Solution
= 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 gives
$$
u+\frac1u=2,\qquad (u-1)^2=0.
$$
Thus $u=1$, so $x=0$ and $y=-x=0$. The unique real solution of the <simultaneous equations> is
$$
\boxed{(x,y)=(0,0).}
$$