An algebra problem by Rahil Sehgal

From the first equation we have that $x + y = 2^{x - y}$ . Substituting into the second equation, we have that

$(x + y)^{x - y} = (2^{x - y})^{x - y} = 2 \Longrightarrow 2^{(x - y)^{2}} = 2 \Longrightarrow (x - y)^{2} = 1 \Longrightarrow x - y = \pm 1$ .

If $x - y = 1$ then from the second equation $x + y = 2$ , giving us $(x,y) = (\frac{3}{2}, \frac{1}{2})$ .

If $x - y = -1$ then $(x + y)^{-1} = 2 \Longrightarrow x + y = \frac{1}{2}$ , giving us $(x,y) = (-\frac{1}{4}, \frac{3}{4})$ .

The desired sum is then $\dfrac{3}{2} + \dfrac{1}{2} - \dfrac{1}{4} + \dfrac{3}{4} = \dfrac{5}{2} = \boxed{2.5}$ .

Note that we didn't really need to solve for $(x,y)$ as the only two cases involved gave us $x + y = 2$ and $x + y = \frac{1}{2}$ , which was all we needed to determine that the desired sum was $2.5$ .