Question

the denominator of a fraction is one more than twice its numerator if the sum of the fraction and its reciprocal is 2 16/21 find the fraction

Answer 1

let the numerator be 'x' then denominator is'2x+1'

given="x / 2x+1" + "2x+1/x"=2 16/21

then="x / 2x+1" + "2x+1/x"=58/21

=x(x)+2x+1(2x+1) / 2x+1(x=58/21

=x2+(2x+1)2  / 2x2+x=58/21

=x2+4x2+4x+1 / 2x2+x=58/21

=5x2 +4x+1 / 2x2+x=58/21

=5x2 +4x+1(21)=58(2x2+x)

=105x2 +84x+21=116x2 +58x

=116x2 +58x-105x2 -84x-21=0

=11x2 - 26x-21=0

by solving using x= -(b)+rootb2 -4ac /2a

we get x=3 so the fraction is 3/7

check

3/7 +7/3 =49 +9 /21=58 /21

hence verified

Answer 2

Question :-

The denominator of a fraction is 1 more than twice its numerator. If the sum of fraction and its reciprocal is 2$\frac{16}{21}$, find the fraction.

$\large\underline{\sf{Solution-}}$

Let assume that

Numerator of a fraction be x.

So,

Denominator of a fraction be 1 + 2x

Thus,

$\sf \: Fraction = \dfrac{x}{1 + 2x}$

According to statement, sum of fraction and its reciprocal is 2$\frac{16}{21}$.

So,

$\sf \: \dfrac{x}{1 + 2x} + \dfrac{2x + 1}{x} = 2 \dfrac{16}{21}$

$\sf \: \dfrac{ {x}^{2} + {(2x + 1)}^{2} }{(1 + 2x)x} = \dfrac{58}{21}$

$\sf \: \dfrac{ {x}^{2} + {4x}^{2} + 4x + 1}{x + 2 {x}^{2} } = \dfrac{58}{21}$

$\sf \: \dfrac{{5x}^{2} + 4x + 1}{x + 2 {x}^{2} } = \dfrac{58}{21}$

$\sf \: 58( {2x}^{2} + x) = 21( {5x}^{2} + 4x + 1)$

$\sf \: {116x}^{2} + 58x ={105x}^{2} + 84x + 21$

$\sf \: {116x}^{2} + 58x - {105x}^{2} - 84x - 21 = 0$

$\sf \: {11x}^{2} - 26x - 21 = 0$

$\sf \: {11x}^{2} - 33x + 7x - 21 = 0$

$\sf \: 11x(x - 3) + 7(x - 3) = 0$

$\sf \: (x - 3)(11x + 7) = 0$

$\sf \: \implies \: x = 3 \: \: \: or \: \: \: x = - \dfrac{7}{11} \: \{rejected \}$

Thus,

$\sf \: \sf \: \implies \: Fraction = \dfrac{x}{1 + 2x} = \frac{3}{7}$

Hence,

$\implies\sf \: \boxed{\sf \: Fraction = \dfrac{3}{7} \: }$