<s>[INST] If $|x + 1| + |x - 3| = 6$, solve for $x$ $|x + 1| + |x - 3| = 6$. Solve for X.
So I know when you have a problem like this: |x| = 6, you solve by doing x=6 and x=-6. That doesn't help us much in the above example. 
You can also solve problems of this fashion by negative the variable portion of the equation. For ex:
$|x+3| = 6. $
You can solve it either by doing:
$x+3 = 6$ or
$x+3 = -6$
$x= 3,-9$
OR
$(x+3) = 6$ so $x=3$
or
$-(x+3) = 6$
$-x -3 =6$
$-x = 9$
$x = -9$
So back to the original problem of $|x + 1| + |x - 3| = 6$
if $X>3$, then $x+1 + x-3 = 6$
$2x=8$
$x=4$
or if $X<3$
$x+1 + x-3 = -6$
$2x=-4$
$x=-2$
Is that correct? Is that all I need to do? I also feel like I got the "critical points" of the equation wrong.
 [/INST] $|x+1| = 6 - |x-3|$
$x + 1 = 6 - |x-3|$ or $x+1 = -6 + |x-3|$
For the first equation: $|x-3| = 5 - x$ so 
$x-3 = 5 -x$ or $ x-3 = x-5$. Only the first part is valid, it gives you the solution $x=4$.
Second equation: $x+7 = |x-3|$. So
$x+7 = x-3$ or $-x - 7 = x-3$. Only the second equation is valid, that gives you $x=-2$. These are the two solutions.
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