Solving a system of equations

Question

Solve the system of equations: $\left\{\begin{array}{l}\sqrt{2y^2-7y+10-x(y+3)}+\sqrt{y+1}=x+1\\\sqrt{y+1}+\dfrac{3}{x+1}=x+2y\end{array}\right.$

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I Wolframalpha.com and get only one solution $(x,y)=(2;0)$. But I can't solve this.

Answer 1

after two hours of calculations i have found an equation for $x$ $$(x-2)(1+8x+x^2)(-1-5x+3x^2-x^3+x^4)=0$$

Answer 2

$u=x+1,v=\sqrt{y+1} \ge0 \implies x=u-1,y=v^2-1,u\ge v \\\left\{\begin{array}{l}2v^4-uv^2-11v^2+2uv-u^2-2u+21=0\\u(u-1)+2u(v^2-1)-uv-3=0\end{array}\right.$

add two equals will have $u=\dfrac{2v^4-11v^2+18}{5-v-v^2}=\dfrac{2(v^2-3)^2+v^2}{5-v-v^2} >0 \implies v^2+v < 5 $

put in 2nd equal, and have some simplify, we have:

$(v-1)(2v^2-2v-7)(v^4-v^3-3v^2+3v+3)=0 \\v-1=0 \implies v=1 \\2v^2-2v-7=0 \implies v=\dfrac{1+\sqrt{15}}{2}>2 \implies u<0 \\v^4-v^3-3v^2+3v+3 >0 $

to prove last one ,we have:

$v^4-v^3-3v^2+3v+3=v^2(v-1)^2+v^3-4v^2+3v+3 \\ v^3-4v^2+3v+3=v(v-3)(v-1)+3=v(v-2)^2+3-v>0$