SOLUTION: Solve the system: 2x + 4y = -8 3x + 2y = 0

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Question 97760: Solve the system:
2x + 4y = -8
3x + 2y = 0

Found 2 solutions by vincy_fyah, Msgenes:
Answer by vincy_fyah(4) About Me  (Show Source):
You can put this solution on YOUR website!
set x as the subject of equation 2 which implies that x=+-2y%2F3,
so you substitute x in equation 2 2%28-2y%2F3%29%2B4y-8=0 this is factorised to give -4y%2B12y-24=0, then 8y=24 and y=3
Take the value of y and substitute it for the y value in equation 1.
2x%2B4%283%29=-8, 2x=-8-12, 2x=-20, x=-20%2F2, therfore x=-10, so when y=3, x=-10
Hope this helps :-)

Answer by Msgenes(10) About Me  (Show Source):
You can put this solution on YOUR website!
We will use the addition method to get one variable by itselft. We will cancel out y first. In this case we will have to multiply the first equation by 2 and the second equation by -4 because we want y to have the same number.

2x + 4y = -8
3x + 2y = 0
2(2x + 4y = -8)
-4(3x + 2y = 0)
4x + 8y = -16
-12x - 8y = 0
-8x = -16
Now we divide by -8 on both sides to get x by itself.
-8x/-8 = -16/-8
x = 2
Now we have to find y. We will use the equation 4x + 8y = -16
4x + 8y = -16
Now we subtitute 2 for x.
4(2) + 8y = -16
8 + 8y = -16
Now we need to get 8y by itself and we do this by subtracting 8 from both sides.
8 + 8y - 8 = -16 - 8
8y = -24
Now we need y by itself and we do this by dividing 8 on both sides.
8y/8 = -24/8
y = 3
The solution is (2,3)
Hope this helps. Good Luck. :)