How do you solve the system of equations $y = 3 x + 12$ $y= 3x + 12$ and $2 x - 2 y = - 4$ $2x - 2y = - 4$ ?

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Answer 1 · 2018-06-20T15:40:18

Substitution.

$y = 3 x + 12 (1)$ $y=3x+12 " " " "(1)$

$2 x - 2 y = - 4$ $2x-2y =-4$

$- 2 y = - 4 - 2 x$ $-2y=-4-2x$

So

$y = \frac{- 4 - 2 x}{- 2}$ $y=(-4-2x)/-2$

$y = 2 + x (2)$ $y= 2+x" " " "(2)$

Sub $(2)$ $(2)$ into $(1)$ $(1)$

$3 x + 12 = 2 + x$ $3x+ 12= 2 +x$

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

$x = - 5$ $x=-5$

Sub $x = - 5$ $x=-5$ into $(1)$ $(1)$ (I would normally sub it into a given equation as I may have done a mistake in my new equation)

$y = 3 (- 5) + 12$ $y= 3(-5)+12$

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

Therefore,

$y = - 3 and x = - 5$ $y= -3 and x= -5$

How do you solve the system of equations y=3x+12y= 3x + 12 and 2x−2y=−42x - 2y = - 4?