Discussion
If \(\frac{1}{2(2x+3y)} + \frac{12}{7(3x-2y)} = \frac{1}{2}\) and \(\frac{7}{2x+3y} + \frac{4}{3x-2y} = 2\) then values of \(x\) and \(y\) are respectively.
Answer: A
The values are \(x\) = 2, \(y\) = 1.
Put \(u = \frac{1}{2x+3y}\) and \(v = \frac{1}{3x-2y}\). The equations become
\[ \begin{aligned} \frac{u}{2} + \frac{12v}{7} &= \frac{1}{2} \\ 7u + 4v &= 2 \end{aligned} \]
Solving gives \(u = \frac{1}{7}\) and \(v = \frac{1}{4}\), so \(2x + 3y = 7\) and \(3x - 2y = 4\), whose solution is \(x = 2\), \(y = 1\).
Check directly: \(\frac{1}{14} + \frac{12}{28} = \frac{1}{2}\) and \(1 + 1 = 2\). The pair 1, 2 swaps the values, and the negative pairs make \(2x + 3y\) negative.
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