Solve the following system of equations - (7).

Question

Solve the following system of equations (over the reals). $$\large \left\{ \begin{aligned} (x + y)^2 &= xy + 3y - 1\\ x + y &= \frac{x^2 + y + 1}{x^2 + 1}\end{aligned} \right.$$

From the system of equations, we have that

$$\left\{ \begin{aligned} (x + y)^2 - 1 &= y(x + 3)\\ x + y - 1&= \frac{y}{x^2 + 1}\end{aligned} \right. \implies \frac{y}{x^2 + 1}\cdot (x + y + 1) = y(x + 3)$$

$$\left[ \begin{aligned} y &= 0\\ x + y + 1 &= (x^2 + 1)(x + 3) \end{aligned} \right.$$

Plugging in $y = 0$ into the first equation, we have that $x^2 = -1$, which is incorrect for $\forall x \in \mathbb R$.

Then $x + y + 1 = (x^2 + 1)(x + 3) \implies y = x^3 + 3x^2 + 2$.

And I am done with my life.

Answer 1

Let $\sigma=x^2+1$, note that $\sigma y\ne 0$. The first equation is equivalent to: $$\sigma+y(x+y)-3y=0\tag1$$ and the second: $$x+y-1=\frac{y}\sigma\tag2$$

$$(1)\iff x+y=3-\frac{\sigma}{y}.$$ Replacing this last one into $(2)$ yields: $$2=\frac{y}\sigma+\frac{\sigma}y\iff (y-\sigma)^2=0.$$ Therefore $y=\sigma=x^2+1=2-x$. I let you take it from here.

Answer 2

Direct substitution method!

From the second: $$x + y = \frac{x^2 + y + 1}{x^2 + 1} \Rightarrow y=\frac{(1-x)(x^2+1)}{x^2}$$ Subbing to the first: $$\left(\frac{x^2-x+1}{x^2}\right)^2=\frac{(x+3)(1-x)(x^2+1)}{x^2}-1 \Rightarrow \\ x^6+2x^5-2x+1=0 \Rightarrow \\ (x^2+1)(x^2+x-1)^2=0 \Rightarrow \\ x^2+x-1=0 \Rightarrow \\ x=\frac{-1\pm \sqrt{5}}{2}.$$ Note: $x^6+2x^5-2x+1=0 \Rightarrow (x^2)^3+1+2x(x^4-1)=0$.