Question

\(\left\{{}\begin{matrix}x^2+xy-2y^2+3y-1=\sqrt{y-1}-\sqrt{x}\\3\sqrt{6-y}+3\sqrt{2x+3y-7}=2x+7\end{matrix}\right.\)

Answer 1

\(\Rightarrow \sqrt{y-1}-\sqrt{x}+(y-1)^{2}-x^{2}+y(y-x-1)=0\)

\(\Leftrightarrow (y-x-1)\left ( \underset{>0,x\geq 0 \& 6\geq y\geq 1}{\underbrace{\frac{1}{\sqrt{y-1}+x}+2y+x-1}} \right )=0\Rightarrow y-x-1=0\Leftrightarrow x=y-1\; \;\)\(3\sqrt{6-y}+3\sqrt{5y-9}=2y+5\;\)

\(\Leftrightarrow (8-y)-3\sqrt{6-y}+3(y-1-\sqrt{5y-9})=0\)

\(\Leftrightarrow \frac{y^{2}-7y+10}{(8-y)+3\sqrt{6-y}}+3.\frac{y^{2}-7y+10}{y-1+\sqrt{5y-9}}=0\)

\(\Leftrightarrow (y^{2}-7y+10)(\underset{>0,\forall \frac{9}{5}\leq y\leq 6}{\underbrace{{\frac{1}{(8-y)+3\sqrt{6-y}}+\frac{3}{y-1+\sqrt{5y-9}}}}})=0\)