Question

Solve the following equation.
$x^2{y}^2-2x+y^2=0,2x^2-4x+3+y^3=0$.

Answer 1

The given system of equations is equivalent to $y^2=2x/(1+x^2),2(x-1{)}^2+1+y^3=0$.
Since $2x/(1+x^2)\le 1$ for all real x, it follows from the first of these equations that $y^2\le 1$ or $-1\le y\le 1$.
Again from the second equation of the system, we get
$y^3=-1-2(x-1{)}^2\le -1$
Now since $y\le -1$ and $y\ge -1$, we must have $y=-1$
Then $x=1$. This the solution is $x=1$, $y=-1$.