Linear Diophantine equation

If \(d=\gcd(a,b)\) divides \(c\), Bézout's identity gives integers \(x_0,y_0\) with \(ax_0+by_0=d\); multiplying by \(c/d\) yields a particular solution. Every other solution has the form \(x=x_0+\frac{b}{d}t\), \(y=y_0-\frac{a}{d}t\) with \(t\in\mathbb{Z}\), and there are no others.

If \(d\) does not divide \(c\), the equation has no integer solutions, even if it has rational ones. This condition is necessary because \(ax+by\) is always a multiple of \(d\). In \(6x+9y=10\) we have \(d=3\), which does not divide \(10\), so there are no integer solutions.

For example, for \(7x+5y=1\) the Euclidean algorithm gives \(7(3)+5(-4)=1\), so \((x_0,y_0)=(3,-4)\) is a particular solution and the general family is \(x=3+5t\), \(y=-4-7t\).

A typical mistake is forgetting to divide by \(d\) when building the general family: for \(a=4\), \(b=6\), \(c=2\), the step in \(x\) is \(6/2=3\) and in \(y\) is \(4/2=2\), not \(6\) and \(4\).