[**Hint:** you may wish to start by multiplying the second equation by \(x+y\).]

Let’s do what the hint suggests, and multiply the second equation by \(x+y\), which yields \[(ax+by)(x+y) = \frac{1}{3}(x+y),\] that is, \[ax^2+by^2+(a+b)xy = \frac{1}{3}(x+y).\]

Substituting in the information from the first and third equations yields \[\frac{1}{5}+xy=\frac{1}{3}(x+y),\] or, rearranging, \[x+y-3xy-\frac{3}{5}=0.\]

Now let’s try the same trick with the third equation; we have \[(ax^2+by^2)(x+y) = \frac{1}{5}(x+y),\] that is, \[ax^3+by^3+xy(ax+by) = \frac{1}{5}(x+y).\]

Substituting in the information from the second and fourth equations, we find that \[\frac{1}{7}+\frac{xy}{3}=\frac{1}{5}(x+y),\] or, rearranging, \[x+y-\frac{5}{3}xy-\frac{5}{7}=0.\]

Now let \(xy = z\). This gives us

\[\begin{equation} x+y-3z-\frac{3}{5}=0, \label{eq:1} \end{equation}\] and \[\begin{equation} x+y-\frac{5}{3}z-\frac{5}{7}=0. \label{eq:2} \end{equation}\]so subtracting these gives us \[-\dfrac{4z}{3}+ \dfrac{4}{35}=0,\] which means that \(z = \dfrac{3}{35}\).

So \(y = \dfrac{3}{35x}\), and thus from (1), \(x+\dfrac{3}{35x} - 3\dfrac{3}{35}-\dfrac{3}{5}=0.\) This is the quadratic equation \[35x^2-30x+3=0.\] Since \(x\) and \(y\) are interchangeable, we take the solution to be \[\begin{align*} x &= \frac{30+4\sqrt{30}}{70}=\frac{3}{7}+\frac{2\sqrt{30}}{35},\\ y &= \frac{30-4\sqrt{30}}{70}=\frac{3}{7}-\frac{2\sqrt{30}}{35}. \end{align*}\]We now wish to find the values of \(a\) and \(b\) for these values of \(x\) and \(y\). The second simultaneous equation with the above values of \(x\) and \(y\) looks like \[\frac{3a}{7}+\frac{2\sqrt{30}a}{35}+\frac{3b}{7}-\frac{2\sqrt{30}b}{35}=\frac{1}{3},\] and substituting in \(b=1-a\) from the first of the simultaneous equations yields \[\frac{4\sqrt{30}a}{35}+\frac{3}{7}-\frac{2\sqrt{30}}{35}=\frac{1}{3}.\] Rearranging, we find that \[a=\frac{1}{2}-\frac{5}{6\sqrt{30}}=\frac{1}{2}-\frac{\sqrt{30}}{36},\] and therefore \[b=1-a=1-\frac{1}{2}+\frac{\sqrt{30}}{36}=\frac{1}{2}+\frac{\sqrt{30}}{36}.\] Notice that if we’d chosen \(x\) and \(y\) to be the other way around then \(a\) and \(b\) would also have been swapped.

How do we know this solution works without substituting our values in to all 4 equations to check?

How would our answer change if we were given the additional constraint that \(ax^4+by^4 = \dfrac{1}{9}\)?

This GeoGebra file helps us to see what is happening.

We are plotting here the three curves \(ax+(1-a)y = \dfrac{1}{3}, ax^2+(1-a)y^2 = \dfrac{1}{5}\) and \(ax^3+(1-a)y^3 = \dfrac{1}{7}\).

As we vary \(a\), can we get the three curves to all pass through the same point?