Solve the following simultaneous equations: 1/(3x + y) + 1/(3x - y) = 3/4 ; 1/(2(3x + y)) - 1/(2(3x - y)) = -1/8

Answer: \((x, y) = (1, 1)\)

Put \(\frac{1}{3x+y} = m\) and \(\frac{1}{3x-y} = n\).

First equation: \(m + n = \frac{3}{4}\) ... (I)

Second equation: \(\frac{m}{2} - \frac{n}{2} = -\frac{1}{8}\). Multiply by 2: \(m - n = -\frac{1}{4}\) ... (II)

Adding (I) and (II): \(2m = \frac{1}{2}\), so \(m = \frac{1}{4}\). Then \(n = \frac{3}{4} - \frac{1}{4} = \frac{1}{2}\).

So \(3x + y = 4\) ... (III) and \(3x - y = 2\) ... (IV).

Adding: \(6x = 6\), \(x = 1\). From (III): \(y = 4 - 3 = 1\).