How do you solve the system of equations $2 x + 4 y = - 12$ $2x + 4y = - 12$ and $5 x + 2 y = - 6$ $5x + 2y = - 6$ ?

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Answer 1 · 2016-11-03T20:20:27

$(x, y) = (0, - 3)$ $(x,y)=(0,-3)$

$2 x + 4 y = - 12$ $2x+4y=-12$
$5 x + 2 y = - 6$ $5x+2y=-6$

To make the coefficient for both y terms equal, we multiply the second equation by 2.

$5 x + 2 y = - 6$ $5x+2y=-6$ |x2
$10 x + 4 y = - 12$ $10x+4y=-12$

Subtract this equation from the first.

2x+4y=-12

$- (10 x + 4 y = - 12)$ $-(10x+4y=-12)$

After subtracting these equations we get:
$(2 x - 10 x) + (4 y - 4 y) = (- 12 + 12)$ $(2x-10x)+(4y-4y)=(-12+12)$

Simplify:
$- 8 x = 0$ $-8x=0$
$x = 0$ $x=0$

Substitute $x$ $x$ back into one of the equations

$2 (0) + 4 y = - 12$ $2(0)+4y=-12$
$4 y = - 12$ $4y=-12$
$y = - 3$ $y=-3$

How do you solve the system of equations 2x+4y=−122x + 4y = - 12 and 5x+2y=−65x + 2y = - 6?