Question

解方程{ 2011x-2013y=2009,2010x-2012y=2008} {8359x+1641y=28359,1

Answer 1

{ 2011x-2013y=2009,(1)

2010x-2012y=2008 (2)

由(1)-(2)得

x-y=1

x=y+1 (3)

将（3）代入（1）得

2011(y+1)-2013y=2009

2011y+2011-2013y=2009

y=1

x=2

{8359x+1641y=28359, (1)

1641x+8359y=21641 (2) }

由(1)-(2)得

6718x-6718y=6718

x-y=1

x=y+1 (3)

将（3）代入（1）得

8359(y+1)+1641y=28359

8359y+8359+1641y=28359

10000y=20000

y=2

x=3