Problem: How many distinct ordered triples (x,y,z) satisfy the equations
x+2y+4zxy+4yz+2xzxyz​=12=22=6?​
Answer Choices:
A. None
B. 1
C. 2
D. 4
E. 6
Solution:
Let u=x/2,v=y and w=2z. Then
u+v+wuv+νw+uwuvw​=6=11=6.​
Consider the polynomial p(t)=(t−u)(t−v)(t−w), where (u,v,w) is a solution to the simultaneous equations above. Then p(t)=t3−6t2+11t -6 and (u,v,w) are the solutions of p(t)=0. Conversely if the solutions of p(t)=0 are listed as a triple in any order, this triple is a solution to the simultaneous equations above. Since p(t)=(t−1)(t−2)(t−3), the six permutations of (1,2,3) are the solutions to the simultaneous equations in u,v and w. Therefore, the given equations in x,y and z have six solutions.