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(A) Let the numerator be $x$. Then the denominator is $2x - 1$. The ratio is $\frac{x}{2x - 1}$.According to the problem,the sum of the ratio and its

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$3/5$

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(A) Let the numerator be $x$. Then the denominator is $2x - 1$. The ratio is $\frac{x}{2x - 1}$.According to the problem,the sum of the ratio and its reciprocal is $2 \frac{4}{15} = \frac{34}{15}$.So,$\frac{x}{2x - 1} + \frac{2x - 1}{x} = \frac{34}{15}$.Let $y = \frac{x}{2x - 1}$. Then $y + \frac{1}{y} = \frac{34}{15}$.$15y^2 - 34y + 15 = 0$.$15y^2 - 25y - 9y + 15 = 0$.$5y(3y - 5) - 3(3y - 5) = 0$.$(5y - 3)(3y - 5) = 0$.So,$y = 3/5$ or $y = 5/3$.If $y = 3/5$,then $\frac{x}{2x - 1} = \frac{3}{5} \implies 5x = 6x - 3 \implies x = 3$. The ratio is $\frac{3}{2(3) - 1} = \frac{3}{5}$.If $y = 5/3$,then $\frac{x}{2x - 1} = \frac{5}{3} \implies 3x = 10x - 5 \implies 7x = 5 \implies x = 5/7$. The ratio is $\frac{5/7}{2(5/7) - 1} = \frac{5/7}{3/7} = \frac{5}{3}$.

The denominator of a non-zero ratio is one less than twice the numerator. If the sum of the ratio and its reciprocal is $2 \frac{4}{15},$ find the ratio.

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