# How do you solve |x - 1| = |2x + 2|?

Sep 21, 2017

See below.

#### Explanation:

$\left\mid x - 1 \right\mid = 2 \left\mid x + 1 \right\mid$ and for $x \ne - 1$

$- 2 = \frac{x - 1}{x + 1} \Leftrightarrow x = - \frac{1}{3}$ and
$2 = \frac{x - 1}{x + 1} \Leftrightarrow x = - 3$

so the solution set is

$x = \left\{- 3 , - \frac{1}{3}\right\}$

Oct 8, 2017

Because this expression involves absolute values there are at least two solutions for the value of $x$.

$x = - 3 \mathmr{and} x = - \frac{1}{3}$

#### Explanation:

How come? We are given: $| x - 1 | = | x + 2 |$

And we know that since there are absolute values involved there will be at least one positive and one negative expression to be calculated. In this case there are two absolute values so let's try all the possibilities.

$| x - 1 | = | 2 x + 2 |$ with both sides positive
$x - 1 = 2 x + 2$
$- 3 = x$

$| x - 1 | = | 2 x + 2 |$ with both sides negative
$- \left(x - 1\right) = - \left(2 x + 2\right)$
$- x + 1 = - 2 x - 2$
$1 = - 3 x$
$- \frac{1}{3} = x$

$| x - 1 | = | 2 x + 2 |$ with LHS positive, RHS negative
$x - 1 = - \left(2 x + 2\right)$
$x - 1 = - 2 x - 2$
$1 = - 3 x$
$- \frac{1}{3} = x$

$| x - 1 | = | 2 x + 2 |$ with RHS positive, LHS negative
$- \left(x - 1\right) = 2 x + 2$
$- x + 1 = 2 x + 2$
$- 1 = 3 x$
$- \frac{1}{3} = x$

To arrive at two answers for the solution of $x$.