# How do you solve 5/8x-1/3=2x-5/6?

Feb 6, 2017

See the entire solution process below:

#### Explanation:

First, multiply each side of the equation by $\textcolor{red}{24}$ to eliminate the fractions. $\textcolor{red}{24}$ is the lowest common denominator of the three fractions and eliminating the fractions will make the problem easier to work:

$\textcolor{red}{24} \left(\frac{5}{8} x - \frac{1}{3}\right) = \textcolor{red}{24} \left(2 x - \frac{5}{6}\right)$

$\left(\textcolor{red}{24} \times \frac{5}{8} x\right) - \left(\textcolor{red}{24} \times \frac{1}{3}\right) = \left(\textcolor{red}{24} \times 2 x\right) - \left(\textcolor{red}{24} \times \frac{5}{6}\right)$

$\left(\cancel{\textcolor{red}{24}} 3 \times \frac{5}{\textcolor{red}{\cancel{\textcolor{b l a c k}{8}}}} x\right) - \left(\cancel{\textcolor{red}{24}} 8 \times \frac{1}{\textcolor{red}{\cancel{\textcolor{b l a c k}{3}}}}\right) = 48 x - \left(\cancel{\textcolor{red}{24}} 4 \times \frac{5}{\textcolor{red}{\cancel{\textcolor{b l a c k}{6}}}}\right)$

$15 x - 8 = 48 x - 20$

Next, subtract $\textcolor{red}{15 x}$ and add $\textcolor{b l u e}{20}$ to each side of the equation to isolate the $x$ term while keeping the equation balanced:

$15 x - 8 - \textcolor{red}{15 x} + \textcolor{b l u e}{20} = 48 x - 20 - \textcolor{red}{15 x} + \textcolor{b l u e}{20}$

$15 x - \textcolor{red}{15 x} - 8 + \textcolor{b l u e}{20} = 48 x - \textcolor{red}{15 x} - 20 + \textcolor{b l u e}{20}$

$0 + 12 = 33 x - 0$

$12 = 33 x$

Now, divide each side of the equation by $\textcolor{red}{33}$ to solve for $x$ while keeping the equation balanced:

$\frac{12}{\textcolor{red}{33}} = \frac{33 x}{\textcolor{red}{33}}$

$\frac{3 \times 4}{\textcolor{red}{3 \times 11}} = \frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{33}}} x}{\cancel{\textcolor{red}{33}}}$

$\frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{3}}} \times 4}{\textcolor{red}{\cancel{3} \times 11}} = x$

$\frac{4}{11} = x$

$x = \frac{4}{11}$