SOLUTION: solve each system by the elimination method. Check each solution
3x-y=-12
x+y=4
2x+y=-15
-x-y=10
5x-y=5
-5x+2y=0
x+y=3
-3x+2y=-19
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Question 583176: solve each system by the elimination method. Check each solution
3x-y=-12
x+y=4
2x+y=-15
-x-y=10
5x-y=5
-5x+2y=0
x+y=3
-3x+2y=-19
Found 2 solutions by spongetatle, solver91311:
Answer by spongetatle(6) (Show Source): You can put this solution on YOUR website!
Check the x and y values in each equation and get one of them to cancel out by multiplying one whole equation equation to make the x or y values cancel out. So pretty much you are doing this with one of the values x + -x = 0. No matter what x equals in that formula they would equal 0. However all of your problems cancel out except the last one. I'll go over the first one, the last one is the trickier one, but if you follow my directions you can do it.
3x-y=-12
x+y=4 The y's already cancel out.
3x=-12
x=4 Add both of these up.
4x=-8 Divide by 4.
x = -2 Plug newly found variable into one of the ORIGINAL equations given.
3(-2)-y=-12 Solve.
-6-y=-12 Add 6
-y=-6 Multiply by -1(or just do this in your head.)
y=6 and x=4 Plug them into the equations and check if they work.
Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!
Since you already have one variable where the coefficients in the two equations are additive inverses, i.e. -1 and 1 on the
variable, simply add the equations:
You should be able to handle it from here. Once you have a value for
, substitute it into either of the given equations and solve for
. Remember to state your answer as an ordered pair
John

My calculator said it, I believe it, that settles it
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