We have, $2^(y-x) (x + y) = 1 implies x + y = (1)/(2^(y-x))$
implies $x + y = 2^(x-y) " ...(1) and " (x + y)^(x-y) = 2 " "....(2)$
$implies (2^(x-y))^(x-y) = 2$
$ implies 2^((x-y)^(2)) = 2^(1)$
$implies x - y = pm 1$
${:("If x -y = 1 then x + y = 2"),("Solving these two, we get"),(x = (3)/(2). y = (1)/(2)):}:|{:("If x - y = - 1 then (x + y)"^(-1)=2),("So, x - y = - 1"),("and x + y ="(1)/(2)),("Solving these two, we get "),(x = - (1)/(4). y = (3)/(4)):}$
$therefore$ ${:(x = (3)/(2)),(y = (1)/(2)):}}$ or ${:(x = - (1)/(4)),(y = (3)/(4)):}}$
