SOLUTION: What are all ordered triples of numbers (x, y, z) which satisfy the system of equations (x + y)(z + 1) = 24, (x + y)(y + 1) = 24, and (y + z)(x + 1) = 24?

Algebra ->  Coordinate Systems and Linear Equations -> SOLUTION: What are all ordered triples of numbers (x, y, z) which satisfy the system of equations (x + y)(z + 1) = 24, (x + y)(y + 1) = 24, and (y + z)(x + 1) = 24?      Log On


   

Question 574555: What are all ordered triples of numbers (x, y, z) which satisfy the system of equations
(x + y)(z + 1) = 24,
(x + y)(y + 1) = 24, and
(y + z)(x + 1) = 24?
Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
%28x+%2B+y%29%28z+%2B+1%29+=+24
%28x+%2B+y%29%28y+%2B+1%29+=+24, and
%28y+%2B+z%29%28x+%2B+1%29+=+24
Combining the first two equations
%28x+%2B+y%29%28z+%2B+1%29+=+%28x+%2B+y%29%28y+%2B+1%29=24
So x%2By cannot be zero and z=y
If z=y, the third equation turns into
%28y+%2B+y%29%28x+%2B+1%29+=+24 ---> 2y%28x%2B1%29=24 --> y%28x%2B1%29=12 --> yx%2By%29=12 which can be substituted into one of the other equations
%28x+%2B+y%29%28y+%2B+1%29+=+24 --> xy%2Bx%2By%5E2%2By=24 --> %28xy%2By%29%2By%5E2%2Bx=24 --> 12%2By%5E2%2Bx=24 --> y%5E2%2Bx=12 --> x=12-y%5E2
Substituting x=12-y%5E2 into y%28x%2B1%29=12 we get
y%2813-y%5E2%29=12 --> 13y-y%5E3=12 --> y%5E3-13y%2B12=0
That can be solve by factoring, starting by factoring out y-1, because it's obvious that y=1 is a solution
y%5E3-13y%2B12=0 --> %28y-1%29%28y%5E2%2By-12%29=0 --> %28y-1%29%28y%2B4%29%28y-3%29=0
SO the solutions are
z=y=1 --> x=12-1%5E2=11
z=y=3 --> x=12-3%5E2=12-9=3
z=y=-4 --> x=12-%28-4%29%5E2=12-16=-4
The solutions would be (11,1,1), (3,3,3) and (-4,-4,-4).