解二元一次方程组(x+3)(y+5)=(x+1)(y+8),(2x-3)(5y+7)=2(5x-6)(y+1)
问题描述：
解二元一次方程组(x+3)(y+5)=(x+1)(y+8),(2x-3)(5y+7)=2(5x-6)(y+1)
答
┏xy+3y+5x+15＝xy+x+y+8
┗10xy+14x-15y-21=10xy-12y+10x-12
┏2y+4x+7=0①
┗4x-3y-9=0②
①-②得
y=-16/5
x=-3/20