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Now at this point we know \(y = 2 + 17k\) will work for any integer \(k\text{.}\) If we haven't made a mistake, we should be able to plug this back into our original Diophantine equation to find \(x\text{:}\)

\begin{equation*} \begin{aligned}17x + 29(2 + 17k) \amp = 41\\ 17x \amp = -17 - 29\cdot 17k\\ x \amp = -1-29k. \end{aligned} \end{equation*}

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