# How do you solve \frac { 6x - 7} { 4} + \frac { 3x - 5} { 7} = \frac { 5x + 7} { 28}?

Jan 27, 2017

See the entire solution process below:$x = \frac{76}{49}$

#### Explanation:

First, multiply both sides for the equation by $\textcolor{red}{28}$ to eliminate the fractions while keeping the equation balanced:

$\textcolor{red}{28} \left(\frac{6 x - 7}{4} + \frac{3 x - 5}{7}\right) = \textcolor{red}{28} \times \frac{5 x + 7}{28}$

$\left(\textcolor{red}{28} \times \frac{6 x - 7}{4}\right) + \left(\textcolor{red}{28} \times \frac{3 x - 5}{7}\right) = \cancel{\textcolor{red}{28}} \times \frac{5 x + 7}{\textcolor{red}{\cancel{\textcolor{b l a c k}{28}}}}$

$\left(\cancel{\textcolor{red}{28}} 7 \times \frac{6 x - 7}{\textcolor{red}{\cancel{\textcolor{b l a c k}{4}}}}\right) + \left(\cancel{\textcolor{red}{28}} 4 \times \frac{3 x - 5}{\textcolor{red}{\cancel{\textcolor{b l a c k}{7}}}}\right) = 5 x + 7$

$7 \left(6 x - 7\right) + 4 \left(3 x - 5\right) = 5 x + 7$

$42 x - 49 + 12 x - 20 = 5 x + 7$

$42 x + 12 x - 49 - 20 = 5 x + 7$

$54 x - 69 = 5 x + 7$

Next, subtract $\textcolor{red}{5 x}$ and add $\textcolor{b l u e}{69}$ to each side of the equation to isolate the $x$ term while keeping the equation balanced:

$54 x - 69 + \textcolor{b l u e}{69} - \textcolor{red}{5 x} = 5 x + 7 + \textcolor{b l u e}{69} - \textcolor{red}{5 x}$

$54 x - \textcolor{red}{5 x} - 69 + \textcolor{b l u e}{69} = 5 x - \textcolor{red}{5 x} + 7 + \textcolor{b l u e}{69}$

$54 x - 5 x - 0 = 0 + 7 + 69$

$49 x = 76$

Now, divide each side of the equation by $\textcolor{red}{49}$ to solve for $x$ while keeping the equation balanced:

$\frac{49 x}{\textcolor{red}{49}} = \frac{76}{\textcolor{red}{49}}$

$\frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{49}}} x}{\cancel{\textcolor{red}{49}}} = \frac{76}{49}$

$x = \frac{76}{49}$