HMMT 二月 2009 · 代数 · 第 8 题

HMMT February 2009 — Algebra — Problem 8

[ 7 ] If a, b, x and y are real numbers such that ax + by = 3, ax + by = 7, ax + by = 16, and 4 4 5 5 ax + by = 42, find ax + by . 4 3 2

解析

[ 7 ] If a , b , x , and y are real numbers such that ax + by = 3, ax + by = 7, ax + by = 16, and 4 4 5 5 ax + by = 42, find ax + by . Answer: 20. 3 3 3 3 Solution: We have ax + by = 16, so ( ax + by )( x + y ) = 16( x + y ) and thus 4 4 2 2 ax + by + xy ( ax + by ) = 16( x + y ) 2 It follows that 42 + 7 xy = 16( x + y ) (1) 2 2 2 2 3 3 2 2 From ax + by = 7, we have ( ax + by )( x + y ) = 7( x + y ) so ax + by + xy ( ax + by ) = 7( x + y ). This simplifies to 16 + 3 xy = 7( x + y ) (2) We can now solve for x + y and xy from (1) and (2) to find x + y = − 14 and xy = − 38. Thus we have 4 4 5 5 3 3 ( ax + by )( x + y ) = 42( x + y ), and so ax + by + xy ( ax + by ) = 42( x + y ). Finally, it follows that 5 5 ax + by = 42( x + y ) − 16 xy = 20 as desired. 4 3 2