SOLUTION: How do I solve this system using elimination? -2x+2y+3z=0 -2x-y+z=-3 2x+3y+3z=5

Algebra ->  Coordinate Systems and Linear Equations -> SOLUTION: How do I solve this system using elimination? -2x+2y+3z=0 -2x-y+z=-3 2x+3y+3z=5      Log On


   

Question 1059326: How do I solve this system using elimination?
-2x+2y+3z=0
-2x-y+z=-3
2x+3y+3z=5
Answer by josgarithmetic(39618) About Me  (Show Source):
You can put this solution on YOUR website!
More than one way.


R1 plus R3:
-2x%2B2y%2B3z%2B2x%2B3y%2B3z=0%2B5
5y%2B6z=5


R2 plus R3:
-2x-y%2Bz%2B2x%2B3y%2B3z=-3%2B5
2y%2B4z=2
y%2B2z=1


Work with the system:
system%285y%2B6z=5%2Cy%2B2z=1%29
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Multiply the second equation here by 5 and subtract from first equation.
system%285y%2B6z=5%2C5y%2B10z=5%29
-
5y%2B6z-5y-10z=5-5
-4z=0
highlight%28z=0%29
-
I would continue using this system to find y, also using elimination step. This time, multiply second equation by 3...

system%285y%2B6z=5%2C3y%2B6z=3%29

5y%2B6z-3y-6z=5-3
2y=2
highlight%28y=1%29

Choosing to continue in Elimination would be more effort than is justified. Take your now found values for y and z, and take either of the original equations and solve for x.