Solve for $x, y \in \mathbb R$: $(x - y)^2 + 5y - 3x + 4 = 2\sqrt{(x + 1)(y - 1)}$ and $\dfrac{3xy - 6x - 5y + 11}{\sqrt{x^2 + 1}} = 5$.

Question

Solve the following system of equation for $x, y \in \mathbb R$. $$\large \left\{ \begin{align} (x - y)^2 + 5x - 3y + 4 = 2\sqrt{(x + 1)(y - 1)}\\ \dfrac{3xy - 6x - 5y + 11}{\sqrt{x^3 + 1}} = 5 \end{align} \right.$$

There should be more practical solutions.

Oh, and, this problem is adapted from a recent competition.

Answer 1

We have that $$(x - y)^2 + 5x - 3y + 4 = 2\sqrt{(x + 1)(y - 1)} $$

$$\iff (x - y)^2 + 4(x - y) + 4 + (x + 1) + (y - 1) - 2\sqrt{(x + 1)(y - 1)} = 0$$

$$\iff (x - y + 2)^2 + (\sqrt{x + 1} - \sqrt{y - 1})^2 = 0 \implies x = y - 2$$

$$$$

Furthermore, $$\frac{3xy - 6x - 5y + 11}{\sqrt{x^3 + 1}} = 5 \iff (3x - 5)(y - 2) + 1 = 5\sqrt{x^3 + 1}$$

$$\iff (3x - 5)x + 1 = 5\sqrt{x^3 + 1} \iff (3x^2 - 5x + 1)^2 = 25(x^3 + 1)$$

$$\iff (x^2 - 5x - 3)(9x^2 - 10x + 8) = 0$$

However, $9x^2 - 10x + 8 = \left(3x - \dfrac{5}{3}\right)^2 + \dfrac{47}{9} > 0, \forall x \in \mathbb R$.

That means $x^2 - 5x - 3 = 0 \iff x = \dfrac{5 \pm \sqrt{37}}{2} \implies (x, y) = \left(\dfrac{5 \pm \sqrt{37}}{2}, \dfrac{9 \pm \sqrt{37}}{2}\right)$.