Here is the problem I was given: Find integers $a, b, c, d, e, f$ and positive integer $n > 1$ for which there is a solution in real numbers $x$ and $y$ for the simultaneous equations $$ax + cy = e,\quad bx + dy = f,$$
and yet there is not a solution in integers $x$ and $y$ for the simultaneous congruences $$ax + cy \equiv e \bmod n,\quad bx + dy \equiv f \bmod n.$$
Here is what I have so far: $ax+cy=e, bx+dy=f$ becomes $(cf-de)x ≡ af – be (\bmod n)$, by multiplying, $ax+cy=e$ by $f$ and $bx+dy=f$ by $e$ and subtracting. A similar process is used to have a similar equation relating to $y$, $(cf-de)y≡bc-ad (\bmod n)$. I don't know where to go from here.
We have$$ \begin{align} ax+cy&=e \tag 1 \\ bx+dy&=f \tag 2 \end{align}$$ Solving this system gives $$ \begin{align} (bc-ad)y &= be-af\\ (bc-ad)x &= cf-de \end{align}$$
So long as $bc-ad\ne 0,$ we can solve these equations for $x,y\in \mathbf R.$ Now, we can follow the same procedure and get two congruences $$ \begin{align} (bc-ad)y &\equiv (be-af) \pmod n\\ (bc-ad)x &\equiv (cf-de) \pmod n\\ \end{align}$$
and we will run into trouble if $bc-ad \equiv 0 \pmod n,$ for then the left-hand side will be $0$ and if the right-hand side is non-zero there is no solution.
Now try to find an example.