# How do you solve \frac { y } { 6} - \frac { 2y } { 5} = 1+ \frac { y } { 3}?

Jun 4, 2018

$y = - \frac{30}{17}$

#### Explanation:

First, when you solve an algebraic equation with a fraction in it, you must always find the HCF (highest common factor) between them.

Therefore, in this case, the HCF between $6$ (or also $3 \cdot 2$), $5$, and $3$ is $30$ (in fact $3 \cdot 2 \cdot 5$ since they are all prime numbers).

After that, you divide $30$ by every denominator of the equation
and the number you get you multiply it by every numerator.

So

$\frac{y}{6} \cdot \frac{5}{5} - \frac{2 y}{5} \cdot \frac{6}{6} = 1 \cdot \frac{30}{30} + \frac{y}{3} \cdot \frac{10}{10}$

$\frac{5 y}{5} - \frac{2 y}{5} = \frac{30}{30} + \frac{10 y}{10}$

$\frac{5 y - 12 y}{30} = \frac{30 + 10 y}{30}$

Then you multiply both sides by $30$. In this way, you get rid of $30$, so now you have an equation without fractions.

$5 y - 12 y = 30 + 10 y$

You move all the number with $y$ on one side,

$5 y - 12 y - 10 y = 30$

$- 17 y = 30$
$y = - \frac{30}{17}$