1. **Stating the problem:** Solve the system of congruences: $$2x + y \equiv 1 \pmod{6}$$ $$x + 3y \equiv 3 \pmod{6}$$ 2. **Recall the rules:** - Congruences mean the expressions are equal modulo 6. - We want integers $x$ and $y$ such that both congruences hold. - We can solve this like a system of linear equations but all operations are modulo 6. 3. **Rewrite the system:** $$2x + y \equiv 1 \pmod{6}$$ $$x + 3y \equiv 3 \pmod{6}$$ 4. **Express $y$ from the first equation:** $$y \equiv 1 - 2x \pmod{6}$$ 5. **Substitute $y$ into the second equation:** $$x + 3(1 - 2x) \equiv 3 \pmod{6}$$ $$x + 3 - 6x \equiv 3 \pmod{6}$$ $$-5x + 3 \equiv 3 \pmod{6}$$ 6. **Simplify:** $$-5x \equiv 0 \pmod{6}$$ Since $-5 \equiv 1 \pmod{6}$, this is: $$x \equiv 0 \pmod{6}$$ 7. **Find $y$ using $x=0$:** $$y \equiv 1 - 2(0) \equiv 1 \pmod{6}$$ 8. **Solution:** $$x \equiv 0 \pmod{6}, \quad y \equiv 1 \pmod{6}$$ This means $x=6k$ and $y=1+6m$ for integers $k,m$. **Final answer:** $$\boxed{x \equiv 0 \pmod{6}, \quad y \equiv 1 \pmod{6}}$$