Solve the system $x^2-3xy+2y^2+x-y=0$, $2x^2-2xy-3y^2-2x+5y=0$ 
Solve the following system of equations:($x,y \in \Bbb R)$
  $$\begin{cases} x^2-3xy+2y^2+x-y=0     \\ 2x^2-2xy-3y^2-2x+5y=0
    \end{cases}$$

It's easily seen that $(0,0)$ and $(1,1)$ are two set of answers,but I can't analyze it to find all answers...
Also,trying to delete $x^2$ gives an equation in terms of $y^2$ whose delta is always positive

$$\left\{\begin{array}{l} x^2-3xy+2y^2+x-y=0     \\ 
2x^2-2xy-3y^2-2x+5y=0
    \end{array}\right.$$
substitute $x=X+1;\;y=Y+1$ in the first equation
$(X+1)^2+-3 (X+1) (Y+1)+2 (Y+1)^2+X+1-Y-1=0$
becomes $X^2 - 3 X Y + 2 Y^2=0$
$X=x-1;\;Y=y-1$
Solve $X=Y;\;X=2Y$ and then $x=y;\;x-1=2y-2\to x=2y-1$
Plug in the second equation and find, for $x=y$
$3 y-3 y^2=0\to y=0;\;y=1$
first two solutions $(0,0);\;(1;\;1)$
for $x=2y-1$
$y^2-5 y+4=0\to y=1;\;y=4$
The other two solutions are $(1,1);\;(7,4)$
