Solve: $3x(x+y-2)=2y$ with $y(x+y-1)=9x$ This is a question from my Regional Mathematics Olympiad. It is from the topic- polynomials.
 A: Multiply both equations,
$$3x(x+y-2)y(x+y-1) = 18yx$$
Cancelling $x$ and $y$, we obtain a solutions $x=0$ and $y=0$
Next assume $x+y=t$ and solve quadratic.
$$(t-2)(t-1)=6$$
$$(t-4)(t+1)=0$$
$$x+y = 4,x+y+1=0$$
Thus the solutions must be all points on the lines $x+y=4$ , $x+y+1=0$ , $x=0$ and $y=0$
Finally, look at the respective equations.
$$3x(x+y-2)=2y$$
Here if we check with the line $x=0$
$$y=0$$
So, one solution is $(0,0)$.
Note we don't need to check with $y=0$ as its already solved here. Also this is satisfying the other given equation.
Next check the same with the other line obtained...(i.e. solve $x+y = 4,-1$ with the given equations) and take the common points as your answer.
I think you can take it from here..
A: First of all,if any one of $x,y$ is $0$,then the other one should also be $0$.
$(x,y)=(0,0)$ is indeed a solution.
Now,multiply them together to arrive at,
$(x+y-1)^2-(x+y-1)-6=0$ which gives $x+y-1=-2,3$
Then,substitute this back into the equation to get the solutions:
$(1,3)$ and $(\frac{2}{7},\frac{-9}{7})$
