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二元一次方程.(x+3)(y+5)=(x+1)(8+y) (2x-3)(5y+7)-2(5x-6)(y+1)=0
人气:167 ℃ 时间:2024-09-30 07:06:30
解答

(x+3)(y+5)=(x+1)(8+y) —xy+5x+3y+15=xy+8x+y+8—3x-2y-7=0
(2x-3)(5y+7)-2(5x-6)(y+1)=0 —10xy+14x-15y-21-10xy-10x+12y+12=0—4x-3y-9=0
解得:
x=3
y=1

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