# How do you solve -4(3+x)+5=4(x+3)?

Jul 24, 2017

See a solution process below:

#### Explanation:

First, expand the terms in parenthesis on each side of the equation by multiplying all the terms within the parenthesis by the term outside the parenthesis:

$\textcolor{red}{- 4} \left(3 + x\right) + 5 = \textcolor{b l u e}{4} \left(x + 3\right)$

$\left(\textcolor{red}{- 4} \times 3\right) + \left(\textcolor{red}{- 4} \times x\right) + 5 = \left(\textcolor{b l u e}{4} \times x\right) + \left(\textcolor{b l u e}{4} \times 3\right)$

$- 12 + \left(- 4 x\right) + 5 = 4 x + 12$

$- 12 - 4 x + 5 = 4 x + 12$

$- 12 + 5 - 4 x = 4 x + 12$

$- 7 - 4 x = 4 x + 12$

Next, add $\textcolor{red}{4 x}$ and subtract $\textcolor{b l u e}{12}$ from each side of the equation to isolate the $x$ term while keeping the equation balanced:

$- 7 - \textcolor{b l u e}{12} - 4 x + \textcolor{red}{4 x} = 4 x + \textcolor{red}{4 x} + 12 - \textcolor{b l u e}{12}$

$- 19 - 0 = \left(4 + \textcolor{red}{4}\right) x + 0$

$- 19 = 8 x$

Now, divide each side of the equation by $\textcolor{red}{8}$ to solve for $x$ while keeping the equation balanced:

$- \frac{19}{\textcolor{red}{8}} = \frac{8 x}{\textcolor{red}{8}}$

$- \frac{19}{8} = \frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{8}}} x}{\cancel{\textcolor{red}{8}}}$

$- \frac{19}{8} = x$

$x = - \frac{19}{8}$