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:

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

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

$2+x+2 =0$


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.

Thank you in advance!

  • $\begingroup$ $|x+3|-|x+1|$ 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 $x = -2$ $\endgroup$ – 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.

  • $\begingroup$ Can you elaborate on what that means? It would really help me if you so me doing that in the process of solving the problem |x+3|−|x+1|+x+2=0, so I can see how to work through with this. Thanks. $\endgroup$ – Ethan Chan Sep 22 '18 at 12:56
  • $\begingroup$ And do you also need to do this for multiplying and dividing multiple absolute value functions? $\endgroup$ – Ethan Chan Sep 22 '18 at 12:56

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