Let $x$ and $y$ be integers such that $4 x+5 y=7$. Determine the smallest value of $5|x|-3|y|$.

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Solution. It is visible that $x=3, y=-1$ is a solution to the equation. Since this is a linear Diophantine equation, all solutions are of the form $x_{k}=3-5 k, y_{k}=-1+4 k$, where $k$ is any integer.

If $k$ is not positive, then $\left|x_{k}\right|>\left|y_{k}\right|$, because $x$ starts from a larger number in absolute value and always increases by a larger amount. If $k=2$, then $\left|x_{k}\right|=\left|y_{k}\right|$, so in this case $5|x|>3|y|(x$ is not 0$)$. If $k>2$, then $\left|x_{k}\right|>\left|y_{k}\right|$ as well, since it starts from the same absolute value and increases by a larger amount. Therefore, for all $k$ not equal to 1: $5|x|-3|y|>0$. For $k=1$, this value is 1, and we cannot find a smaller value, since the difference is always a positive integer, and 1 is the smallest positive integer.

Thus, the minimum of the expression is 1, and it is achieved when $x=-2, y=3$.