# How do you solve abs(3-x)-1=abs(4x+2)?

Jul 17, 2015

$\textcolor{red}{x = 0}$

#### Explanation:

We have two absolute conditions, so we must write four different equations without the absolute value symbols and solve for $x$.

$| 3 - x | - 1 = | 4 x + 2 |$

So these equations would be

(1): $\left(3 - x\right) - 1 = \left(4 x + 2\right)$
(2): $\left(3 - x\right) - 1 = - \left(4 x + 2\right)$
(3): $- \left(3 - x\right) - 1 = \left(4 x + 2\right)$
(4): $- \left(3 - x\right) - 1 = - \left(4 x + 2\right)$

Solve Equation 1:

$\left(3 - x\right) - 1 = \left(4 x + 2\right)$
$3 - x - 1 = 4 x + 2$
$2 - x = 4 x + 2$
$- x = 4 x$
$0 = 5 x$
$x = 0$

This is a possible solution, which we will check later.

Solve Equation 2.

$\left(3 - x\right) - 1 = - \left(4 x + 2\right)$
$3 - x - 1 = - 4 x - 2$
$2 - x = - 4 x - 2$
$- x = - 4 x$
$0 = - 3 x$
$x = 0$

This is a possible solution, which we will check later.

Solve Equation 3.

$- \left(3 - x\right) - 1 = \left(4 x + 2\right)$
$- 3 + x - 1 = 4 x + 2$
$x - 4 = 4 x + 2$
$3 x = - 6$
$x = - 2$

This is a possible solution, which we will check later.

Solve Equation 4.

$- \left(3 - x\right) - 1 = - \left(4 x + 2\right)$
$- 3 + x - 1 = - 4 x - 2$
$x - 4 = - 4 x - 2$
$5 x = 2$
$x = \frac{2}{5}$

This is a possible solution, which we will check later.

Check Solutions 1 and 2

If $x = 0$

$| 3 - x | - 1 = | 4 x + 2 |$
|3-0|-1 = |4×0+2|
$| 3 | - 1 = | 0 + 2 |$
$3 - 1 = | 2 |$
$2 = 2$

Check Solution 3

If $x = - 2$

$| 3 - x | - 1 = | 4 x + 2 |$
$| 3 - \left(- 2\right) | - 1 = | 4 \left(- 2\right) + 2 |$
$| 3 + 2 | - 1 = | - 8 + 2 |$
$| 5 | - 1 = | - 6 |$
$5 - 1 = 6$
$4 = 6$

This is an impossible result. $x = - 2$ is not a solution.

Check Solution 4

If $x = \frac{2}{5}$

$| 3 - x | - 1 = | 4 x + 2 |$
|3-2/5|-1 = |4×2/5+2|
$| \frac{15}{5} - \frac{2}{5} | - 1 = | \frac{8}{5} + \frac{10}{5} |$
$| \frac{13}{5} | - 1 = | \frac{18}{5} |$
13/5 – 1 = 18/5
$\frac{13}{5} - \frac{5}{5} = \frac{18}{5}$
$\frac{8}{5} = \frac{18}{5}$

This is an impossible result. $x = \frac{2}{5}$ is not a solution.

The only solution is $x = 0$.