Absolute value equation infinite solutions $$|3-x|+4x=5|2+x|-13$$
One of the solutions is $[3,\infty)$
I'm not familiar with interval solutions for absolute equations. 
How to solve for this interval?
 A: Do case work lucky guess start with $3-x\geq 0$ then $|3-x|=x-3$ and $|2+x|=2+x$ so
$$x-3+4x=10+5x-13\\5x=5x\\0=0$$
Since $0=0$ is always true we have that for each $x\geq 3$ you have a solution.
If for example you've got $1=2$ which is always false you wouldn't have any solutions.
For other solution check the other two cases $-2<x<3$ and $x\leq -2$
A: One way to solve a problem like this is by graphing both sides of the equation.
You can do this in, for example, Desmos here. An image of the linked graph, showing that the two lines coincide for $x \geq 3$ (and showing that they coincidence at one other $x$-value, too):

Another way to solve these sorts of problems is by analyzing different cases: I see now that kingW3 has already provided an answer in this direction, so I'll curtail my response here. 
A: The first thing to do is to notice that if $x$ is large and positive, then
$|3-x|$ is the same as $x-3$, so the equation becomes
$$
(x-3)+4x = 5(x+2) -13 \\
5x-3 = 5x -3
$$ 
and so any time $x$ is that large, the equation is automatically true.  How large is "that large"?
Well, as long as $x\geq 3$ the essential property that $|3-x| = x-3$ holds. So the equation is autmatically satisfied for $x\geq 3$, that is, on the intervale $[3,+\infty)$.
(By the way, the square brace on the left means $x$ is greater than or equal to $3$; the parenthesis on the right means $x < \infty$. 
If $x<3$ then $|3-x| = 3-x$ and if $x \geq -2$  the equation becomes
$$
(3-x)+4x = 5(x+2)-13\\
3x+3 = 5x -3 \\
x=3
$$
and that adds nothing new.  But if $x<-2$ the equation becomes
$$
(3-x)+4x = -5(x+2)-13\\
3x+3 = -5x -23 \\
8x=-26 \\
x= -\frac{13}{4}
$$
and that is an isolated additional solution:
$$
|3-(-\frac{13}{4})| +4(-\frac{13}{4})= \frac{25}{4}-13= -\frac{27}{4}\\
5|2+(-\frac{13}{4})|-13 = 5(\frac54)-13 = -\frac{27}{4}
$$
