Question

the denominator of positive fraction is one more than twice the numerator .if the sum of the fraction and its reciprocal is 2.9 find the fraction

Answer 1

$\textbf{Answer}$

Suppose the numerator of required fraction is x
$\textbf{Since,}$
Denominator of a fraction is one more than twice the numerator,
=> Denominator = (2x + 1)
So the required fraction is (x)/(2x+1)

We know that $\textbf{Reciprocal}$ of any given fraction $\textbf{a/b is b/a}$.

$\textbf{According to the question,}$

(x)/(2x+1) + (2x+1)/x = 2.9

=> { x^2 + (2x+1)^2 } / (x)(2x+1) = 2.9

=> x^2 + 4x^2 + 4x + 1 = 2.9(x)(2x+1)

=> 5x^2 + 4x + 1 = 5.8x^2 + 2.9x

=> 5.8x^2 - 5x^2 + 2.9x - 4x - 1 = 0

=> .8x^2 - 1.1x - 1 = 0

$\textbf{Multiplying both sides by 10,}$

=> 8x^2 - 11x - 10 = 0

=> 8x^2 - 16x + 5x - 10 = 0

=> 8x(x-2) + 5(x-2) = 0

=> (x-2) (8x+5) = 0

=> x = 2 or x = -5/8

Value of x can not be negative as it is a positive fraction.

=> $\textbf{x = 2}$

=> $\textbf{Denominator,}$

2x + 1 = (2×2) + 1
=> 2x + 1 = 5

$\textbf{So the required fraction is 2/5}$

$\textbf{Hope My Answer Helped}$

$\textbf{Thanks}$

Answer 2

Let the numerator be x,
So, denominator will be ( 2x + 1 )



Fraction = x / ( 2x + 1 )



Given that sum of the fraction and Its reciprocal is 2.9




According to the question :


$\frac{x}{2x + 1} + \frac{2x + 1}{x} = 2.9 \\ \\ \\ => \frac{ {x}^{2} + {(2x + 1)}^{2} }{x(2x + 1)} = \frac{29}{10 } \\ \\ \\ => \frac{ {x}^{2} + 4 {x}^{2} + 1 + 4x}{2 {x}^{2} + x} = \frac{29}{10} \\ \\ \\ => 10(5 {x}^{2} + 4x + 1) = 29(2 {x}^{2} + x) \\ \\ \\ = > 50 {x}^{2} + 40x + 10 = 58 {x}^{2} + 29x \\ \\ \\ => 0 = 8 {x}^{2} - 11x - 10 \\ \\ \\ => 0 = 8 {x}^{2} - 16x + 5x - 10 \\ \\ \\ => 0 = 8x(x - 2) + 5(x - 2) \\ \\ \\ => 0 = (x - 2)(8x + 5)$


By Zero Product Rule, we get,


x = 2 or x = -5/8



Taking + value, x = 2





Original fraction = 2 / { 2(2) + 1 }

Original fraction = 2 / 5