Solve the simultaneous equations \[\frac{x}{3} - \frac{y}{10} = \frac{5}{6}, \quad x(y-2) = 2y + 3.\]

Let’s start by multiplying the first equation by \(30\), which is the lowest common multiple of the denominators; in this way, we will have no more fractions. The equation thus becomes \[10x-3y=25.\]

We could rearrange this to find \(x = \dfrac{3y+25}{10}\) and substitute this into the second equation, but this reintroduces the fractions and so seems somewhat ugly.

Alternatively, we could rearrange the second equation to get \(x=\dfrac{2y+3}{y-2}\) and substitute this into the first equation, which we can then rearrange and solve to find \(y\). This, though, also has fractions in it, and this time they are algebraic.

A somewhat nicer approach is to multiply the first equation by \(y-2\), to get \[10x(y-2)-3y(y-2)=25(y-2).\] Since we know \(x(y-2)=2y+3\), this now becomes \(10(2y+3)-3y(y-2)=25(y-2)\), which we can rearrange to get \[3y^2-y-80=0.\]

This equation factorises into \((y+5)(3y-16)=0\), thus \(y = -5\) or \(y=\dfrac{16}{3}\). (Alternatively, we could use the quadratic formula to obtain these solutions.)

The corresponding values for \(x\) (using either of the original equations or their rearranged variants) are \(x=1\) and \(x=\dfrac{41}{10}\). We check that our answers are correct by substituting these \((x,y)\) pairs into the other equation.

Therefore there are two solutions: \((x,y) = (1, -5)\) or \((x,y)= \left(\dfrac{41}{10}, \dfrac{16}{3}\right)\).

Looking at this geometrically, \(10x-3y=25\) is a straight line which intersects the axes at \((\frac52,0)\) and \((0,-\frac{25}{3})\), while we can rearrange \(x(y-2) = 2y + 3\) to get \[y=\frac{2x+3}{x-2}=\frac{2(x-2)+7}{x-2}=2+\frac{7}{x-2}\] so this is a stretched and translated \(y=\frac{1}{x}\) graph (which is a rectangular hyperbola).

The straight line and other graph intersect twice, as we have discovered:

the line crosses the hyperbola in the solution points