2014 UNM-PNM Statewide High School Mathematics Contest II Problems/Problem 6

Problem

How many triples $(x, y, z)$ of rational numbers satisfy the following system of equations? \begin{align*}  x + y + z &= 0\\ xyz + 4z &= 0\\ xy + xz + yz + 2y &= 0 \\ \end{align*}

Solution

If $z=0$, then $x+y = 0$ and $xy + 2y = 0$. We can rearrange and solve:

$xy + 2y = 2x + 2y \Rightarrow xy = 2x \Rightarrow x=0, y =2$.

This results in solutions: $(x,y,z) = (0,0,0), (-2,2,0)$.

If $z\neq 0$, $x+y+z = 0$, $xyz +4z = 0 \Rightarrow xy = -4$ and $xy+yz+xz + 2y = 0$. We can set up a 3rd degree polynomial $f(t)$ with roots $x,y,z$, so that $f(t) = t^3 -2yt +4z = 0$.

See also

2014 UNM-PNM Contest II (ProblemsAnswer KeyResources)
Preceded by
Problem 5
Followed by
Problem 7
1 2 3 4 5 6 7 8 9 10
All UNM-PNM Problems and Solutions
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