SOLUTION: Solving linear systems in three variables by eliminating variables. x+y+z=4 x-y+4z=18 2x+y+z=5

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Question 768644: Solving linear systems in three variables by eliminating variables.
x+y+z=4
x-y+4z=18
2x+y+z=5
Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!

x%2By%2Bz=4........eq.1
x-y%2B4z=18.......eq.2
2x%2By%2Bz=5........eq.3
______________________
x%2By%2Bz=4........eq.1
x-y%2B4z=18.......eq.2
____________________________add
x%2By%2Bz%2Bx-y%2B4z=4%2B18
2x%2B5z=22......solve for x
2x=22-5z
x=11-%285%2F2%29z......eq.1a

2x%2By%2Bz=5........eq.3 plug in x
2%2811-%285%2F2%29z%29%2By%2Bz=5........eq.3
22-5z%2By%2Bz=5.......solve for y
y=5z-z%2B5-22
y=4z-17.............eq.3a
go to eq.1 and substitute x from eq.1a and y from eq.3a

11-%285%2F2%29z%2B4z-17%2Bz=4........eq.1...solve for z

-%285%2F2%29z%2B5z-6=4
-%285%2F2%29z%2B5z=4%2B6...multiply by 2
-5z%2B10z=20
5z=20
highlight%28z=4%29
now find x and y
x=11-%285%2F2%29z......eq.1a
x=11-%285%2F2%29%2A4
x=11-5%2A2
x=11-10
highlight%28x=1%29
y=4z-17.............eq.3a
y=4%2A4-17
y=16-17
highlight%28y=-1%29