Question

In system of equations 2^(y-x) (x+y)=1 and (x+y)^(x-y)=2, find the value of xy

Answer 1

2^(y-x) (x+y)=1

(x+y)^(x-y)=2

2^(y-x) = 1/(x+y) = (x+y)^-1
(x+y)^(x-y) = 2
Now multiply the bottom by the top and you have
(x+y)^(x-y-1) / 2^(1+y-x) = 2^(-(x-y-1))
so now we have
(x+y)^(x-y-1) = 1/2 (x-y-1)
x+y = 1/2