1. Problem: Find all solutions for the system of linear congruences using the Chinese Remainder Theorem (CRT) or substitution method as specified. 2. For part (a): Given: $$x \equiv 1 \pmod{4},\quad x \equiv 2 \pmod{5},\quad x \equiv 3 \pmod{7}$$ - The moduli 4, 5, and 7 are pairwise coprime. - The product of moduli is $$N = 4 \times 5 \times 7 = 140$$. 3. Compute partial products: $$N_1 = \frac{N}{4} = 35,\quad N_2 = \frac{N}{5} = 28,\quad N_3 = \frac{N}{7} = 20$$ 4. Find the modular inverses $M_i$ such that: $$N_i M_i \equiv 1 \pmod{n_i}$$ - For $N_1=35$ mod 4: $$35 \equiv 3 \pmod{4}$$ Find $M_1$ with $3 M_1 \equiv 1 \pmod{4}$, so $M_1=3$. - For $N_2=28$ mod 5: $$28 \equiv 3 \pmod{5}$$ Find $M_2$ with $3 M_2 \equiv 1 \pmod{5}$, so $M_2=2$. - For $N_3=20$ mod 7: $$20 \equiv 6 \pmod{7}$$ Find $M_3$ with $6 M_3 \equiv 1 \pmod{7}$, so $M_3=6$. 5. Compute solution: $$x \equiv a_1 N_1 M_1 + a_2 N_2 M_2 + a_3 N_3 M_3 \pmod{N}$$ $$x \equiv 1 \times 35 \times 3 + 2 \times 28 \times 2 + 3 \times 20 \times 6 \pmod{140}$$ $$x \equiv 105 + 112 + 360 = 577 \pmod{140}$$ 6. Simplify modulo 140: $$577 \equiv 577 - 4 \times 140 = 577 - 560 = 17 \pmod{140}$$ 7. Final solution for (a): $$\boxed{x \equiv 17 \pmod{140}}$$ This means all solutions are of the form $x = 17 + 140k$ for integer $k$.