Item 2.

Here, since we have numbers larger than the modulus, we can reduce them prior to applying any algebra. We have \(84 \equiv 9\text{,}\) \(38 \equiv 8\) and \(79 \equiv 4\text{.}\) Thus,

\begin{equation*} \begin{aligned}84x - 38 \amp \equiv 79 \pmod{15} \\ 9x - 8 \amp \equiv 4 \pmod{15} \\ 9x \amp \equiv 12 \pmod{15} \\ 9x \amp \equiv 72 \pmod{15}. \end{aligned} \end{equation*}

We got the 72 by adding \(0 \equiv 60 \pmod{15}\) to both sides of the congruence. Now divide both sides by 9. However, since \(\gcd(9, 15) = 3\text{,}\) we must divide the modulus by 3 as well:

\begin{equation*} x \equiv 8 \pmod 5\text{.} \end{equation*}

So the solutions are those values which are congruent to 8, or equivalently 3, modulo 5. This means that in some sense there are 3 solutions modulo 15: 3, 8, and 13. We can write the solution:

\begin{equation*} x \equiv 3 \pmod{15}; ~~ x \equiv 8 \pmod{15}; ~~x \equiv 13 \pmod{15}\text{.} \end{equation*}
in-context