How do you solve the following system?: $- 2 x + 12 y = - 20, y = \frac{1}{3} x$ $-2x+12y=-20 , y=1/3x$

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Answer 1 · 2016-08-23T10:43:13

$x = - 10$ $x=-10$

$y = - \frac{10}{3}$ $y=-10/3$

$- 2 x + 12 y = - 20$ $-2x+12y=-20$

or

$2 x - 12 y = 20$ $2x-12y=20$

or

$x - 6 y = 10$ $x-6y=10$

$y = \frac{x}{3}$ $y=x/3$

or

$x = 3 y$ $x=3y$

Plugging the value $x = 3 y$ $x=3y$ in the equation $x - 6 y = 10$ $x-6y=10$

We get

$3 y - 6 y = 10$ $3y-6y=10$

or

$- 3 y = 10$ $-3y=10$

or

$y = - \frac{10}{3}$ $y=-10/3$ --------------------------Ans $1$ $1$

Plugging the value
$y = - \frac{10}{3}$ $y=-10/3$ in the equation $x = 3 y$ $x=3y$ we get

$x = 3 \times (- \frac{10}{3})$ $x=3times(-10/3)$

or

$x = - 10$ $x=-10$ -----------------------------Ans $2$ $2$

How do you solve the following system?: -2x+12y=-20 , y=1/3x −2x+12y=−20,y=13x-2x+12y=-20 , y=1/3x