Algebra - System of Equations and Inequalities

Solve the following system of equations: $\;$ $\sqrt{\dfrac{2x - 1}{y + 2}} + \sqrt{\dfrac{y + 2}{2x - 1}} = 2$, $\;$ $x + y = 2$

Given system of equations:

$\sqrt{\dfrac{2x - 1}{y + 2}} + \sqrt{\dfrac{y + 2}{2x - 1}} = 2$ $\;\;\; \cdots \; (1)$, $\;\;$ $x + y = 2$ $\;\;\; \cdots \; (2)$

We have from equation $(2)$, $\;\;\;$ $x = 2 - y$ $\;\;\; \cdots \; (3)$

In view of equation $(3)$, equation $(1)$ becomes

$\sqrt{\dfrac{2 \left(2 - y\right) - 1}{y + 2}} + \sqrt{\dfrac{y + 2}{2 \left(2 - y\right) - 1}} = 2$

i.e. $\;$ $\sqrt{\dfrac{3 - 2y}{y + 2}} + \sqrt{\dfrac{y + 2}{3 - 2y}} = 2$

i.e. $\;$ $\dfrac{3 - 2y + y + 2}{\sqrt{\left(y + 2\right) \left(3 - 2y\right)}} = 2$

i.e. $\;$ $5 - y = 2 \sqrt{6 - y - 2y^2}$ $\;\;\; \cdots \; (4)$

Squaring equation $(4)$ gives

$25 - 10y + y^2 = 24 - 4y -8y^2$

i.e. $\;$ $9y^2 -6y + 1 = 0$

i.e. $\;$ $\left(3y - 1\right)^2 = 0$

i.e. $\;$ $3y - 1 = 0$

i.e. $\;$ $y = \dfrac{1}{3}$

Substituting $\;$ $y = \dfrac{1}{3}$ $\;$ in equation $(3)$ gives $\;\;$ $x = 2 - \dfrac{1}{3} = \dfrac{5}{3}$

$\therefore \;$ The solution to the given system of equations is $\left(x, y\right) = \left\{\left(\dfrac{5}{3}, \dfrac{1}{3}\right) \right\}$