# How do you solve 3/5(15x+20)+1/6(18x-12)=38?

Jan 6, 2017

See the full process below in the Explanation

#### Explanation:

Step 1) Expand the terms within the parenthesis:

$\textcolor{red}{\frac{3}{5}} \left(15 x + 20\right) + \textcolor{b l u e}{\frac{1}{6}} \left(18 x - 12\right) = 38$

$\left(\textcolor{red}{\frac{3}{5}} \times 15 x\right) + \left(\textcolor{red}{\frac{3}{5}} \times 20\right) + \left(\textcolor{b l u e}{\frac{1}{6}} \times 18 x\right) - \left(\textcolor{b l u e}{\frac{1}{6}} \times 12\right) = 38$

$\left(\textcolor{red}{\frac{3}{\cancel{5}}} \times \textcolor{red}{\cancel{\textcolor{b l a c k}{15}}} 3 x\right) + \left(\textcolor{red}{\frac{3}{\cancel{5}}} \times \textcolor{b l a c k}{\cancel{\textcolor{red}{20}}} 4\right) + \left(\textcolor{b l u e}{\frac{1}{\cancel{6}}} \times \textcolor{b l u e}{\cancel{\textcolor{b l a c k}{18}}} 3 x\right) - \left(\textcolor{b l u e}{\frac{1}{\cancel{6}}} \times \textcolor{b l u e}{\cancel{\textcolor{b l a c k}{12}}} 2\right) = 38$

$\left(\textcolor{red}{3} \times 3 x\right) + \left(\textcolor{red}{3} \times 4\right) + \left(\textcolor{b l u e}{1} \times 3 x\right) - \left(\textcolor{b l u e}{1} \times 2\right) = 38$

$9 x + 12 + 3 x - 2 = 38$

We can now group and consolidate like terms:

$9 x + 3 x + 12 - 2 = 38$

$\left(9 + 3\right) x + 10 = 38$

$12 x + 10 = 38$

We can next subtract $\textcolor{red}{10}$ form each side of the equation to isolate the $x$ term while keeping the equation balanced:

$12 x + 10 - \textcolor{red}{10} = 38 - \textcolor{red}{10}$

$12 x + 0 = 28$

$12 x = 28$

Now we can divide each side of the equation by $\textcolor{red}{12}$ to solve for $x$ and keep the equation balanced:

$\frac{12 x}{\textcolor{red}{12}} = \frac{28}{\textcolor{red}{12}}$

$\frac{\textcolor{b l a c k}{\cancel{\textcolor{red}{12}}} x}{\cancel{\textcolor{red}{12}}} = \frac{4 \times 7}{4 \times 3}$

$x = \frac{\cancel{4} \times 7}{\cancel{4} \times 3}$

$x = \frac{7}{3}$