# Thread: Solving for x when x's cancel out?

1. ## Solving for x when x's cancel out?

(I also posted this is the Algebra forum, but got a lot of views, but no replies, so I'm trying here instead...)

I'm sure the answer to this is incredibly obvious, but something is slipping my mind:

│x-3│= x + 2

if you move the x out of the absolute value braket, the x's cancel... so how do you solve it?

2. $|x-3|=x+2$

There are 3 possible cases.

Case 1:
$x-3 > 0$
$x > 3$
If it's positive, we can say that $x-3 = x+2$ which doesn't have a root.

Case 2:
$x-3<0$
$x<3$
It's negative, we can rewrite it as $x-3 = -(x+2)$
$2x = 1$
$\boxed{x = \frac{1}{2}}$ --! We found a root

Case 3:
$x-3 = 0$
$x=3$
For $x=3$, $|x-3|\neq x+2$ so we don't have roots in this region.

Conclusion: The equation has only the root $x=\frac{1}{2}$.

Graph:
We can rewrite the equation as $|x-3|-x-2=0$ and plot this function.

$y=|x-3|-x-2$
See where this graph crosses the x axis.