Solve the following simultaneous equations.
\[\frac{1}{2(3x+4y)} + \frac{1}{5(2x-3y)} = \frac{1}{4}\]\[\frac{5}{3x+4y} - \frac{2}{2x-3y} = -\frac{3}{2}\]Answer: \((x, y) = (2, 1)\)
Put \(\frac{1}{3x+4y} = m\), \(\frac{1}{2x-3y} = n\).
First: \(\frac{m}{2} + \frac{n}{5} = \frac{1}{4}\). Multiply by 20: \(10m + 4n = 5\) ... (I)
Second: \(5m - 2n = -\frac{3}{2}\). Multiply by 2: \(10m - 4n = -3\) ... (II)
Adding: \(20m = 2\), \(m = \frac{1}{10}\). From (I): \(4n = 5 - 1 = 4\), \(n = 1\).
So \(3x + 4y = 10\) ... (III) and \(2x - 3y = 1\) ... (IV).
(III)×3 + (IV)×4: \(9x + 8x = 30 + 4\), so \(17x = 34\) and \(x = 2\). From (IV): \(3y = 4 - 1 = 3\), \(y = 1\).
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