Question 18

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

Solution:

Let “x/y” be the required fraction. Here x is known as numerator and y is known as denominator

Here the denominator is one more than twice the numerator

Then y = 2 x + 1

x/(2 x + 1)

We have to consider the given fraction in the question as x/(2 x + 1)

The sum of the fraction and its reciprocal is 58/21

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

[x² + (2 x + 1) 2]/[x (2 x + 1)] = 58/21

[x² + 4 x² + 2 (2 x) (1) + 1²]/[2 x² +  x] = 58/21

[5 x² + 4 x + 1]/[2 x² +  x] = 58/21

21 [5 x² + 4 x + 1] = 58[2 x² +  x]

105 x² + 84 x + 21 = 116 x² + 58 x

116 x² - 105 x² + 58 x – 84 x – 21 = 0

11 x² – 26 x – 21 = 0

11 x² + 7 x – 33 x – 21 = 0

x (11 x + 7) – 3 ( 11 x + 7) = 0

(x – 3) (11 x + 7) = 0

x – 3 = 0                           11 x + 7 = 0

x = 3                                     11 x = -7

x = -7/11

The negative value of x is not possible. So we can take 3 for x.

Here x represents the numerator of the fraction. Since the denominator is one more than twice the numerator

By applying x = 3 in y = 2x + 1

We get y = 2(3) + 1

y = 7

Therefore the required fraction is 3/7

Verification:

The sum of the fraction and its reciprocal is 58/21

(3/7) + (7/3) = 58/21

(9 + 49)/21 = 58/21

58/21 = 58/21