Solve for $x, y \in \mathbb R$: $(xy + 1)(x + 1) = 6$ and $x^2(y^2 + y + 1) = 7$.

Question

Solve the system of equations for $x, y \in \mathbb R$: $$\large \left\{ \begin{align} (xy + 1)(x + 1) = 6\\ x^2(y^2 + y + 1) = 7 \end{align} \right.$$

And again,( for approximately the $356th$ time,) this problem is adapted from a recent competition. I have provided a solution below, although it is a little bit clunky-sounding, I hope that it can help you a little bit.

Answer 1

It is evident that $\left\{ \begin{align} (xy + 1)(x + 1) = 6\\ x^2(y^2 + y + 1) = 7 \end{align} \right.$$\iff \left\{ \begin{align} x^2y + xy + x + 1 = 6\\ x^2(y^2 + 2y + 1) = x^2y + 7 \end{align} \right.$

$\iff \left\{ \begin{align} x^2y + x(y + 1) + 7 = 12\\ [x(y + 1)]^2 = x^2y + 7 \end{align} \right. \implies [x(y + 1)]^2 + x(y + 1) - 12 = 0$

$\iff [x(y + 1) + 4][x(y + 1) - 3] = 0 \iff \left[ \begin{align} x(y + 1) = -4\\ x(y + 1) = 3 \end{align} \right.$

$\implies \left[ \begin{align} x(y + 1) = -4 &\text{ and } x^2y = 9\\ x(y + 1) = 3 &\text{ and } x^2y = 2\end{align} \right.$.

Using the Vièta formulas, we have that $x$ and $xy$ are roots of $z^2 + 4z + 9 = 0$ or $z^2 - 3z + 2 = 0$.

However, $z^2 + 4z + 9 = (z + 2)^2 + 5 \ge 5, \forall z \in \mathbb R$.

$\implies z^2 - 3z + 2 = 0 \iff (z - 1)(z - 2) = 0 \iff \left[ \begin{align} z = 1\\ z = 2 \end{align} \right.$$\implies \left[ \begin{align} x = 1 &\text{ and } xy = 2\\ x = 2 &\text{ and } xy = 1 \end{align} \right.$

$\iff (x, y) \in \left\{(1, 2), \left(2, \dfrac{1}{2}\right)\right\}$.

Answer 2

From the first equation we obtain $$y=\frac{5-x}{x(x+1)}$$ and after substitution on the second we obtain: $$(x-1)(x-2)(x^2+4x+9)=0$$ and the rest is smooth.

Answer 3

$$(\!xy+ x- 3\!)\!(\!xy+ x+ 4\!)\!=\!\left (\!x^{\!2}\!(\!y^{\!2}+ y+ 1\!)\!-\!7\!\right )\!+\!\left (\!(\!xy+ 1\!)\!(\!x+ 1\!)\!-\!6\!\right )\!\therefore\!(\!xy+ x- 3\!)\!(\!xy+ x+ 4\!)\!=\!0$$ Then, my solution will be as same as yours! The solution is $\{\!x= 1,\,y= 2\!\}\!\bigcap\!\{\!x= 2,\,y= 1\!/\!2\!\}$ / Q.E.D