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.
 A: First let us do it without algebra. Draw a number line. Then $|x+1|+|x-3|$ is the sum of the distances of $x$ from $-1$ and from $3$. It is clear that if $x$ is between $-1$ and $3$, the sum of the distances is $4$. The sum of the distances will be $6$ if we are $1$ unit to the right of $3$, or $1$ unit to the left of $-1$.
Now with algebra, essentially like you approached it.
If $x\ge 3$, then $|x-3|=x-3$ and $|x+1|=x+1$. So we want $(x-3)+(x+1)=6$. giving $x=4$.
If $x\gt -1$, and $x\lt 3$, then $|x+1|=x+1$ and $|x-3|=-x+3$, so we want $(x+1)+(-x+3)=6$, impossible.
If $x\lt -1$, then $|x+1|=-x-1$ and $|x-3|=-x+3$, so we want $-2x+2=6$, giving $x=-2$. 
A: Divide the domain to
(1) $x \le -1$
(2) $-1 \le x \le 3$
(3) $x \ge 3$
solve the equation in each domain
A: $|x+1| = 6 - |x-3|$
$x + 1 = 6 - |x-3|$ 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.
