解:方程组为x²-y²=7,y²+2xy=8;化为x=(8-y²)/(2y),有[(8-y²)/(2y)]²-y²=7,(8-y²)²/(4y²)-y²=7;再设y²=z,方程化为(8-z)²/(4z)-z=7,(8-z)²-4z²=28z,64-16z+z²-4z²-28z=0,64-44z-3z²=0,3z²+44z-64=0,(3z-4)(z+16)=0,得:z=4/3或-16(舍去),有y²=4/3,x²=25/3;当y=2/√3时,x=5/√3,当y=-2/√3时,x=-5/√3
解二元二次方程组
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