$\begin{cases} 2x(4x^2-y+2)+4x^2(y-1)=\sqrt{y-2}-2x+y^2-3y+2\\ (\sqrt{y-2}-1)\sqrt{2x+1}=8x^3-13(y-2)+82x-29\end{cases}$

Question

$$\begin{cases} 2x(4x^2-y+2)+4x^2(y-1)=\sqrt{y-2}-2x+y^2-3y+2\\ (\sqrt{y-2}-1)\sqrt{2x+1}=8x^3-13(y-2)+82x-29\end{cases}$$

My attempt:

$2x(4x^2-y+2)+(y-1)(4x^2-y+2)=\sqrt{y-2}-2x$ $2x(4x^2-y+2)+(y-1)(4x^2-y+2)=\dfrac{y-2-4x^2}{\sqrt{y-2}+2x}$

So we have $4x^2-y+2=0$ or $2x+y-1-\dfrac{1}{\sqrt{y-2}+2x}=0$,

and I can continue with $4x^2-y+2=0$ but the $2x+y-1-\dfrac{1}{\sqrt{y-2}+2x}=0$.

Can you help me? Thank you very much!

Answer 1

$2x(4x^2-y+2)+(y-1)(4x^2-y+2)=\dfrac{y-2-4x^2}{\sqrt{y-2}+2x}$

so we have $4x^2-y+2=0$ or $2x+y-1-\dfrac{1}{\sqrt{y-2}+2x}=0$

This is incorrect.

We have $$4x^2-y+2=0\tag1$$ or $$2x+y-1\color{red}{+}\frac{1}{\sqrt{y-2}+2x}=0\tag2$$

I can continue with $4x^2-y+2=0$

In the following, let us see that there are no $(x,y)\in\mathbb R$ such that $(2)$ holds in the system.

First of all, since we have $\sqrt{2x+1}$ and $\sqrt{y-2}$ in the system, we have to have $$2x+1\ge 0\quad\text{and}\quad y-2\ge 0\iff x\ge -\frac 12\quad\text{and}\quad y\ge 2$$

Also, if $x\ge 0$, we see that the LHS of $(2)$ is positive.

So, we have to have $$-\frac 12\le x\lt 0\tag3$$

Now $$(2)\implies (2x+y-1)\sqrt{y-2}=-1-2x(2x+y-1)$$ Since the LHS is non-negative, we have $$-1-2x(2x+y-1)\ge 0\iff y\ge -2x+1-\frac{1}{2x}$$ The RHS is increasing for $(3)$, we get $$y\ge -2\left(-\frac 12\right)+1-\frac{1}{2\left(-\frac 12\right)}=3\tag4$$

Finally, with $(3)(4)$, we see that the LHS of $(2)$ is positive.

Therefore, there are no $(x,y)\in\mathbb R$ such that $(2)$ holds in the system.