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

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

Hi there, 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. Originally Posted by Lisa87654
Hi there, 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?
$\mid x-3 \mid = x + 2$

You can see that the two sides are going to have alternative signs, meaning either (x-3) will be negative or (x+2) will be negative so cancel the absolute value, and make either side negative:
$-(x-3)= x + 2$

Distribute the negative sign
$-x+3= x + 2$

Subtract (x+3) from both sides
$-x+3 -(x+3)= x + 2 -(x+3)$

Distribute the negatives
$-x+3 -x -3= x + 2 -x-3$

Simplify
$-2x= -1$

Divide both sides by -2
$x= \frac {-1}{-2}$

Simplify
$x= \frac 12$

3. Originally Posted by Lisa87654
Hi there, 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?
Use the definition of the absolute value:

$|x-3| = \left\{\begin{array}{l}x-3 ~,~ x-3 \geq 0~\implies~x \geq 3\\ -(x-3) ~,~ x-3 < 0~\implies~x < 3 \end{array} \right.$

That means you get 2 different equations:

$x-3 = x+2~,~x \geq 3$ This equation has no real roots.

$-(x-3) = x+2~,~x < 3~\implies~ x = \frac12$