Solve the system of equations
$$\begin{matrix}
2x^{5}- 2x^{3}y- xy^{2}+ 10x^{3}+ y^{2}- 5y= 0\\
\left ( x+ 1 \right )\sqrt{y+ 5}= y- 3x^{2}+ x- 2
\end{matrix}$$
My try
$$\left ( 1- x \right )y^{2}- \left ( 5+ 2x^{3} \right )y+ 2x^{5}+ 10x^{3}=0\\
\Delta _{y}= \left ( 5+ 2x^{3} \right )^{2}- 4\left ( 1- x \right )\left ( 2x^{5}+ 10x^{3} \right )\geq 0$$
Then I used Wolfram Alpha, the inequality plot is
I can' t continue. I need the help. Thanks!
On
The graphics of the system is
Therefore the solution are
$x=-1, y=6$ and $x=2-\frac{\sqrt{39}}{3}, y=\frac{40}{3}-\frac{4*\sqrt{39}}{3}$
On
COMMENT.-After some handling we have two quadratic equations in $y$: $$y^2-(5-7x^2)y+(9x^4-6x^3+8x^2-14x-1)=0\\(1-x)y^2-(5+2x^3)y+(2x^5+10x^3)=0$$ A way to find out the real roots of this system is a bit of work after plotting the corresponding curves. We have the six following approximate real roots: $$(x,y)=(-0.0686,0),(0.9604,1.5639),(-0.2063,4.148),(0.0592,5.314),(-4.962,-60288),(2.6401,-33.506)$$ The other non-real roots are four of difficult calculation.
Solving your second equation for $y$ we get: $$y_1=3\,{x}^{2}+ \left( x/2+1/2+1/2\,\sqrt {13\,{x}^{2}-2\,x+29} \right) x- x/2+5/2+1/2\,\sqrt {13\,{x}^{2}-2\,x+29} $$ or $$y_2=3\,{x}^{2}+ \left( x/2+1/2-1/2\,\sqrt {13\,{x}^{2}-2\,x+29} \right) x- x/2+5/2-1/2\,\sqrt {13\,{x}^{2}-2\,x+29} $$ plugging this into the first equation we get the following polynomial (after squaring again) $$157\,{x}^{10}-466\,{x}^{9}+281\,{x}^{8}-663\,{x}^{7}+345\,{x}^{6}-405 \,{x}^{5}-242\,{x}^{4}+121\,{x}^{3}-190\,{x}^{2}+x+1 =0$$ which can only be solved by a numerical way.