Question

A positive integer is 1 less than another. If the sum of the reciprocal of the smaller and twice the reciprocal of the larger is 19/42, then find the two integers

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Gerhard

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Given that a positive integer is 1 less than another, we can assume the smaller integer to be x-1 and the larger integer to be x.

Let's consider the sum of the reciprocal of the smaller and twice the reciprocal of the larger:

\frac{1}{(x-1)} + 2\cdot\frac{1}{x} = \frac{19}{42}

Multiplying through by42x(x-1) to clear the fractions:

42x + 84(x-1) = 19x(x-1)

Expanding and simplifying:

42x + 84x - 84 = 19x^2 - 19x

126x - 84 = 19x^2 - 19x

Rearranging terms:

19x^2 - 145x + 84 = 0

Now, we need to factor the quadratic equation or use the quadratic formula to solve for x.

After factoring or solving with the quadratic formula, we find that the two integers are 6 and 7

\boxed{x=7} and \boxed{x-1=6}

Let's consider the sum of the reciprocal of the smaller and twice the reciprocal of the larger:

Multiplying through by

Expanding and simplifying:

Rearranging terms:

Now, we need to factor the quadratic equation or use the quadratic formula to solve for x.

After factoring or solving with the quadratic formula, we find that the two integers are 6 and 7

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