2009x+2011y=-2000 (1)
2011x+2009y=2000 (2)
(1)+(2)=4020x+4020y=0,即x+y=0,亦即y=-x带入(1)式,
得,2009x-2011x=-2000,即x=1000,
又x=-y,所以y=-1000.
综上可得:x=1000;y=-1000.
2009x+2011y=-2000 (1)
2011x+2009y=2000 (2)
(1)+(2)=4020x+4020y=0,即x+y=0,亦即y=-x带入(1)式,
得,2009x-2011x=-2000,即x=1000,
又x=-y,所以y=-1000.
综上可得:x=1000;y=-1000.