<s>[INST] Solve the system of equations by variable estimation Solve the system of equations: $\left\{\begin{array}{l}(x-1)\sqrt{x-y^2}=y(x-2y+1)\\y\sqrt{x-1}+3\sqrt{x-y^2}=2x+y-1\end{array}\right.$

I guess there is only one solution $(x;y)=(2;1)$.
This is my try

Condition: $x\ge 1;x\ge y^2$.
We have: $x-2y+1\ge x-2|y|+1\ge x-(y^2+1)+1=x-y^2\ge 0 \Rightarrow y\ge 0$
Applying AM-GM inequality, we have:
$2(2x+y-1)=2y\sqrt{x-1}+6\sqrt{x-y^2}\le(y^2+x-1)+3(1+x-y^2)=4x-2y^2+2$
$\Rightarrow y^2+y-2\le0 \Rightarrow 0\le y\le 1$
Squaring both side of first equation, we get: $(x-2y^2+2y-1)(x^2-x+2y^2-2xy)=0$
If $x-2y^2+2y-1=0$, we get $\dfrac{1}{2}\le x\le 1$. Thus, we get $x=1$ and $y=0$ or $y=1$. Two pairs of $(x,y)$ don't satisfy the system of equations.
If $x^2-x+2y^2-2xy=0$, ...

Who can help me finish it?
 [/INST] since you have
$$x^2-x+2y^2-2xy=0\Longrightarrow x-y^2=x^2-2xy+y^2=(x-y)^2$$
Note $x>y$,so have
$$\sqrt{x-y^2}=x-y$$
take the second equation we have
$$y\sqrt{x-1}+3(x-y)=2x+y-1$$
so we have
$$y=\dfrac{x+1}{4-\sqrt{x-1}}\in[0,1]\Longrightarrow 1\le x\le 2$$
take
$$x^2-x+2y^2-2xy=0\Longrightarrow x^2-x+2\left(\dfrac{x+1}{4-\sqrt{x-1}}\right)^2-2x\dfrac{x+1}{4-\sqrt{x-1}}=0,1\le x\le 2$$
$$\Longrightarrow \dfrac{x^3-6\sqrt{x-1}x^2+8x^2+10\sqrt{x-1}x-19x+2}{(\sqrt{x-1}-4)^2}=0$$
take $\sqrt{x-1}=t$,then we have
$$\Longrightarrow (t^2+1)^3-6t(t^2+1)^2+8(t^2+1)^2+10t(t^2+1)-19(t^2+1)+2=0$$
$$\Longrightarrow (t-1)(t^5-5t^4+6t^3+4t^2+4t+8)=0,0\le t\le 1$$
since
$$f(t)=t^5-5t^4+6t^3+4t^2+4t+8=t^5+6t^3+4t^2+4t+(8-5t^4)>0,0\le t\le 1$$
so $t=1$
then you have $x=2$
</s>