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$$|3-x|+4x=5|2+x|-13$$

One of the solutions is $[3,\infty)$

I'm not familiar with interval solutions for absolute equations.

How to solve for this interval?

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  • $\begingroup$ I don't think that interval can be a solution: you wrote an equation, not an inequality ... $\endgroup$ – DonAntonio Jun 14 '17 at 22:35
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    $\begingroup$ @DonAntonio One simple example is $|x-1| + |x+1| = 2$. All $x\in[-1,1]$ are solutions. $\endgroup$ – peterwhy Jun 14 '17 at 22:41
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    $\begingroup$ @DonAntonio No...every $x$ in that interval is a solution. Indeed, in that range $|3-x|=x-3$ and $|2+x|=2+x$ so the equation reads $x-3+4x=10+5x-13$ which is true. $\endgroup$ – lulu Jun 14 '17 at 22:42
  • $\begingroup$ @peterwhy Excellent point, thanks. $\endgroup$ – DonAntonio Jun 14 '17 at 22:42
  • $\begingroup$ @lulu Excellent comment, thank you. $\endgroup$ – DonAntonio Jun 14 '17 at 22:42
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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: $$ |3-(-\frac{13}{4})| +4(-\frac{13}{4})= \frac{25}{4}-13= -\frac{27}{4}\\ 5|2+(-\frac{13}{4})|-13 = 5(\frac54)-13 = -\frac{27}{4} $$

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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 $x\leq -2$

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  • $\begingroup$ I get it now, every time the solution is an identity, ex. 5=5, it means that all x-es from the defined intervals are the solutions. Thanks a lot! $\endgroup$ – Sal Khanov Jun 14 '17 at 23:00
  • $\begingroup$ @user455156 Exactly, no problem. $\endgroup$ – kingW3 Jun 14 '17 at 23:00
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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):

enter image description here

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.

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