SOLUTION: I am having some trouble with this one. Solve the system of equations. Let z be the parameter. x+2y+3z=11 2x-y+z=2

Algebra ->  Coordinate Systems and Linear Equations  -> Lessons -> SOLUTION: I am having some trouble with this one. Solve the system of equations. Let z be the parameter. x+2y+3z=11 2x-y+z=2      Log On


   

Question 332612: I am having some trouble with this one.
Solve the system of equations. Let z be the parameter.
x+2y+3z=11
2x-y+z=2
Answer by Edwin McCravy(20056) About Me  (Show Source):
You can put this solution on YOUR website!
Solve the system of equations. Let z be the parameter. 

system%28x%2B2y%2B3z=11%2C%0D%0A2x-y%2Bz=2%29

Get the constants and parameter variable terms on the right

system%28x%2B2y=11-3z%2C%0D%0A2x-y=2-z%29

Eliminate y by multiplying the second equation through by 2:

system%28x%2B2y=11-3z%2C%0D%0A4x-2y=4-2z%29


Add corresponding terms:

5x=15-5z

Divide through by 5:

5x%2F5=15%2F5-5z%2F5

x=3-z

Substitute in 

x%2B2y=11-3z


%283-z%29%2B2y=11-3z

3-z%2B2y=11-3z

2y=8-2z

Divide through by 2:

2y%2F2=8%2F2-2z%2F2

y=4-z

So the solution is (x,y) = (3-z,4-z)

Edwin