Solving for x when x's cancel out?

**Lisa87654** · February 11th 2008, 10:58 AM

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?

**angel.white** · February 11th 2008, 11:11 AM

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
$x= \frac 12$

**earboth** · February 11th 2008, 11:11 AM

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.

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

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