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    Topics || Problems

    The sum of the reciprocals of two numbers is 7. The larger reciprocal exceeds the smaller by \(\frac{7}{ 3}\). Find the numbers. Solution:

    Let \(x,y\) be the numbers.

    \(\frac{1}{x} + \frac{1}{y} = 7\), let \(\frac{1}{x}\) be the larger reciprocal.

    \(\frac{1}{x} = \frac{1}{y} + \frac{7}{3}\)


    \(\frac{1}{y} + \frac{7}{3} + \frac{1}{y} = 7\)

    \(\frac{2}{y} = \frac{14}{3}\)

    \(\frac{3}{7} = y\)


    \(\frac{1}{x} =\frac{7}{3} + \frac{7}{3}\)

    \(\frac{1}{x} = \frac{14}{3}\)

    \(\frac{3}{14} = x\)

    Thus the numbers are \(\frac{3}{7} and \frac{3}{14}\)