1. **State the problem:** Solve the system of congruences: $$x + 5y \equiv 3 \pmod{9}$$ $$4x + 5y \equiv 1 \pmod{9}$$ 2. **Understand the goal:** We want to find integers $x$ and $y$ such that both congruences hold true modulo 9. 3. **Step 1: Subtract the first congruence from the second to eliminate $y$:** $$ (4x + 5y) - (x + 5y) \equiv 1 - 3 \pmod{9} $$ $$ 4x - x + 5y - 5y \equiv -2 \pmod{9} $$ $$ 3x \equiv -2 \pmod{9} $$ 4. **Simplify the right side modulo 9:** $$ -2 \equiv 7 \pmod{9} $$ So, $$ 3x \equiv 7 \pmod{9} $$ 5. **Solve for $x$:** We need to find the inverse of 3 modulo 9. However, $\gcd(3,9) = 3 \neq 1$, so 3 is not invertible modulo 9. 6. **Check if the congruence $3x \equiv 7 \pmod{9}$ has solutions:** A linear congruence $ax \equiv b \pmod{m}$ has solutions if and only if $\gcd(a,m)$ divides $b$. Here, $\gcd(3,9) = 3$, but 3 does not divide 7. 7. **Conclusion:** No solution exists for $x$ in the congruence $3x \equiv 7 \pmod{9}$, so the original system has no solution. **Final answer:** There is no solution to the system of congruences because the necessary divisibility condition is not met.