# How do you solve \frac{y}{6} + \frac{y}{3} = \frac{5}{8}?

May 17, 2017

See a solution process below:

#### Explanation:

First, multiply the $\frac{y}{3}$ term by the appropriate form of $1$ to create a common denominator with the other $y$ fraction and the add the fractions:

$\frac{y}{6} + \left(\frac{2}{2} \times \frac{y}{3}\right) = \frac{5}{8}$

$\frac{y}{6} + \frac{2 \times y}{2 \times 3} = \frac{5}{8}$

$\frac{y}{6} + \frac{2 y}{6} = \frac{5}{8}$

$\frac{y + 2 y}{6} = \frac{5}{8}$

$\frac{1 y + 2 y}{6} = \frac{5}{8}$

$\frac{\left(1 + 2\right) y}{6} = \frac{5}{8}$

$\frac{3 y}{6} = \frac{5}{8}$

$\frac{3 y}{3 \times 2} = \frac{5}{8}$

$\frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{3}}} y}{\textcolor{red}{\cancel{\textcolor{b l a c k}{3}}} \times 2} = \frac{5}{8}$

$\frac{y}{2} = \frac{5}{8}$

Now, multiply each side of the equation by $\textcolor{red}{2}$ to solve for $y$ while keeping the equation balanced:

$\textcolor{red}{2} \times \frac{y}{2} = \textcolor{red}{2} \times \frac{5}{8}$

$\cancel{\textcolor{red}{2}} \times \frac{y}{\textcolor{red}{\cancel{\textcolor{b l a c k}{2}}}} = \textcolor{red}{2} \times \frac{5}{\left(2 \times 4\right)}$

$y = \cancel{\textcolor{red}{2}} \times \frac{5}{\left(\textcolor{red}{\cancel{\textcolor{b l a c k}{2}}} \times 4\right)}$

$y = \frac{5}{4}$

Or

$y = 1.25$