5. Consider equations of the form $a x+b=c$, where $a, b$ and $c$ are integers such that one is the sum of the other two and $a$ is non-zero. What are the possible integer values of $x$ ?
Solution
We have three cases to consider: $c=b+a, b=c+a$ and $a=b+c$.
For $c=b+a$ :
In this case the given equation can be written as $a x+b=b+a$, which can be simplified to give $a x=a$. As we were told in the question that $a$ is non-zero we may divide the previous equation by $a$ to give $x=1$.

For $b=c+a$ :
In this case the given equation can be written as $a x+c+a=c$, which can be rearranged and simplified to give $a x=-a$. Again, as we were told in the question that $a$ is non-zero we may divide the previous equation by $a$ to give $x=-1$.
For $a=b+c$ :
In this case the given equation can be written as $(b+c) x+b=c$, which can be rearranged to give $x=\frac{c-b}{b+c}$.
This expression for $x$ is more complicated than the previous two cases and it is not clear what integer values it can take.
To simplify things we will let $a=2$. This means that $b+c=2$, which can be rearranged to give $c=2-b$.
We can now rewrite the expression for $x$ in terms of $b$ only, giving $x=\frac{2-b-b}{2}$ which after simplifying leads to $x=1-b$.

We may set $b$ to be any integer we wish, which means that $x=1-b$ can also equal any integer we wish.

If we want $x=n$, then we set $a=2, b=1-n$ and $c=1+n$. Each of these three expressions will give integers if $n$ is an integer and they satisfy $a=b+c$.
The result of the final case means we know that under the conditions given in the question, any integer is a possible value for $x$.