HMMT 二月 2014 · 代数 · 第 1 题

HMMT February 2014 — Algebra — Problem 1

Given that x and y are nonzero real numbers such that x + = 10 and y + = , find all possible y x 12 values of x .

解析

Given that x and y are nonzero real numbers such that x + = 10 and y + = , find all possible y x 12 values of x . 1 1 1 5 1 1 x + z 10 Answer: 4 , 6 OR 6 , 4 Let z = . Then x + z = 10 and + = . Since + = = , we y x z 12 x z xz xz 2 have xz = 24. Thus, x (10 − x ) = 24, so x − 10 x + 24 = ( x − 6)( x − 4) = 0, whence x = 6 or x = 4. 5 Alternate solution: Clearing denominators gives xy + 1 = 10 y and yx + 1 = x , so x = 24 y . Thus we 12 2 x 5 want to find all real (nonzero) x such that + 1 = x (and for any such x , y = x/ 24 will satisfy the 24 12 original system of equations). This factors as ( x − 4)( x − 6) = 0, so precisely x = 4 , 6 work.