Solve the system of equations: $ \sqrt{y^2 - 8x + 9} - \sqrt[3]{xy - 6x + 12} = 1$ and $\sqrt{2(x - y)^2 + 10x - 6y + 12} - \sqrt{y} = \sqrt{x + 2}$.

Question

Solve the following system of equations $$\large \left\{ \begin{align} \sqrt{y^2 - 8x + 9} - \sqrt[3]{xy - 6x + 12} = 1\\ \sqrt{2(x - y)^2 + 10x - 6y + 12} - \sqrt{y} = \sqrt{x + 2}\end{align} \right.$$

This system of equations uses inequalities to solve. The following solution is not created by me, as I would have had a different approach to the problem.

Answer 1

Let $x' = x + 1$ and $y' = y - 1$, the system of equations becomes

$$\left\{ \begin{align} \sqrt{y'^2 - 8x' + 2y' + 18} - \sqrt[3]{x'y' - 5x' - y' + 17} = 1\\ \sqrt{2(x' - y')^2 + 2x' + 2y' + 4} - \sqrt{y' + 1} = \sqrt{x' + 1}\end{align} \right.$$

Solving the second equation, we have that $$\sqrt{2(x' - y')^2 + 2x' + 2y' + 4} \ge \sqrt{2[(x' + 1) + (y' + 1)]} \ge \sqrt{x' + 1} + \sqrt{y' + 1}$$

The equality sign occurs when $x' = y'$. Substituting $x' = y'$ into the first equation, we have that

$$\sqrt{x'^2 - 6x' + 18} - \sqrt[3]{x'^2 - 6x' + 17} = 1$$

Let $\sqrt[3]{x'^2 - 6x' + 17} = t (t \ge 2)$, we have that $$\sqrt{t^3 + 1} - t = 1 \iff \sqrt{t^2 - t + 1} = \sqrt{t + 1} \iff t^2 - 2t = 0 \implies t = 2$$

$$x'^2 - 6x' + 9 = 0\implies x' = y' = 3 \implies (x, y) = (2, 4)$$.

Answer 2

Hint: Squaring two times and factorizing your second equation we get $$ \left( 4\,{x}^{2}-8\,xy+4\,{y}^{2}+20\,x-12\,y+25 \right) \left( 2+x -y \right) ^{2} =0$$ This can be solved for $x$ or $y$ and you can eliminate one variable.

Answer 3

By C-S $$\sqrt{x+2}+\sqrt{y}=\sqrt{2(x-y)^2+10x-6y+12}=\sqrt{2((x-y)^2+4(x-y)+4+x+y+2)}=$$ $$=\sqrt{2((x-y+2)^2+x+y+2)}\geq\sqrt{(1^2+1^2)(x+2+y)}\geq\sqrt{x+2}+\sqrt{y},$$ which says that all inequalities here they are equalities, which occurs for $$x-y+2=0$$ and $$(1,1)||(\sqrt{x+2},\sqrt{y}),$$ which gives $$y=x+2$$ and the rest is smooth.