# How do you solve abs(8x+8) = abs(9x-4)?

Jul 11, 2015

Split into cases for intervals $\left(- \infty , - 1\right)$, $\left[- 1 , \frac{4}{9}\right)$ and $\left(\frac{4}{9} , \infty\right)$, solving the linear equations that result to find:

$x = - \frac{4}{17}$ or $x = 12$

#### Explanation:

Split into cases:

Case 1: $x \in \left(- \infty , - 1\right)$

$8 x + 8 < 0$, so $\left\mid 8 x + 8 \right\mid = - \left(8 x + 8\right) = - 8 x - 8$
$9 x - 4 < 0$, so $\left\mid 9 x - 4 \right\mid = - \left(9 x - 4\right) = - 9 x + 4$

Equation becomes:

$- 8 x - 8 = - 9 x + 4$

Add $9 x + 8$ to both sides to get:

$x = 12$

This lies outside $\left(- \infty , - 1\right)$ so is not valid for this case.

Case 2: $x \in \left[- 1 , \frac{4}{9}\right)$

$8 x + 8 \ge 0$, so $\left\mid 8 x + 8 \right\mid = 8 x + 8$
$9 x - 4 < 0$, so $\left\mid 9 x - 4 \right\mid = - \left(9 x - 4\right) = - 9 x + 4$

Equation becomes:

$8 x + 8 = - 9 x + 4$

Add $9 x - 8$ to both sides to get:

$17 x = - 4$

Divide both sides by $17$ to get:

$x = - \frac{4}{17}$

This does lie in $\left[- 1 , \frac{4}{9}\right)$ so is a valid solution.

Case 3: $x \in \left[\frac{4}{9} , \infty\right)$

$8 x + 8 \ge 0$, so $\left\mid 8 x + 8 \right\mid = 8 x + 8$
$9 x - 4 \ge 0$, so $\left\mid 9 x - 4 \right\mid = 9 x - 4$

Equation becomes:

$8 x + 8 = 9 x - 4$

Add $- 8 x + 4$ to both sides to get:

$x = 12$

This does lie in $\left[\frac{4}{9} , \infty\right)$ so is a valid solution.

Jul 11, 2015

$\left\mid u \right\mid = \left\mid v \right\mid$ if and only if either $u = v$ or $u = - v$.

#### Explanation:

Two numbers have the same absolute value when the numbers are equal or they are opposites (negatives) of each other.

$\left\mid 8 x + 8 \right\mid = \left\mid 9 x - 4 \right\mid$

if and only if:

$8 x + 8 = 9 x - 4$ $\textcolor{w h i t e}{\text{xx}}$ or $\textcolor{w h i t e}{\text{xx}}$ $8 x + 8 = - \left(9 x - 4\right)$

Solving $8 x + 8 = 9 x - 4$ , we get

$8 + 4 = 9 x - 9 x$ so $12 = x$, that is: $x = 12$

Solving $8 x + 8 = - \left(9 x - 4\right)$. we get

$8 x + 8 = - \left(9 x - 4\right)$, so $8 x + 8 = - 9 x + 4$

and then $8 x + 9 x = 4 - 8$, so $17 x = - 4$, and finally $x = - \frac{4}{17}$

The solutions are: $12$ and $- \frac{4}{17}$