1. **State the problem:** Solve the system of linear congruences. Since the user did not specify the system, let's consider a general example: $$\begin{cases} x \equiv a_1 \pmod{m_1} \\ x \equiv a_2 \pmod{m_2} \end{cases}$$ 2. **Formula and rules:** To solve such a system, we use the Chinese Remainder Theorem (CRT) if $m_1$ and $m_2$ are coprime (i.e., $\gcd(m_1,m_2)=1$). 3. **Steps to solve:** - Compute $M = m_1 \times m_2$. - Compute $M_1 = \frac{M}{m_1}$ and $M_2 = \frac{M}{m_2}$. - Find the modular inverses $y_1$ and $y_2$ such that: $$M_1 y_1 \equiv 1 \pmod{m_1}$$ $$M_2 y_2 \equiv 1 \pmod{m_2}$$ - The solution is: $$x \equiv a_1 M_1 y_1 + a_2 M_2 y_2 \pmod{M}$$ 4. **Explanation:** The CRT guarantees a unique solution modulo $M$ when $m_1$ and $m_2$ are coprime. The modular inverses ensure that each term aligns correctly with its respective congruence. 5. **Example:** Solve $$\begin{cases} x \equiv 2 \pmod{3} \\ x \equiv 3 \pmod{5} \end{cases}$$ - $M = 3 \times 5 = 15$ - $M_1 = 15/3 = 5$, $M_2 = 15/5 = 3$ - Find $y_1$ such that $5 y_1 \equiv 1 \pmod{3}$: - $5 \equiv 2 \pmod{3}$, so solve $2 y_1 \equiv 1 \pmod{3}$ - $y_1 = 2$ because $2 \times 2 = 4 \equiv 1 \pmod{3}$ - Find $y_2$ such that $3 y_2 \equiv 1 \pmod{5}$: - $3 y_2 \equiv 1 \pmod{5}$ - $y_2 = 2$ because $3 \times 2 = 6 \equiv 1 \pmod{5}$ - Compute: $$x \equiv 2 \times 5 \times 2 + 3 \times 3 \times 2 = 20 + 18 = 38 \pmod{15}$$ - Simplify: $$38 \equiv 8 \pmod{15}$$ **Final answer:** $$x \equiv 8 \pmod{15}$$ This means all integers congruent to 8 modulo 15 satisfy the system.