Question

Find x and y if 3^x+y= 1/3 and 4^2x-y= 1/16

Answer 1

$3^{x+y}=1/3\\3^{x+y}=3^{-1}$

equate the exponents since the bases are equal
$x+y = -1 ~~~~~\star$

$4^{2x-y} = 1/16\\4^{2x-y} = 4^{-2}\\2x-y=-2~~~~~\diamond$

you can solve $\star,~\diamond$ for $x,~y$

Answer 2

$3^{x+y}=1/3=3^{-1}\ \ \ \ \ 4^{2x-y}=1/16=4^{-2}\\\\x+y=-1\\2x-y=-2\\add\ the\ equations\\\\3x=-3,\ \ \ \ x=-1\\hence,\ \ y=-1-x=0$