Since both RHS = y is equivalent to the single equation. $$ 2^x = 3-x. $$
If you haven't found a way to solve this algebraically, I don't blame yourself. There is no way to 'solve for x' in algebra. This is a transcendental function and must be solved graphically.
If you graph $2^x$ and $3-x$, you'll see they intersect once. By inspection $x =1$ is a solution, so that's the one real solution.
(additionally there are complex solutions in terms of the Lambert W-function)