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Question 144967: solve the following systems of linear equations using the elimination method show your solution as an ordered triple
x-2y+3=-4
3x+y-z=0
2x+3y-5z=1
Answer by seema230480(3)   (Show Source):
You can put this solution on YOUR website! x-2y+3=-4--------------1
3x+y-z=0--------------2
2x+3y-5z=1------------3
From 2, z=3x+y
So 3 becomes, 2x+3y-5(3x+y)=1
Solving the above we get, -13x-2y=1---------4
From 1 we have, x-2y=-7
Solving 1 and 4 we have
-13x-2y=1-----------4
x-2y=-7-------------5
Multiply 13 on both sides of 5, we have now 13x-26y=-91
Subtracting one from the other, we have
13x-26y=-91
-13x-2y=1
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0-28y=-90
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y=90/28=45/14
x=-7+2*45/14
x=(-49+45)/7=-4/7
z=3*(-4/7)+45/14=3/2
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