HMMT 二月 2009 · 代数 · 第 8 题

英文解析

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 .
   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 )
   It follows that 2
   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