I want to prove
$$\left ( x+ 1 \right )\sqrt{y- 5}= y- 3x^{2}+ x- 2\Leftrightarrow y= 6x^{2}$$
Who can help me? If this problem be solved, I can solve the system was posted here:
Solve the system of equations $2x^{5}- 2x^{3}y- xy^{2}+ 10x^{3}+ y^{2}- 5y= 0$ ...
It's not true. We can very easily see this by actually setting $y=6x^2$, since then we get
$$\left ( x+ 1 \right )\sqrt{6x^2+ 5}= 6x^2- 3x^{2}+ x- 2$$
since now on the right we have a polynomial and on the left we don't. We can even prove this further by setting $x=0$, to see that $\sqrt5 = -2$.
Even after the edit, the same argument applies. We can still set $x=2$ to see $3\sqrt{19}=12$.