Question

Solve the following system of equations.
$x+y=4+\sqrt{y^2+2},\log x-2\log 2=\log (1+\frac{1}{2}y)$

Answer 1

$x+y=4+\sqrt{y^2+2}$

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

$x^2+y^2+16+2xy-8x-8y=y^2+2$

$x^2+2xy-8x-8y+14=0\to 1$

$\log x-2\log 2=\log (1+\frac{1}{2}y)$

$\log \left(\frac{x}{4}\right)=\log (1+\frac{1}{2}y)$

$\frac{x}{4}=1+\frac{1}{2}y$

$\left(\frac{x-4}{2}\right)=y\to 2$

From $1$ and $2$

$x^2+x(x-4)-8x-4(x-4)+14=0$

$x^2+x^2-4x-8x-4x+16+14=0$

$2x^2-16x+30=0$

$x^2-8x+15=0$

$(x-3)(x-5)=0$

$x=3,5$

From $2$; $y=\left(\frac{x-4}{2}\right)y=\left(\frac{3-4}{2}\right)=-\frac{1}{2}y=\left(\frac{5-4}{2}\right)=\frac{1}{2}$

$(x,y):(3,-\frac{1}{2}),(5,\frac{1}{2})$