# How do you solve  abs(2x - 6) = abs(x +7) - 3 ?

Jun 30, 2018

The solutions are $S = \left\{\frac{2}{3} , 10\right\}$

#### Explanation:

The equation is

$| 2 x - 6 | = | x + 7 | - 3$

$\implies$, $| 2 x - 6 | - | x + 7 | + 3 = 0$

The points to be considered are when

$\left\{\begin{matrix}2 x - 6 = 0 \\ x + 7 = 0\end{matrix}\right.$

$\implies$, $\left\{\begin{matrix}x = 3 \\ x = - 7\end{matrix}\right.$

There are $3$ cases to consider

In the interval $\left(- \infty , - 7\right)$

$2 x - 6 - \left(- x - 7\right) + 3 = 0$

$\implies$, $2 x - 6 + x + 7 + 3 = 0$

$\implies$, $3 x + 4 = 0$

$\implies$, $x = - \frac{4}{3}$

This solution is not possible since $x = - \frac{4}{3}$ $\notin$ to $\left(- \infty , - 7\right)$

In the interval $\left(- \infty , - 7\right)$

$- 2 x + 6 - \left(x + 7\right) + 3 = 0$

$\implies$, $- 2 x + 6 - x - 7 + 3 = 0$

$\implies$, $- 3 x + 2 = 0$

$\implies$, $x = \frac{2}{3}$

This solution is possible since $x = \frac{2}{3}$ $\in$ to $\left(- 7 , 3\right)$

In the interval $\left(3 , + \infty\right)$

$2 x - 6 - \left(x + 7\right) + 3 = 0$

$\implies$, $2 x - 6 - x - 7 + 3 = 0$

$\implies$, $x - 10 = 0$

$\implies$, $x = 10$

This solution is possible since $x = 10$ $\in$ to $\left(3 , + \infty\right)$

The solutions are $S = \left\{\frac{2}{3} , 10\right\}$

graph{|2x-6|-|x+7|+3 [-18.02, 18.03, -9.01, 9.01]}