Solution: The above system of equations is set to formula ① and formula ② can be changed to: {15[(x-y)^2+2xy]-34xy}(x+3)^2-15=0[15 （x-y）^2-4xy]（x+3）^2-15=0…③② can be changed to: 2x(x-y)+6(x-y)-2=0∴x(x-y)+3(x-y)- 1=0∴(x-y)(x+3

Solution: The above system of equations is set from top to bottom to equation ① and equation ②

Equation ① can be changed to:

{15[(x-y)^2+2xy]-34xy}(x+3)^2-15=0

[15(x-y)^2-4xy](x+3)^2-15=0…③

② formula can be changed to:

2x(x-y)+6(x-y)-2=0

3 ∴x(x-y)+3(x-y)-1=0

∴(x-y)(x+3)=1…④

③-extended: 15(x-y)^2(x+3)^2-4xy(x+3) ^2-15=0…⑤

Substitute ④ into ⑤

15-4xy(x+3)^2-15=0

∴xy(x+3)=0

∴xy(x+3)=0

∴x=0 or y=0 or x=-3

x1=0 , y1=-1/3; x2= (-3+√13)/2 , y2=0; x3= (-3-√13)/2, y3=0; [x4=-3, (x-y)( x+3)=1 does not hold true, ∴Give up]