Question

Question:-



The denominator of a fraction is one more than twice the numerator. if the sum of the fraction and its reciprocal is 58/21 find the fraction.​

Answer 1

$\boxed{\mathsf{\green{A} \blue{n} \red{s} \orange{w} \pink{e} \purple{r} :-}}$

$\mathsf{ \dfrac{3}{7}}$

$\boxed{\mathsf{\green{E} \blue{x} \red{p} \orange{l} \pink{a} \purple{n} \green{a} \red{t}\orange{i}\pink{o} \purple{n} :-}}$

Given:-

• DN of a fraction is one more than, twice the numerator

• Sum of the fraction and reciprocal = 58/21

To Find:-

The fraction

Solution:-

let

✭The numerator of the fraction is x

✭The denominator of the fraction is x+1

NOW,

according to the question

$\mathsf{\dfrac{x}{x+1} +\dfrac{x+1}{x} = \dfrac{58}{21} }$

$\mathsf{\dfrac{x(x) + (x+1)(x+1) }{(x+1)(x)} = \dfrac{58}{21} }$

$\mathsf{\dfrac{x^2 + x^2 + 2x + 1 }{x^2 + x} = \dfrac{58}{21} }$

$\mathsf{\dfrac{2 x^2 + 2x + 1 }{x^2 + x} = \dfrac{58}{21} }$

$\mathsf{42 x^2 + 42x + 21 = 58 x^2 + 58x}$

$\mathsf{ 16 x^2 +16x -21 = 0}$

$\mathsf{ 16 x^2 +28x - 12x -21 = 0}$

$\mathsf{ 4x(4x +7) - 3 (4x+7)= 0}$

$\mathsf{ (4x+7) (4x-3) = 0}$

Therefore

x = $\dfrac{-7}{4}$

x = $\dfrac{3}{4}$

Considering, x = $\dfrac{3}{4}$,

The fraction is,

$\mathsf{\implies \dfrac{x}{x+1} }$

$\mathsf{\implies \dfrac{\dfrac{3}{4}}{\dfrac{3}{4} +1} }$

$\mathsf{\implies \dfrac{\dfrac{3}{4}}{\dfrac{3+4}{4} } }$

$\mathsf{\implies \dfrac{\dfrac{3}{4}}{\dfrac{7}{4}} }$

$\mathsf{\implies \dfrac{3}{7}}$