# How do you solve the system x/3+y/5=2 and x/3 - y/2= -1/3?

Mar 25, 2016

I did this in a hurry so you need to check the calculations!
$y = \frac{40}{21}$

$x = \frac{51}{4}$

#### Explanation:

My first reaction was to consider getting rid of the fractions. That was, until I spotted the coefficients of $x$ are the same in both equation.

Given:

$\frac{x}{3} + \frac{y}{5} = 2$..............................(1)

$\frac{x}{3} - \frac{y}{2} = - \frac{1}{3}$...........................(2)

Subtract equation (2) from (1)

$0 + \frac{y}{5} + \frac{y}{2} = 2 + \frac{1}{3}$

$\frac{2 y + 5 y}{10} = \frac{4}{3}$

$7 y = \frac{40}{3}$

$y = \frac{40}{21}$

Now all you have to do is substitute

Substitute in equation (1) giving

$\frac{x}{3} + \left(\frac{1}{5} \times \frac{40}{21}\right) = 2$.........................(1_a)

$\frac{x}{3} = 2 - \frac{8}{21} = \frac{34}{8}$

$x = \frac{3 \times 34}{8} = 12 \frac{3}{4} = \frac{51}{4}$