SOLUTION: I am having some difficulty with an equation. I worked it out, atleast I thought I did, and I got a totaly different answer then the real answer. 4x-3y+7z=38 x+4y+z=16 5y-z

Algebra ->  Rational-functions -> SOLUTION: I am having some difficulty with an equation. I worked it out, atleast I thought I did, and I got a totaly different answer then the real answer. 4x-3y+7z=38 x+4y+z=16 5y-z      Log On


   

Question 44373: I am having some difficulty with an equation. I worked it out, atleast I thought I did, and I got a totaly different answer then the real answer.
4x-3y+7z=38
x+4y+z=16
5y-z=6
my final was x=45, y=-2, z=-21
According to my assignments, it was totaly different.
Can you help me with this equation please.
Answer by Earlsdon(6294) About Me  (Show Source):
You can put this solution on YOUR website!
Solve the system of equations:
1) 4x-3y%2B7z+=+38
2) x%2B4y%2Bz+=+16
3) 5y-z+=+6
Rewrite equation 3) as:
3a) z+=+5y-6 and substitute into equations 1) and 2) to eliminate the variable z.
1a) 4x-3y%2B7%285y-6%29+=+38
2a) x%2B4y%2B%285y-6%29+=+16
Simplify 1a) and 2a)
1b) 4x%2B32y+=+80
2b) x%2B9y+=+22
Multiply equation 2b) by 4 and subtract equation 1b) from it.
2b) 4x%2B36y+=+88
1b) -4x%2B32y+=+80
4) 4y+=+8 Divide bothe sides by 4.
y+=+2 Substitute this into equation 3a) and solve for z.
z+=+5%282%29-6
z+=+10-6
z+=+4
Finally, substitute y=2 and z=4 into equation 2 (or into equation 1) and solve for x.
2) x%2B4%282%29%2B4+=+16
x%2B12+=+16
x+=+4
Check: Substitute x = 4, y = 2, and z = 4 into the three original equations.
1) 4%284%29-3%282%29%2B7%284%29+=+16-6%2B28 = 38 It checks!
2) 4%2B4%282%29%2B4+=+4%2B8%2B4 = 16 It checks!
3) 5%282%29-4+=+10-4 = 6 It checks!