Question

The denominator of a fraction is one more than twice the numerator. If the sum of the fraction and it's reciprocal is
$2 \frac{16}{21}$
find the fraction.

Answer 1

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!

Answer 2

Hi,

Please see the attached file!


Thanks