Modulus Problem: $|x+1| - |x| + 3|x-1| -2|x-2| = x+2$

Question

I do not understand how to solve such a question:

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

How would you go about all the possibilities with which sign the modulus could take? Appreciate any help!

Answer 1

Addendum: The first case.

For $x\le 1$ as mentioned ALL of the expressions in the $|\cdot|$ are negative, so the expression takes the form:

$$-(x+1)+x-3(x-1)+2(x-2)$$ $=-x-2$ after tidying up. This is valid as long as $-\infty<x\le -1$ - which the diagram confirms.

It is easy to see that $x+2$ and $-x-2=-(x+2)$ intersect at $x=-2$ - the first one is found for you

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A picture is worth 1000 words:

All you have to do is check at the cases where these functions switch. So for example with $x\le -1$ ALL of the modulus parts are the negative of their internal values.

For $x\in[0,1]$ the first one is just $x+1$ and the other 3 are all negatives of their internal values,

carry forward with this process.

You are in effect finding the line segments in the diagram below.