the denominator of a fraction is one more than twice the numerator if the sum of the fraction and its reciprocal is 2 and 16/21. find the fraction
Answers
Answer:
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
Also,
the sum of the fraction and it's reciprocal is
Reciprocal of the original fraction
Their sum is 58/21
So,
the following balanced equation will be formed
Taking LCM as x(2x + 1),
we have,
Multiplying the terms,
we've
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
3/7
Answer:
Required fraction = 3/7
Step by step explanation :
Let the numerator of the required fraction be x .
Then , it's denominator = (2x+1)
Therefore ,
rre = x/ ( 2x+1)
And it's reciprocal = (2x+1) / x .
Therefore ,
x/ ( 2x+1) + (2x+1) / x = 58 / 21 .
=> 21× [ x^2 + ( 2x+1) ^2 ]= 58x ( 2x+1)
=> 21× [5x^2 +4x +1] = 116x^2+58 x
=> 11 x^2 - 26 x -21 = 0
=> 11 x ^2 - 33 x + 7 x - 21 = 0
=> 11x ( x - 3 ) 7 (x - 3) = 0
=>( x - 3) (11x+7) =0
=> x= 3 or x = - 7 / 11
Therefore ,
x = 3 ( since numerator cannot be a negative
fraction )
Therefore, required fraction = x/ ( 2x+ 1)= 3/7.
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