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

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Answer 1 · 2017-02-01T09:02:15

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

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

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

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

$x + 3 y = 0$ $x+3y=0$

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

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

Open the brackets and simplify.

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

$2 x - 30 = 0$ $2x-30=0$

Add $30$ $30$ to each side.

$2 x = 30$ $2x=30$

Divide both sides by $2$ $2$ .

$x = 15$ $x=15$

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

$3 y = x - 30$ $3y=x-30$

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

$3 y = - 15$ $3y=-15$

Divide both sides by $3$ $3$ .

$y = - 5$ $y=-5$

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