<s>[INST] Solve the following system of equations - (7). 
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.
 [/INST] 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: $x^6+2x^5-2x+1=0 \Rightarrow (x^2)^3+1+2x(x^4-1)=0$.
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