Prove Proposition (3.24).
Proposition (3.24) is as follows.
Let $n_1, n_2, n_3, \ldots , n_r$ be positive integers which are pairwise prime. Also, integers $c_k$’s satisfy $\gcd (c_k, n_k) = 1$ for $k= 1, 2, \ldots , n$. Then the simultaneous linear congruences
$c_1 x ≡ b_1 \pmod {n_1}$
$c_2 x ≡ b_2 \pmod {n_2}$
⋮
$c_r x ≡ br \pmod {n_r}$
have a solution satisfying all these equations.
Moreover, the solution is unique modulo $n_1 × n_2 × n_3 × ⋯ × n_r$.
We're given that $\gcd(c_k, n_k)=1$ which means a linear congruence of the form $c_k x \equiv b_k \pmod {n_k}$ has a unique solution. This is Corollary (3.19), a specialisation of Proposition (3.16).
This means the simultaneous linear congruences are of the form
$ x \equiv d_1 \pmod {n_1}$
$ x \equiv d_1 \pmod {n_2}$
$\vdots$
$ x \equiv d_r \pmod {n_r}$
Since the modulii are all pair-wise corpime, then the Chinese Remainder Theorem can be used to find a unique solution modulo $n_1 \times n_2 \times \ldots \times n_r$.