the denominator is one more than twice the numerator , if the sum of the fraction and its recoprocal is 2 16/21 . find the fraction

let the numerator be 'x' then denominator is'2x+1'

given="x / 2x+1" + "2x+1/x"=2 16/21

then="x / 2x+1" + "2x+1/x"=58/21

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

=x^{2}+(2x+1)^{2} / 2x^{2}+x=58/21

=x^{2}+4x^{2}+4x+1 / 2x^{2}+x=58/21

=5x^{2} +4x+1 / 2x^{2}+x=58/21

=5x^{2} +4x+1(21)=58(2x^{2}+x)

=105x^{2} +84x+21=116x^{2} +58x

=116x^{2} +58x-105x^{2} -84x-21=0

=11x^{2 }- 26x-21=0

by solving using x= -(b)+rootb^{2 }-4ac /2a

we get x=3 so the fraction is 3/7

__check__

3/7 +7/3 =49 +9 /21=58 /21

hence verified