1. **Problem statement:** Find the set of solutions for the linear congruences: i. $x \equiv 3 \pmod{5}$ 2. **Formula and rules:** A linear congruence $ax \equiv b \pmod{m}$ has solutions if and only if $\gcd(a,m)$ divides $b$. The solutions are given by $$x \equiv x_0 + \frac{m}{d}k \pmod{m}$$ where $d = \gcd(a,m)$ and $x_0$ is a particular solution. 3. **Solve i:** Given $x \equiv 3 \pmod{5}$, this means $x$ leaves a remainder 3 when divided by 5. 4. **Solution i:** The set of solutions is all integers of the form $$x = 3 + 5k, \quad k \in \mathbb{Z}$$ 5. **Solve ii:** Given $2x \equiv 5 \pmod{9}$. 6. **Check gcd:** $\gcd(2,9) = 1$, which divides 5, so solutions exist. 7. **Find inverse of 2 mod 9:** We want $2y \equiv 1 \pmod{9}$. Testing values: $$2 \times 5 = 10 \equiv 1 \pmod{9}$$ So inverse of 2 mod 9 is 5. 8. **Multiply both sides by inverse:** $$x \equiv 5 \times 5 = 25 \equiv 7 \pmod{9}$$ 9. **Solution ii:** The set of solutions is all integers of the form $$x = 7 + 9k, \quad k \in \mathbb{Z}$$ **Final answer:** - i. $\{x \mid x \equiv 3 \pmod{5}\}$ - ii. $\{x \mid x \equiv 7 \pmod{9}\}$