HMMT 二月 2014 · 代数 · 第 1 题

英文解析

1. Given that x and y are nonzero real numbers such that x + = 10 and y + = , find all possibley 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 + = = , wey x z 12 x z xz xzhave xz = 24. Thus, x (10 − x ) = 24, so x − 10 x + 24 = ( x − 6)( x − 4) = 0, whence x = 6 or x = 4.2
   Alternate solution: Clearing denominators gives xy + 1 = 10 y and yx + 1 = x , so x = 24 y . Thus we 5
   x 5212
   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.