 Solution

The original PDF puts both $x+y$ in the exponent in the second equation. Since the exponential function $3^t$ is injective, $3^{x+y}=1$ gives $x+y=0$. Put $u=3^x>0$, so $3^y=1/u$. The first equation then givesThus $u=1$, so $x=0$ and $y=-x=0$. The unique real solution of the simultaneous equations is