# How do you solve the following system?: 5x +5y =27, 9x -5y = 0

Nov 12, 2015

By substitution.

#### Explanation:

We have two equations, $5 x + 5 y = 27$ and $9 x - 5 y = 0$. In order to use substitution, we need to solve for one variable. The easiest one to solve for would be $9 x - 5 y = 0$

$9 x - 5 y = 0$
$9 x = 5 y$

Now looking at the equation we need to substitute in for, it would be easier for us to plug something in with a denominator of 5 so they would cancel out and we can do a little less math. So we shall solve for y in this equation.

$y = \frac{9}{5} x$

Now we plug this y into the other equation.

$5 x + 5 \left(\frac{9}{5} x\right) = 27$ and simplifying this would give us
$5 x + 9 x = 27$ now we need to combine like terms
$14 x = 27$ and finally
$x = \frac{27}{14}$

Now we plug this into the other equation.

$9 \left(\frac{27}{14}\right) - 5 y = 0$

$\frac{243}{14} - 5 y = 0$

$\frac{243}{14} = 5 y$

$\frac{243}{70} = y$

My calculator isn't working right now so I can't quite give you decimal values but those fractions should be correct.