初一数学题,要详细的过程,求代数式的值一直x^2+xy=3 xy+y^2=2,求下列代数式的值,要过程.①x^2-y^2②x^2+4xy+3y^2③2x^2-xy-3y^2

(x^2+xy)-(xy+y^2)=x^2-y^2=1
x^2+xy+3(xy+y^2)=x^2+4xy+3y^2=3+6=9
2(x^2+xy)-3(xy+y^2)=2x^2-xy-3y^2=0

x^2-y^2=(3-xy)-(2-xy)=1
x^2+4xy+3y^2==(3-xy)+4xy+3(2-xy)=9
2x^2-xy-3y^2=2(3-xy)-xy-3(2-xy)=0

x^2+xy=3
xy+y^2=2，
x^2-y^2=1
①x^2-y^2=1
②x^2+4xy+3y^2=x^2+xy+3xy+3y^2=3+2*3=9
③2x^2-xy-3y^2=2x^2+2xy-3xy-3y^2=6-6=0

x^2+xy=3-----------A
xy+y^2=2-----------B
①A-B得：(x^2+xy)-(xy+y^2)=x^2-y^2=3-2=1
②A+3*B得：(x^2+xy)+3*(xy+y^2)=x^2+4xy+3y^2=3+3*2=9
③2*A-3*B得：2*(x^2+xy)-3*(xy+y^2)=2x^2-xy-3y^2=2*3-3*2=0

(1)已知x^2+xy=3 xy+y^2=2,两式相减得x^2-y^2=1
（2）x^2+4xy+3y^2=x^2+xy+3（xy+y^2）=3+3*2=9
（3）2x^2-xy-3y^2=2（x^2+xy）-3（xy+y^2）=2*3-3*2=0