Question 186539: Solve the system.
2x+2y+3z=10
3x+y-z=0
x+y+2z=6
x=
y=
z=
Answer by uday1435(57)   (Show Source):
You can put this solution on YOUR website! 2x+2y+3z=10 (1)
3x+y-z=0 (2)
x+y+2z=6 (3)
Multiply eqn (2) by 2 and subtract from eqn(1)
2x+2y+3z=10 (1)
6x+2y - 2z=0 (2) x 2
- 4x + 5z = 10 (4) (1) – (2) x 2
Multiply eqn (3) by 2 and subtract from eqn(1)
2x+2y+3z=10 (1)
2x+2y+4z=12 (5) (3) x 2
- z = - 2
So z = 2
Plugging in the value of z in eqn(4)
- 4x + 5(2) = 10; -4x = 10-10 =0; so x = 0
Now plug-in the values of x (=0) ad z (=2) in eqn (3)
x + y + 2z = 6 (3)
0 + y +2(2) =6
y + 4 =6; so y = 6-4 = 2
So the answers are x =0, y=2, z = 2
If you need further clarification, please email: udayakumar.t.r@gmail.com
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