Solve the following simultaneous equations: 2/x + 2/(3y) = 1/6 ; 3/x + 2/y = 0

Answer: \((x, y) = (6, -4)\)

Put \(\frac{1}{x} = m\), \(\frac{1}{y} = n\): \(2m + \frac{2}{3}n = \frac{1}{6}\). Multiply by 6: \(12m + 4n = 1\) ... (I). Second: \(3m + 2n = 0\) ... (II).

Multiply (II) by 2: \(6m + 4n = 0\). Subtract from (I): \(6m = 1\), \(m = \frac{1}{6}\).

From (II): \(2n = -\frac{1}{2}\), \(n = -\frac{1}{4}\).

So \(x = \frac{1}{m} = 6\) and \(y = \frac{1}{n} = -4\).

Check: \(\frac{2}{6} + \frac{2}{-12} = \frac{1}{3} - \frac{1}{6} = \frac{1}{6}\), which is true.