已知x2+xy=4,xy+y2=12,求代数式x2-y2与x2+2xy+y2的值各为多少格式正确

x2+xy=4 xy+y2=12
两式相加为 （x2+xy=4）+（xy+y2=12）=x2+2xy+y2=16
两式相减为 （x2+xy=4）-（xy+y2=12）=x2-y2=-8

x^2+xy=4----(1)
xy+y^2=12----(2)
(1)-(2)
(x^2+xy)-(xy+y^2)=4-12
x^2-y^2=-8
(1)+(2)
(x^2+xy)+(xy+y^2)=4+12
x^2+2xy+y^2=16

X2+xy-(xy+y2)=4-12
x2+xy-xy-y2=-8
x2-y2=-8
x2+xy+xy+y2=4+12
x2+2xy+y2=16