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 $f(t) = t^3 -2yt +4z = 0$ .

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

Page

Toolbox

Search

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

Problem

Solution

See also