1$$ doesn't exist. At first we have: $$x\geq -3$$ and $$x\geq -1$$ so we get $$x+3-x-1+x+2=0$$ and we obtain $$x=-4$$ which is impossible, since we he $$x\geq -1$$ and so on.