How to solve these equations for x and y.. equations are 

$(x-y)(x+2y)(2x+y) = 20$ 

and

$x^2+xy+y^2 = 7$

i want the METHOD not the solutions 

You are given:$$(x-y)(x+2y)(2x+y)=20\tag{1}$$$$x^2+xy+y^2=7\tag{2}$$Expand the last two terms in (1) to get:$$(x+2y)(2x+y)=2x^2+5xy+2y^2=2(x^2+xy+y^2)+3xy\tag{3}$$Substitute (2) into (3) to get:$$(x+2y)(2x+y)=2(7)+3xy=14+3xy\tag{4}$$Substitute (4) into (1) to get:$$(x-y)(14+3xy)=20\tag{5}$$Then use the hint given by @Leox, i.e. let:$$a=x-y\tag{6}$$$$b=xy\tag{7}$$To get:$$a(14+3b)=20$$$$\therefore 14+3b=\frac{20}{a}\tag{8}$$If it is integer solutions that you are looking for then $a$ must be evenly divisible into $20$ which means $a$ can only be from the set: $1,-1,2,-2,4,-4,5,-5,10,-10,20,-20$
Use each value of $a$ in (8) to see which ones give an integer solution for $b$.
Finally use (6) and (7) to then find $x$ and $y$.
