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

Apr 26, 2017

See the solution process below:

#### Explanation:

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

$4 - \textcolor{red}{2} \left(2 + 4 x\right) = x - 3$

$4 - \left(\textcolor{red}{2} \cdot 2\right) - \left(\textcolor{red}{2} \cdot 4 x\right) = x - 3$

$4 - 4 - 8 x = x - 3$

$0 - 8 x = x - 3$

$- 8 x = x - 3$

Next, subtract $\textcolor{red}{x}$ from each side of the equation to isolate the $x$ term while keeping the equation balanced:

$- \textcolor{red}{x} - 8 x = - \textcolor{red}{x} + x - 3$

$- 1 \textcolor{red}{x} - 8 x = 0 - 3$

$\left(- 1 - 8\right) x = - 3$

$- 9 x = - 3$

Now, divide each side of the equation by $\textcolor{red}{- 9}$ to solve for $x$ while keeping the equation balanced:

$\frac{- 9 x}{\textcolor{red}{- 9}} = \frac{- 3}{\textcolor{red}{- 9}}$

$\frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{- 9}}} x}{\cancel{\textcolor{red}{- 9}}} = \frac{- 3 \times 1}{\textcolor{red}{- 3 \times 3}}$

$x = \frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{- 3}}} \times 1}{\textcolor{red}{\textcolor{b l a c k}{\cancel{\textcolor{red}{- 3}}} \times 3}}$

$x = \frac{1}{3}$

Apr 26, 2017

$x = \frac{1}{3}$

#### Explanation:

Firstly, distribute the bracket.

$\Rightarrow \cancel{4} \cancel{- 4} - 8 x = x - 3$

$\text{subtract x from both sides}$

$- 8 x - x = \cancel{x} \cancel{- x} - 3$

$\Rightarrow - 9 x = - 3$

$\text{divide both sides by - 9}$

$\frac{\cancel{- 9} x}{\cancel{- 9}} = \frac{- 3}{- 9}$

$\Rightarrow x = \frac{1}{3}$

$\textcolor{b l u e}{\text{As a check}}$

Substitute this value into the equation and if the left side equals the right side then it is the solution.

$4 - 2 \left(2 + \frac{4}{3}\right) = 4 - 2 \left(\frac{10}{3}\right) = 4 - \frac{20}{3} = - \frac{8}{3}$

$\frac{1}{3} - 3 = \frac{1}{3} - \frac{9}{3} = - \frac{8}{3} \leftarrow \textcolor{red}{\text{ right side}}$

$\Rightarrow x = \frac{1}{3} \text{ is the solution}$