Why arent there 4 cases to |$a+b$|=|$x+y$| With absolute value equations, such as $|x-4| = 3$ , to turn it into a normal equation we take the 2 cases of +- from the absolute value, which I understand.
What I dont however, is why for say $|x-3| = |2x+5|$, we only do $x+3 = \pm(2x+5)$. Shouldn't we have to do $\pm(x+3) = \pm (2x+5)$, as both sides expand out into a $\pm$ sign (because you would have the postive on the left matching to 2 cases on the right, and negative on the left matching to 2 on the right).
When I searched this up, it stated that there were only 2 cases, but that greatly confused me.
 A: $$ 2x+5 =x-3$$ is the same equation as $$ -(2x+5)=-(x-3)$$
and $$(2x+5)=-(x-3) $$ is the same equation as $$-(2x+5)=x-3. $$
In each case, one is obtained from the other by multiplying both sides by $-1.$
A: Think about it this way, how could we have $|m|=|n|$? As you stated, $\pm m=\pm n$. Just take a look at the cases: We could have $m=n$ - that's obvious. Now if we have $m=-n$ then multiplication by $-1$ gives $n=-m$, one of the other cases. So checking one of them gives the other. Same with the final case: $-m=-n$ if and only if $m=n$. I guess this is a matter of how you think about it: there are $4$ cases but $2$ end up being the same. So are there really $4$ or $2$? Maybe one should talk about unique cases but then what does unique mean? Life is probably too short to care about if its $4$ or $2$ so long as we understand that we only need check the case where $m=n$ and $m=-n$.
We could also be more sly with this by asking when is $|m|=|n|$? Well, if this is true then 
$$
\begin{split}
|m|&=|n| \\
|m|^2&=|n|^2 \\
|m|^2-|n|^2&=0 \\
|m^2|-|n^2|&\stackrel{*}{=}0 \\
m^2-n^2&\stackrel{**}{=}0 \\
(m-n)(m+n)&=0
\end{split}
$$
where ($*$) follows from the fact that $|m|^2=|m^2|$ (try this with something like $m=-4$ and $m=5$ to see if you can see why this works for any positive even power) and ($**$) follows from the fact that $u^2 \geq 0$ for any real number $u$. This last equation always gives $m=n$ or $m= -n$, which is of course $m=\pm n$. So we begin with $|m|=|n|$ and end up seeing that it must be $m= \pm n$.
A: I think we have here two cases only:


*

*$a+b=x+y$; 2. $a+b=-x-y$.


If you want to write $-a-b=x+y$ then it's $a+b=-x-y$, which was already.
If you want to write $-a-b=-x-y$ then it's $a+b=x+y$, which again was already.
