Question

解几道二元一次方程

{9x-5y=1 {2002x-2003y=2004 {18(x-3)+5(3/2+y)=5
6x-7y=2 2001x-2002y=2003 6(3/2+y)-7(x-3)=6
{3x-2y=11
2x+3y=16

Answer 1

1.
{9x-5y=1
6x-7y=2
{9x-5y=1
9x-(21/2)y=3

9x-(21/2)y-(9x-5y)=3-1
-(11/2)y=2
y=-(4/11)
x=-(1/11)
2.
{3x-2y=11
2x+3y=16
{6x-4y=22
6x+9y=48

(6x+9y)-(6x-4y)=48-22
6x+9y-6x+4y=26
13y=26
y=2
x=5
3.
{2002x-2003y=2004
2001x-2002y=2003
(2002x-2003y)-(2001x-2002y)=2004-2003
2002x-2003y-2001x+2002y=1
x-y=1
2001(x-y)-y=2003
-y=2003-2001
y=-2
x=-1
4.
{18(x-3)+5(3/2+y)=5
6(3/2+y)-7(x-3)=6
{108(x-3)+30(2/3+y)=30
30(3/2+y)-35(x-3)=30
[108(x-3)+30(3/2+y)]-[30(3/2+y)-35(x-3)]=30-30
108(x-3)+30(3/2+y)-30(3/2+y)-35(x-3)=0
143(x-3)=0
143x-429=0
x=3
y=-(1/2)

Answer 2

(1) 9x-5y=1 6x-7y=2
解得x=-1/11 y=-4/11
(2)2002x-2003y=2004 2001x-2002y=2003
解得x=-1 y=-2
(3)18(x-3)+5(3/2+y)=5 6(3/2+y)-7(x-3)=6
解得:x=3 y=-0.5
(4)3x-2y=11 2x+3y=16
解得x=5 y=2