<s>[INST] How to solve $|a+b|+|a-b|=c$? It is intuitive that $a=\pm \frac{c}{2}$, with $-\frac{c}{2}\leq b\leq \frac{c}{2}$ or vice-versa are solutions to the problem. Can I get to these solutions without dividing the expression in all the cases (i.e. $a$ and $b$ being positive or negative and $|a|$ being greater than or less than $|b|$)? Nothing wrong with that, I'm just looking for other way of solving this problem.
I tried using the fact that $|x|=\sqrt{x^2}$ but the expression quickly becomes very complex with the two variables.
 [/INST] There are several standard methods: intervals (cases), squaring, graphical. 
Let's do squaring:
$$(|a+b|+|a-b|)^2=c^2 \iff \\
2a^2+2b^2+2|a^2-b^2|=c^2 \iff \\
(2|a^2-b^2|)^2=(c^2-2(a^2+b^2))^2 \iff \\
4a^2+4b^2-8a^2b^2=c^4-4(a^2+b^2)c^2+4a^2+4b^2+8a^2b^2 \iff \\
c^4-4(a^2+b^2)c^2+16a^2b^2=0 \Rightarrow \\
c^2=2(a^2+b^2)\pm \sqrt{(2(a^2+b^2))^2-16a^2b^2}=2a^2+2b^2\pm \sqrt{4(a^2-b^2)^2} \Rightarrow \\
c^2=2a^2+2b^2\pm 2|a^2-b^2|\stackrel{WLOG: \  |a|\ge |b|}{=} 4a^2;4b^2 \Rightarrow \\
a=\pm \frac c2; |b|\le |a| \Rightarrow -\frac c2\le b\le \frac c2.$$
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