Given that \(a\) and \(b\) are non-zero constants, and that the equations
\[\begin{align*} a^2x+aby&=a+4b\\ abx+b^2y&=a+b \end{align*}\]have an infinite number of solutions for \(x\) and \(y\), express \(a\) in terms of \(b\).
For given values of \(a\) and \(b\), \(a^2x+aby=a+4b\) and \(abx+b^2y=a+b\) represent straight lines in the \(x-y\) plane.
Approach 1
Multiply the first equation by \(b\), and the second equation by \(a\). Now subtract — we find that
\[ ba^2x + ab^2y - a^2bx -ab^2y = ab+4b^2 - a^2 - ba. \]
So all the \(x\) and \(y\) terms cancel, giving us that (if the equations are to be consistent) \(4b^2-a^2=0\).
So the set of equations has either no solutions (the lines are parallel, and the equations are inconsistent) or an infinite number of solutions (the lines are on top of each other, and the equations are consistent).
So for an infinite number of solutions we need \(0=4b^2 - a^2\) to be true, which leads us to \[a=\pm2b.\]
Approach 2
We can re-write each equation in the form \(y = mx + c\) ( dividing by \(a\) and \(b\) is not a problem, since they are non-zero). We have
\[y = -\frac{ax}{b}+\frac{4}{a}+\frac{1}{b}, \quad y=-\frac{ax}{b}+\frac{a}{b^2}+\frac{1}{b}.\]
We require these straight lines to coincide (be on top of each other). Their gradients are equal, so this can only happen if their \(y\)-intercepts are the same.
This means we need \[\frac{4}{a}+\frac{1}{b}=\frac{a}{b^2}+\frac{1}{b},\]
which yields \[ a=\pm2b. \]
