Solve the following system of equations: $\begin{cases} x + y = 4 + \sqrt{y^2 + 2} \\ \log x - 2 \log 2 = \log \left(1 + \dfrac{1}{2} y\right) \end{cases}$
Given system of equations:
$x + y = 4 + \sqrt{y^2 + 2}$ $\;\;\; \cdots \; (1)$
$\log x - 2 \log 2 = \log \left(1 + \dfrac{1}{2} y\right)$ $\;\;\; \cdots \; (2)$
We have from equation $(2)$,
$\log x - \log 4 = \log \left(1 + \dfrac{y}{2}\right)$
i.e. $\;$ $\log \left(\dfrac{x}{4}\right) = \log \left(1 + \dfrac{y}{2}\right)$ $\;\;\; \cdots \; (3)$
Taking antilog on both sides of equation $(3)$ gives
$\dfrac{x}{4} = 1 + \dfrac{y}{2}$
i.e. $\;$ $x - 4 = 2y$ $\;\;\; \cdots \; (4)$
In view of equation $(4)$ equation $(1)$ becomes
$2y + y = \sqrt{y^2 +2}$
i.e. $\;$ $3y = \sqrt{y^2 + 2}$
i.e. $\;$ $9y^2 = y^2 + 2$
i.e. $\;$ $8y^2 = 2$
i.e. $\;$ $y^2 = \dfrac{1}{4}$ $\implies$ $y = \pm \dfrac{1}{2}$
When $y = \dfrac{-1}{2}$, we have from equation $(4)$
$x = 4 - 2 \times \dfrac{1}{2} = 3$
Substituting $\left(x, y\right) = \left(3, \dfrac{-1}{2}\right)$ in equation $(1)$ gives
$3 - \dfrac{1}{2} = 4 + \sqrt{\dfrac{1}{4} + 2}$
i.e. $\;$ $\dfrac{5}{2} = 4 + \dfrac{3}{2}$
i.e. $\;$ $\dfrac{5}{2} = \dfrac{11}{2}$ $\;$ which is not true.
$\implies$ $\left(3, \dfrac{-1}{2}\right)$ is not a valid solution.
When $y = \dfrac{1}{2}$, we have from equation $(4)$
$x = 4 + 2 \times \dfrac{1}{2} = 5$
Substituting $\left(x, y\right) = \left(5, \dfrac{1}{2}\right)$ in equation $(1)$ gives
$5 + \dfrac{1}{2} = 4 + \sqrt{\dfrac{1}{4} + 2}$
i.e. $\;$ $\dfrac{11}{2} = 4 + \dfrac{3}{2}$
i.e. $\;$ $\dfrac{11}{2} = \dfrac{11}{2}$ $\;$ which is true.
$\therefore \;$ The solution to the given system of equations is $\;\;$ $x = \left\{\left(5, \; \dfrac{1}{2}\right) \right\}$