SOLUTION: 14+2z=2(x-3y) 4(x-3y-z)=28 3(2x+y)+4z=-22 The answer supposedly is -3,-4,2 but I can't get it to go.??? Please help by showing steps as there are several questions like this i

Algebra ->  Systems-of-equations -> SOLUTION: 14+2z=2(x-3y) 4(x-3y-z)=28 3(2x+y)+4z=-22 The answer supposedly is -3,-4,2 but I can't get it to go.??? Please help by showing steps as there are several questions like this i      Log On


   



Question 431962: 14+2z=2(x-3y)
4(x-3y-z)=28
3(2x+y)+4z=-22
The answer supposedly is -3,-4,2 but I can't get it to go.??? Please help by showing steps as there are several questions like this in my assignment with no examples in book or on-line homework. When I use the rref f(x) my top row comes out 1, 0, .428......,second row 0,1,.4761.....,third row 0,0,0. Scratches head. Argggh!

Answer by rwm(914) About Me  (Show Source):
You can put this solution on YOUR website!
If x equals -3 then y does equal -4 and z equals 2
part of your problem might be that equation one and equation two are the same equations
divide the first by 2
7+z=x-3y
divide the second by 4
x-3y-z=7