Question

The denominator of fraction is 3 more than the numerator. When it is added to its reciprocal, the result is 31/40. What is the fraction?

Answer 1

51.6

1. fraction

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Answer 2

${\huge{\red{Ãñs}wêr}} \\ </p><p>[tex]let \: the \: denominator \: fraction \: is = \frac{x}{y} \\ then \: according \: to \: given \: question . \\ y = x + 3 \: equation(1) \\ again = \frac{x}{y} + \frac{y}{x} = \frac{31}{40} \\ \frac{ {x}^{2} + {y}^{2} }{xy} = \frac{31}{40}$

${40x}^{2} + {40y}^{2} = 31xy \: equation(1) \\ put \: the \: value \: of \: y \: in \: equation \: (2) . \\ {40x}^{2} + 40(x + 3)^{2} = 31x(x + 3) \\ {40x}^{2} + 40( {x}^{2} + 9 + 6x) = {31x}^{2} + 93x \\ 40x^{2} + 40 {x}^{2} + 360 + 240x = 31 {x}^{2} + 93x \\ 80 {x}^{2} - 31 {x}^{2} + 240x - 93x + 360 = 0 \\ 49 {x }^{2} + 147x + 360 = 0 \\ 7(7 {x}^{2} + 21x + 51 ) = 0 \\ (7 {x}^{2} + 21x + 51) = 0 \\ (x + 17)(x - 3) by \: factorise \\ (x + 17) = 0 \: ( x - 3) = 0 \\ x = - 17 \: x = 3 \\ put \: x = 3 \: in \: equation \: (1) \\ y = 3 + 3 = 6 \\ fraction = \frac{x}{y} = \frac{6}{5} </p><p>$