Solve the simultaneous equations \[x+ \frac{1}{y} =2, \qquad 2y + \frac{1}{x}=3.\]

We can obviously not have \(x=0\) or \(y=0\). Rearranging the first equation gives: \[x=2-\frac{1}{y}=\frac{2y-1}{y}\] so that \[\frac{1}{x}=\frac{y}{2y-1}.\]

We can then substitute into the second equation to get: \[2y+\frac{y}{2y-1}=3.\]

Multiplying by \(2y-1\) gives: \[2y(2y-1)+y=3(2y-1)\] so expanding the brackets and rearranging, we get \[4y^2-7y+3=0.\]

This can now be factorised, giving: \[(4y-3)(y-1)=0\] so \(y=1\) or \(y=\dfrac{3}{4}\).

Since \(x=2-\dfrac{1}{y}\), we can substitute these \(y\) values in to this to get the solutions:

\[x=1,\ y=1\qquad\text{or}\qquad x=\frac{2}{3},\ y=\frac{3}{4}.\]

Alternatively, we could have begun by rearranging the first equation to get \[\frac{1}{y}=2-x\] so that \[y=\frac{1}{2-x}\] and then substituted this into the second equation.

We could also have rearranged the second equation and then substituted this into the first one.