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 
 A: Method: Express the equations in terms of $xy$, $x-y$ and then change variables.
A: 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 $y$.
A: My suggestion for this problem is to first complete the square in the second equation and then to change variables to get an ellipse with half axes parallel to the coordinate axes. Thus,
$$
(x+\tfrac{1}{2}y)^2+\tfrac{3}{4}y^2=7.
$$
Notice that $x+\tfrac{1}{2}y=\tfrac{1}{2}(2x+y)$ fits well with one factor in the first equation.
Thus, set $u=x+\tfrac{1}{2}y$ and $v=y$. The equations become (after a small simplification)
$$
u^2+\tfrac{3}{4}v^2=7,\quad\text{and}\quad 2u^3-\tfrac{9}{2}uv^2=20.
$$
Here, we could easily solve the first equation for $v^2$ and insert it into the second one. This will give a qubic equation in $u$, that can be solved rather easily. Inserting back into the first one and changing back to $x$ and $y$ will give a total of six pairs of solutions $(x,y)$ to the original equations.
Notice, that this is not a general approach, but a method to solve your system of equations.
A: Equation 2 is the equation of an ellipse. Basically, any equation
\begin{eqnarray}
Ax^2 + Bxy + Cy^2 + Dx + Ey + C = 0
\end{eqnarray}
can be translated, by a change of variable, into:
\begin{eqnarray}
\left(\frac{\tilde{x}}{a}\right)^2 + \left(\frac{\tilde{y}}{b}\right)^2 = 1
\end{eqnarray}
Leading to:
\begin{eqnarray}
\tilde{x} & = & a\cos(\theta) \\
\tilde{y} & = & b\sin(\theta) \\
\tilde{y} & = & b\sqrt{1 - \frac{\tilde{x}^2}{a^2}}
\end{eqnarray}
Now, for the transformation from $(x, y)$ to $(\tilde{x}, \tilde{y})$, we need to find the rotation angle $\Theta$ and translation $(0,0) \mapsto (x_c, y_c)$ so that the ellipse is centered on $0$ and its minor and major axis are aligned with the axis of our base. Back to the generic equation of an ellipse:
\begin{eqnarray}
A & = & a^2 \sin(\Theta)^2 + b^2 \cos(\Theta)^2 \\
B & = & 2(b^2 - a^2)\sin(\Theta)\cos(\Theta) \\
C & = & a^2 \cos(\Theta)^2 + b^2 \sin(\Theta)^2 \\
D & = & -2Ax_c - By_c \\
E & = & -Bx_c - 2 Cy_c \\
F & = & Ax_c^2 + Bx_cy_c + Cy_c^2 - a^2b^2 \\
\end{eqnarray}
With $D$ and $E$, you can find $(x_c, y_c)$. Here, there are no term $D$ or $E$, because your ellipse is already centered on $0$. $A+C = a^2 + b^2$ and $F = a^2b^2$, so you deduce $a$ and $b$. Then $B= (b^2-a^2)sin(2\Theta)$, and you find $\theta$. So 
\begin{eqnarray}
\tilde{x} & = & x\cos(\Theta) - y\sin(\theta) \\
\tilde{y} & = & x\sin(\Theta) + y\cos(\theta)
\end{eqnarray}
A: 1073177
Use Sylvester's dialytic eliminant:
$\begin{vmatrix}
-2&-7x&3x^2&2x^3-20&0\\
0&-2&-7x&3x^2&2x^3-20\\
1&x&x^2-7&0&0\\
0&1&x&x^2-7&0\\
0&0&1&x&x^2-7\\
\end{vmatrix}$
$(=79x^6-1022x^4-80x^3+2891x^2+1120x-972)$
to get rid of the $y$s. Equate it to $0$, solve it for $x$, then substitute each value into one of the original equations (preferably the simpler one), then solve for the corresponding $y$.
A: Do the operations to result to $$2x^3 + 3x^2y- 3xy^2 - 2y^3 - 20 = 0 \iff \\ 2(x^3 - y^3) +3xy(x-y) - 20 = 0 \iff \\ 2(x-y)(x^2 + xy + y^2) + 3xy(x-y) - 20 = 0 \iff \\ 14(x-y)+3xy(x-y) - 20 = 0$$
That simplifies the solution so keep substituting (2) into (1) to simplify the system
A: Drawing the two associated curves, we see that there are $6$ simple solutions (the maximum according to Bezout theorem). Let $s=x+y,p=xy$. If $(x,y)$ is a solution, then $(-y,-x)$ is also a solution. Thus the solutions in $p$ have multiplicity $2$ and $p$ is any root of a polynomial of degree $3$. In the same way, the solutions in $s$ are pairwise opposite.
We obtain $s^2=p+7$ and $(x-y)(2s^2+p)=2$ or $(x-y)(3p+14)=20$ that implies $(s^2-4p)(3p+14)^2=400$ or $(7-3p)(3p+14)^2=400$. The solutions are $p=-6,-3,2$ ; the corresponding values of $s$ are $\pm 1,\pm 2,\pm 3$. The associated solutions in $(x,y)$ are $(-2,3),(-3,2)$ and $(3,-1),(1,-3)$ and $(-1,-2),(2,1)$.
