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solve each system using elimination.
2x+3y=-12
-2x+y=4

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[tex]2x+3y=-12 \\ \underline{-2x+y=4 \ \ \ } \\ 3y+y=-12+4 \\ 4y=-8 \\ y=-2 \\ \\ 2x+3y=-12 \\ 2x+3 \times (-2)=-12 \\ 2x-6=-12 \\ 2x=-12+6 \\ 2x=-6 \\ x=-3 \\ \\ \boxed{(x,y)=(-3,-2)}[/tex]
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2x + 3y = -12
-2x + y = 4

First, you have to pick a variable to cancel out (eliminate)
the 'y' would be the easiest

Multiply the bottom equation by a -3

2x + 3y = -12
6x - 3y = -12

Then, you practically add them together (the y's will cancel out)

8x = -24

Divide both sides by 8

x = -3

Then, plug your x into either equation

-2 (-3) + y = 4

6 + y = 4

Subtract your 6 over to the 4

y = -2

(You can check this buy plugging in both your x and y into both equations)

(x, y)
(-3, -2)
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