Question

x² + y² + 2xy - 2008x - 2008y - 2009 = 0

Find the number of positive ordered pair integer solutions.

A. 2008
B. 2008
C.2010
D.2007

Show the necessary calculations!

Answer 1

using the concept of number of positive integral solution of combination we can solve easily....

answer is 2008

_________-----________________

hope it will help u

Answer 2

x²+y²+2xy-2008x-2008y-2009=1

(x+y)²-2008(x+y)=2009

taking x+y as common,

(x+y)[(x+y)-2008]=2009×1

Comparing LHS and RHS,

x+y=2009 & x+y-2008=1=>x+y=2009

So,
x+y=2009

No. of positive ordered pair integer solutions,
2009-1 C 2-1
=2008 C 1
=(2008)!/(2008-1)!
=2008!/2007!
=2008×2007!/2007!
=2008

So, answer is 2008.

hope it helps