Difference between revisions of "2014 UNM-PNM Statewide High School Mathematics Contest II Problems/Problem 6"
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If <math>z=0</math>, then <math>x+y = 0</math> and <math>xy + 2y = 0</math>. We can rearrange and solve: | 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>. | + | <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>. | This results in solutions: <math>(x,y,z) = (0,0,0), (-2,2,0)</math>. | ||
− | If <math>z\neq | + | 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 |