Olympiad Maths Prep

Track / Stage 4 / 186 of 340 #446 of 2000

Problem 446

AMC 12 late, AIME early
Algebra Difficulty 4.8 Find the answer

66. (10-11) Solve the system of equations {xy(x+y)=30x3+y3=35.\left\{\begin{array}{l}x y(x+y)=30 \\ x^{3}+y^{3}=35 .\end{array}\right.

This one wants a proof. Work it on paper, read the official solution, then mark yourself honestly — the ladder only means something if the record is true.

Official solution

66. Tripling the members of the first equation and adding to the second, the given system can easily be reduced to two systems:

 a) {x+y=5xy=6 \text { a) }\left\{\begin{array}{l} x+y=5 \\ x y=6 \end{array}\right.

{x+y=5±532xy=121±3 \left\{\begin{array}{l} x+y=\frac{-5 \pm 5 \sqrt{3}}{2} \\ x y=\frac{12}{-1 \pm \sqrt{3}} \end{array}\right.

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.