Question

The denominator of a fraction is one more than twice it's numerator.if the sum of the fraction and it's reciprocal is 2 16/21,find the fraction.??

Answer 1

Hi there !!

Let the numerator be x
Given,
The denominator of a fraction is one more than twice it's numerator,
So, denominator = 2x + 1

The original fraction will be

$\frac{x}{2x + 1}$

Also,

the sum of the fraction and it's reciprocal is
$2 \frac{16}{21} = \frac{58}{21}$

Reciprocal of the original fraction

$= \frac{2x + 1}{x}$

Their sum is 58/21

So,
the following balanced equation will be formed


$\frac{x}{2x + 1} + \frac{2x + 1}{x} = \frac{58}{21}$

Taking LCM as x(2x + 1),
we have,

$\frac{x(x) + (2x + 1)(2x + 1)}{x(2x + 1)} = \frac{58}{21}$
Multiplying the terms,
we've

$\frac{x {}^{2} + {4x}^{2} + 4x + 1 }{2x {}^{2} + x } = \frac{58}{21}$

21(x² + 4x² + 4x + 1) = 58(2x² + x)

21(5x² + 4x + 1) = 58(2x² + x)

105x² + 84x + 21 = 116x² + 58x

0 = 116x² - 105x² + 58x - 84x - 21

0 = 11x² - 26x - 21

Splitting the middle term
and using the rule where ab = 11 × -21 = -231 = -33 × 7 and a + b = -33 + 7 = -26
we have,

0 = 11x² - 33x + 7x - 21

Taking out common factors,
we've,

0 = 11x(x - 3) + 7(x - 3)

0 = (11x + 7)(x - 3)

11x + 7 = 0 x - 3 = 0
11x = -7 x = 0 + 3 = 3
x = -7/11.

Since the denominator of a fraction can't be taken as -7/11 because it can't be negative

we'll take x = 3

Numerator = x = 3
Denominator = 2x + 1 = 3 × 2 + 1 = 6 + 1 = 7


Thus,
the fraction is
$\frac{3}{7}$

Answer 2

Answer: 3/7

Step-by-step explanation:

Let the numerator  be  x and  denominator  be  y

According  to  first  condition, denominator of the fraction is one more than twice it's numerator.it  means,

y = 2x + 1

y - x = x +1  ................i)

According to second  condition,

sum of the fraction and it's reciprocal is $2\frac{16}{21}$

it means,

$\frac{x}{y}+\frac{y}{x}=2\frac{16}{21}\\\;\\\frac{x}{y}+\frac{y}{x}=\frac{42+16}{21}\\\;\\\frac{x}{y}+\frac{y}{x}=\frac{58}{21}\\\;\\\frac{x^2+y^2}{xy}=\frac{58}[12}\\\;\\\frac{x^2+y^2-2xy+2xy}{xy}=\frac{58}{12}\\\;\\\frac{(y-x)^2+2xy}{xy}=\frac{58}{12}\\\;\\\frac{(x+1)^2+2xy}{xy}=\frac{58}{12}\;\;\;(using\;eq\;i)\\\;$

Putting y = 2x+1 ,

$\frac{(x+1)^2+2x(2x+1)}{x(2x+1)}=\frac{58}{12}\\\;\\\frac{(x+1)^2}{x(2x+1)}+\frac{2x(2x+1)}{x(2x+1)}=\frac{58}{12}\\\;\\\frac{x^2+1+2x}{2x^2+x}+2=\frac{58}{21}\\\;\\\frac{x^2+1+2x}{2x^2+x}=\frac{58}{21}-2\\\;\\\frac{x^2+1+2x}{2x^2+x}=\frac{58-42}{21}\\\;\\\frac{x^2+1+2x}{2x^2+x}=\frac{16}{21}\\\;\\21x^2+21+42x=32x^2+16x\\\;\\0=32x^2-21x^2+16x-42x-21\\\;\\0=11x^2-26x-21\\\;\\11x^2-26x-21=0\\\;\\11x^2-33x+7x-21=0\\\;\\11x(x-3)+7(x-3)=0\\\;\\(x-3)(11x+7)=0$

$x=3\;\;or\;\;x=-\frac{7}{11}(Negligible)\\\;\\when\;x=3\;\;from\;eq.\;i)\\\;\\y=2\times3+1\\\;\\y=6+1\\\;\\y=7\\\;\\Hence\\\;\\\text{fraction is}=\frac{3}{7}$