If denominator of fraction is 1 more than twice a numerator . If the sum of the fraction and it's reciprocal is 58/21. Find the fraction.
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Solution :-
☆ 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
☆ ☆ ☆ Hop its helpful ☆ ☆ ☆
☆ Regards :- ♡♡《 Nitish kr singh 》♡♡
Solution :-
☆ 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
☆ ☆ ☆ Hop its helpful ☆ ☆ ☆
☆ Regards :- ♡♡《 Nitish kr singh 》♡♡
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