Solve the following simultaneous equations: 7/(2x + 1) + 13/(y + 2) = 27 ; 13/(2x + 1) + 7/(y + 2) = 33

Answer: \((x, y) = \left(-\frac{1}{4}, -1\right)\)

Put \(\frac{1}{2x+1} = m\), \(\frac{1}{y+2} = n\): \(7m + 13n = 27\) ... (I), \(13m + 7n = 33\) ... (II).

Adding: \(20m + 20n = 60\), so \(m + n = 3\) ... (III).
Subtracting (I) from (II): \(6m - 6n = 6\), so \(m - n = 1\) ... (IV).

Hence \(m = 2\), \(n = 1\).

\(2x + 1 = \frac{1}{2}\), so \(2x = -\frac{1}{2}\) and \(x = -\frac{1}{4}\).
\(y + 2 = 1\), so \(y = -1\).