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Then to divide both sides by 3, we first add 0 to both sides. Of course, on the right-hand side, we want that 0 to be a 10 (yes, \(10\) really is 0 since they are congruent modulo 5). This gives,

\begin{equation*} 3x \equiv 12 \pmod{5}\text{.} \end{equation*}

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