Solve $x^2+y^2+\frac {2xy}{x+y}=1$, $\sqrt {x+y}=x^2-y$

Question

Solve the following system :-

$$x^2+y^2+\frac {2xy}{x+y}=1\tag 1$$

$$\sqrt {x+y}=x^2-y \tag 2$$

My try follows:-

I have tried many idea's including squaring the second equation , but i have got stuck . Is there any reference to be more familiar with these idea's.

Any help is appreciated thank you

Answer 1

We have \begin{eqnarray*} x^2+y^2+\frac{2xy}{x+y}=1 \\ \sqrt{x+y}=x^2-y \end{eqnarray*} Now rearrange $(2)$ \begin{eqnarray*} x+y=x^4-2x^2y+y^2 \\ y^2-(2x^2+1)y+x^4-x=0 \\ y= \frac{2x^2+1 \pm (2x+1)}{2} \end{eqnarray*} Now taking the positive root $y=x^2+x+1$ and subtituting in $(1)$ ... \begin{eqnarray*} x(x+2)((x+x)^2+(2x+1)^2+1)=0 \end{eqnarray*} For the negative root $y=x^2-x$ ... \begin{eqnarray*} x^2(x+1)(x-1)((x-1)^2+2)=0 \end{eqnarray*} We shall reject the solution $(0,0)$ as it causes division by zero in equation $(1)$. So to summarise $\color{red}{(-2,3),(-1,2),(0,1),(1,-1)}$.

Answer 2

Let $\sqrt{x+y}=t$.

Hence, $y=t^2-x$ and $t=x^2-y=x^2-t^2+x$, which says $$t^2+t=x^2+x$$ or $$(t-x)(t+x+1)=0.$$ 1. $t=x$.

Thus, $\sqrt{x+y}=x$, $y=x^2-x$, where $x\geq0$ and we can substitute it in the first equation.

Since $x+y\neq0$, we obtain $x>0$ and we get $$x^2+(x^2-x)^2+\frac{2x(x^2-x)}{x^2}=1$$ or $$x^4-2x^3+2x^2+2x-3=0$$ or $$x^4-x^3-x^3+x^2+x^2-x+3x-3=0$$ or $$(x-1)(x^3-x^2+x+3)=0$$ and since $x^3-x^2+x+3>0$ for $x>0$, we get $x=1$;

2. $t=-x-1$.

Thus, $\sqrt{x+y}=-x-1$, where $x\leq-1$.

Hence, $y=x^2+x+1$ and we get $$x(x^5+4x^4+9x^3+14x^2+10x+4)=0$$ or $$x^5+4x^4+9x^3+14x^2+10x+4=0$$ or $$x^5+2x^4+2x^4+4x^3+5x^3+10x^2+4x^2+8x+2x+4=0$$ or $$(x+2)(x^4+2x^3+5x^2+4x+2)=0$$ and since $x^4+2x^3+5x^2+4x+2>0$, we get $x=-2$ and the answer: $$\{(1,0),(-2,3)\}.$$ Done!