The denominator of a fraction is one more than twice the numerator. If the sum of the fraction and it's reciprocal is
[tex]2 \frac{16}{21} [/tex]
find the fraction.
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The denominator of a fraction is one more than twice the numerator. If the sum of the fraction and it's reciprocal is
[tex]2 \frac{16}{21} [/tex]
find the fraction.
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Let the numerator be x.Given that denominator of a fraction is one more than twice the numerator = 2x + 1.
Therefore, the original fraction = x/2x + 1. ----- (1)
Given that sum of the fraction and its reciprocal = 58/21
Now,
= > (x/2x + 1) + (2x + 1/x) = 58/21
= > x^2 + (2x + 1)^2/x * (2x + 1) = 58/21
= > x^2 + 4x^2 + 1 + 4x/(2x^2 + x) = 58/21
= > 21(x^2 + 4x^2 + 1 + 4x) = 58(2x^2 + x))
= > 21(5x^2 + 4x^2 + 1) = 58(2x^2 + x)
= > 105x^2 + 84x^2 + 21 = 116x^2 + 58x
= > 11x^2 - 26x - 21 = 0
= > 11x^2 - 33x + 7x - 21 = 0
= > 11x(x - 3) + 7(x - 3) = 0
= > (11x + 7)(x - 3) = 0
= > x = 3 (or) x = -7/11.
Therefore the value of x = 3.
Therefore the fraction is 3/2(3) + 1
= > 3/7.
Hope this helps!