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由條件(1)(x+1)(y-3)=0

得(x+1)=0 或 (y-3)=0

若(x+1)=0﹐則x=-1

代入(2)

y=5-2x=5+2=7

x=-1,y=7

若(y-3)=0﹐則y=3

代入(2)

x=(1/2)(5-y)=(1/2)(5-3)=1

x=1,y=3

(x+1)(y-3) = 0...(1)
2x+y = 5...(2)

From (1)
x+1 = 0 or y - 3 = 0
x = -1 or y = 3

when x = -1
2x + y = 5...(2)
2(-1) + y = 5
y - 2 = 5
y = 7
therefore, x = -1, y = 7//

when y = 3
2x + y = 5...(2)
2x + 3 = 5
x = 1
therefore, x = 1, y = 3//

from (2)
y = 5 - 2x ---- (3)

sub (3) to (1)
(x+1)[(5-2x)-3]=0
(x+1)(2-2x)=0
2(x+1)(1-x)=0
x+1 = 0 or 1-x = 0
x = -1 or x = 1