Solve for $x, y \in \mathbb R$: $(\sqrt{x^2 + 1} + x)(\sqrt{y^2 + 1} - y) = 1$ and $3\sqrt{x + 2y - 2} + x\sqrt{x - 2y + 6} = 10$.

Question

Solve for the following system of equations for $x, y \in \mathbb R$: $$\large \left\{ \begin{align} \left(\sqrt{x^2 + 1} + x\right) \cdot \left(\sqrt{y^2 + 1} - y\right) = 1\\ 3\sqrt{x + 2y - 2} + x\sqrt{x - 2y + 6} = 10 \end{align} \right.$$

$\left(x \ge -2, \dfrac{2 - x}{2} \le y \le \dfrac{x + 6}{2}\right)$

We have that $$\left(\sqrt{x^2 + 1} + x\right) \cdot \left(\sqrt{y^2 + 1} - y\right) = 1 \iff \left(\sqrt{x^2 + 1} - x\right) \cdot \left(\sqrt{y^2 + 1} + y\right) = 1$$

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

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

$$\iff 3\sqrt{x \pm 2x - 2} + x\sqrt{x \mp 2x + 6} = 10$$

Then I don't know what to do next.

Answer 1

You have $$\sqrt{x^2+1}+x=\sqrt{y^2+1}+y$$ by multiplying both sides in first equation by $\sqrt{y^2+1}+y$. Also $$\sqrt{x^2+1}-x=\sqrt{y^2+1}-y$$ by multiplying both sides in first equation by $\sqrt{x^2+1}-x$. Now, subtract the second equation from first to get $$x=y$$ So now we need to solve $$3\sqrt{3x-2}+x\sqrt{6-x}=10$$ which can easily be solved by squaring. The required solution is $x=2$.

Answer 2

Hint

Let $x=\cot2A,0<2A<\pi$

$\implies\sqrt{x^2+1}+x=+\csc2A+\cot2A=\cot A$

Similarly let $y=\cot2B,0<2B<\pi,$

$$\sqrt{y^2+1}-y=\tan B$$

$\implies\cot A=\cot B$

$\implies x=\cot2A=\cot2B=y$