SOLUTION: I have the equations 3x+ y+2z=6 x+ y+4z=3 2x+3y+2z=2 I'm supposed to solve using elimination. I've tried eliminating every variable but the answer is never right when i check

Question 55978: I have the equations
3x+ y+2z=6
x+ y+4z=3
2x+3y+2z=2
I'm supposed to solve using elimination. I've tried eliminating every variable but the answer is never right when i check it. Please help.
Answer by funmath(2933) (Show Source): You can put this solution on YOUR website!
Solve by elimination:
E1) 3x+ y+2z=6
E2) x+ y+4z=3
E3) 2x+3y+2z=2
Eliminate the y in the E1) and E2), multiply E2) by -1 and add them together:
-1(x+y+4z)=-1(3)
---------------
3x+y+2z=6
-x-y-4z=-3
__________
2x+0y-2z=3
E4) 2x-2z=3
+++++++++++++++++++++++++++++++++++++++++++++++++++++++++++
Eliminate the y in E2) and E3), multiply E2 by -3 and add them together:
-3(x+y+4z)=-3(3)
---------------
-3x-3y-12z=-9
2x+3y+2z=2
____________
-x+0y-10z=-7
E5) -x-10z=-7
++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++
Eliminate the x in E4) and E5), multiply E5) by 2 and add them together:
2(-x-10z)=2(-7)
------------------
2x-2z=3
-2x-20z=-14
_____________
0x-22z=-11
-22z=-11

z=1/2
+++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++
Sustitute z=1/2 into E5) and solve for x:

-x-5=-7
-x-5+5=-7+5
-x=-2
-(-x)=-(-2)
x=2
+++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++
Substitute z=1/2 and x=2 into E1,E2, or E3 and solve for y. I'm going with E2.
(2)+y+4(1/2)=3
2+y+2=3
y+4=3
y+4-4=3-4
y=-1
Therefore the solution is (x,y,z)=(2,-1,1/2)
I'll leave it to you to check. If you can't do that then let me know. I checked it with a calculator.
Happy Calculating!!!
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