# How do you solve 3y=x-30 and 3x+9y=0 using substitution?

Feb 1, 2017

$x = 15$ and $y = - 5$

#### Explanation:

$3 y = x - 30$
$3 x + 9 y = 0$

In the second equation, divide all terms by $3$.

$3 x + 9 y = 0$

$x + 3 y = 0$

In this reduced equation, substitute $3 y$ with $\textcolor{red}{\left(x - 30\right)}$, the value from the first equation.

$x + \textcolor{red}{\left(x - 30\right)} = 0$

Open the brackets and simplify.

$x + \textcolor{red}{x - 30} = 0$

$2 x - 30 = 0$

Add $30$ to each side.

$2 x = 30$

Divide both sides by $2$.

$x = 15$

In the first equation, substitute $x$ with $\textcolor{b l u e}{15}$.

$3 y = x - 30$

$3 y = \textcolor{b l u e}{15} - 30$

$3 y = - 15$

Divide both sides by $3$.

$y = - 5$