Question

Solve the following system of equations.
$0.5.{\log }_{2}x-{\log }_{4}y=0,x^2-5y^2+4=0.$

Answer 1

$05\cdot {\log }_{2}x-{\log }_{4}y=0$

$\Rightarrow \frac{1}{2}{\log }_{2}x-\frac{1}{2}{\log }_{2}y=0$

$\Rightarrow {\log }_{2}x={\log }_{2}y$

$\Rightarrow x=y$ (equation $1$)

$x^2-5y^2+y=0$

From $1$

$\Rightarrow x^2-5x^2+4=0$

$\Rightarrow -4x^2+4=0$

$\Rightarrow (x^2-1)=0$

$\therefore x=-1,1$

From $1$, $y=+1,-1$

So, $(x,y):(1,1),(-1,-1)$