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

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Answer 1 · 2017-02-01T09:02:15

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

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

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

$3x+9y=0$

$x+3y=0$

In this reduced equation, substitute $3y$ with $color(red)((x-30))$ , the value from the first equation.

$x+color(red)((x-30))=0$

Open the brackets and simplify.

$x+color(red)(x-30)=0$

$2x-30=0$

Add $30$ to each side.

$2x=30$

Divide both sides by $2$ .

$x=15$

In the first equation, substitute $x$ with $color(blue)(15)$ .

$3y=x-30$

$3y=color(blue)(15)-30$

$3y=-15$

Divide both sides by $3$ .

$y=-5$

How do you solve 3y=x-303y=x-30 and 3x+9y=03x+9y=0 using substitution?