# How do you solve 7| x + 3| = 21?

May 2, 2017

See the solution process below:

#### Explanation:

First, divide each side of the equation by $\textcolor{red}{7}$ to isolate the absolute value function while keeping the equation balanced:

$\frac{7 \left\mid x + 3 \right\mid}{\textcolor{red}{7}} = \frac{21}{\textcolor{red}{7}}$

$\frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{7}}} \left\mid x + 3 \right\mid}{\cancel{\textcolor{red}{7}}} = 3$

$\left\mid x + 3 \right\mid = 3$

The absolute value function takes any negative or positive term and transforms it to its positive form. Therefore, we must solve the term within the absolute value function for both its negative and positive equivalent.

Solution 1)

$x + 3 = - 3$

$x + 3 - \textcolor{red}{3} = - 3 - \textcolor{red}{3}$

$x + 0 = - 6$

$x = - 6$

Solution 2)

$x + 3 = 3$

$x + 3 - \textcolor{red}{3} = 3 - \textcolor{red}{3}$

$x + 0 = 0$

$x = 0$

The solution is: $x = - 6$ and $x = 0$

May 2, 2017

$x = 0 \mathmr{and} x = - 6$

#### Explanation:

Begin by dividing by $7$ on both sides:

$\cancel{\frac{7}{7}} | x + 3 | = \frac{21}{7} \to | x + 3 | = 3$

Set up the two stipulations and solve for $x$:

$\textcolor{b l u e}{x + 3 = 3}$

$\textcolor{b l u e}{x + \cancel{3 - 3} = 3 - 3}$

$\textcolor{b l u e}{x = 0}$

----------------------------------------------$\mathmr{and}$----------------------------------------------

$\textcolor{red}{x + 3 = - 3}$

$\textcolor{red}{x + \cancel{3 - 3} = - 3 - 3}$

$\textcolor{red}{x = - 6}$