SOLUTION: I am having trouble with this problem. SOLVE THIS SYSTEM OF EQUATIONS, WHEN POSSIBLE... x+y+z=4 x-y+2z=8 2x+y-z=3 I used the elimination method and broke this down into

Algebra ->  Coordinate Systems and Linear Equations -> SOLUTION: I am having trouble with this problem. SOLVE THIS SYSTEM OF EQUATIONS, WHEN POSSIBLE... x+y+z=4 x-y+2z=8 2x+y-z=3 I used the elimination method and broke this down into       Log On


   

Question 24526: I am having trouble with this problem.
SOLVE THIS SYSTEM OF EQUATIONS, WHEN POSSIBLE...
x+y+z=4
x-y+2z=8
2x+y-z=3
I used the elimination method and broke this down into two equations with only two variables. I try to eliminate another variable, but cannot do it! HELP PLEASE!!
Answer by Earlsdon(6294) About Me  (Show Source):
You can put this solution on YOUR website!
Solve:
1) x%2By%2Bz+=+4
2) x-y%2B2z+=+8
3) 2x%2By-z+=+3 Add equations 1) and 2) to get:
4) 2x%2B3z+=+12 Add equations 2) and 3) to get:
5) 3x%2Bz+=+11 Multiply equation 5) by 3 to get:
6) 9x%2B3z+=+33 Now subtract equation 4) from equation 6) to get:
7) 7x+=+21 Divide both sides by 7.
x+=+3 Substitute this into equation 5) and solve for z.
3%283%29%2Bz+=+11
9%2Bz+=+11 Subtract 9 from both sides.
z+=+2 Finally, substitute x=3 and z=2 into equation 1) and solve for y.
3%2By%2B2+=+4
y%2B5+=+4 Subtract 5 from both sides.
y+=+-1
Solution:
x = 3
y = -1
z = 2
Check by substituting these values into the three equations, 1), 2), and 3)