Difference between revisions of "2014 UNM-PNM Statewide High School Mathematics Contest II Problems/Problem 6"
(Created page with "== Problem == How many triples <math>(x, y, z)</math> of rational numbers satisfy the following system of equations? <cmath>\begin{align*} x + y + z &= 0\\ xyz + 4z &= 0\\ xy +...") |
(→Solution) |
||
(3 intermediate revisions by the same user not shown) | |||
Line 9: | Line 9: | ||
== Solution == | == Solution == | ||
+ | If <math>z=0</math>, then <math>x+y = 0</math> and <math>xy + 2y = 0</math>. We can rearrange and solve: | ||
+ | |||
+ | <math>xy + 2y = 2x + 2y \Rightarrow xy = 2x \Rightarrow x=0, y =2</math>. | ||
+ | |||
+ | This results in solutions: <math>(x,y,z) = (0,0,0), (-2,2,0)</math>. | ||
+ | |||
+ | If <math>z\neq 0</math>, <math>x+y+z = 0</math>, <math>xyz +4z = 0 \Rightarrow xy = -4</math> and <math>xy+yz+xz + 2y = 0</math>. We can set up a 3rd degree polynomial <math>f(t)</math> with roots <math>x,y,z</math>, so that <math>f(t) = t^3 -2yt +4z = 0</math>. | ||
== See also == | == See also == |
Latest revision as of 20:57, 25 July 2016
Problem
How many triples of rational numbers satisfy the following system of equations?
Solution
If , then and . We can rearrange and solve:
.
This results in solutions: .
If , , and . We can set up a 3rd degree polynomial with roots , so that .
See also
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 |