Solution: Solve $x \equiv 3 \pmod7$ and $x \equiv 5 \pmod9$. Let $x = 7k + 3$. Substitute into the second congruence: $7k + 3 \equiv 5 \pmod9 \Rightarrow 7k \equiv 2 \pmod9$. The modular inverse of 7 modulo 9 is 4 (since $7 imes 4 = 28 \equiv 1 \pmod9$), so $k \equiv 2 imes 4 = 8 \pmod9$. Thus, $k = 9m + 8$, and $x = 7(9m + 8) + 3 = 63m + 59$. The solutions are $59, 122, 185, \dots$. Within 1–150, only 59 and 122 satisfy. The count is $oxed2$. - Crosslake