1. **State the problem:** Solve the system of congruences: $$3x + 13y \equiv 8 \pmod{55}$$ $$5x + 21y \equiv 34 \pmod{55}$$ 2. **Recall the method:** We want to find integers $x$ and $y$ such that both congruences hold modulo 55. 3. **Rewrite the system:** $$3x + 13y \equiv 8 \pmod{55} \implies 3x \equiv 8 - 13y \pmod{55}$$ 4. **Express $x$ in terms of $y$:** We need the inverse of 3 modulo 55. Since $3 \times 37 = 111 \equiv 1 \pmod{55}$ (because $111 - 2 \times 55 = 1$), the inverse of 3 mod 55 is 37. Thus, $$x \equiv 37(8 - 13y) \equiv 37 \times 8 - 37 \times 13y \equiv 296 - 481y \pmod{55}$$ 5. **Simplify modulo 55:** $$296 \equiv 296 - 5 \times 55 = 296 - 275 = 21 \pmod{55}$$ $$481y \equiv (481 - 8 \times 55) y = (481 - 440) y = 41y \pmod{55}$$ So, $$x \equiv 21 - 41y \pmod{55}$$ 6. **Substitute $x$ into the second congruence:** $$5x + 21y \equiv 34 \pmod{55}$$ Substitute $x$: $$5(21 - 41y) + 21y \equiv 34 \pmod{55}$$ $$105 - 205y + 21y \equiv 34 \pmod{55}$$ $$105 - 184y \equiv 34 \pmod{55}$$ 7. **Simplify constants modulo 55:** $$105 \equiv 105 - 1 \times 55 = 50 \pmod{55}$$ So, $$50 - 184y \equiv 34 \pmod{55}$$ 8. **Rearrange:** $$-184y \equiv 34 - 50 = -16 \pmod{55}$$ Multiply both sides by $-1$: $$184y \equiv 16 \pmod{55}$$ 9. **Reduce 184 modulo 55:** $$184 - 3 \times 55 = 184 - 165 = 19$$ So, $$19y \equiv 16 \pmod{55}$$ 10. **Find inverse of 19 modulo 55:** Using the Extended Euclidean Algorithm, inverse of 19 mod 55 is 29 because $19 \times 29 = 551 \equiv 1 \pmod{55}$. 11. **Solve for $y$:** $$y \equiv 29 \times 16 = 464 \equiv 464 - 8 \times 55 = 464 - 440 = 24 \pmod{55}$$ 12. **Find $x$ using $y=24$:** $$x \equiv 21 - 41 \times 24 = 21 - 984 = -963 \pmod{55}$$ Calculate $-963 \pmod{55}$: $$-963 + 18 \times 55 = -963 + 990 = 27$$ 13. **Final solution:** $$x \equiv 27 \pmod{55}, \quad y \equiv 24 \pmod{55}$$ This means the solutions are all integers $x, y$ such that: $$x = 27 + 55k, \quad y = 24 + 55m$$ for integers $k, m$.