Question

The numerator of a positive fraction is one more than twice the denominator. If the sum of the fraction and its reciprocal is $2.9$; find the fraction.

• A. $\frac{2}{5}$
• B. $\frac{25}{12}$
• C. $\frac{5}{2}$
• D. $\frac{9}{4}$

Answer 1

Answer: (C)

$\frac{5}{2}$

$letthefractionbe\left(\frac{x}{y}\right)x=2y+1and\left(\frac{x}{y}\right)+\left(\frac{y}{x}\right)=2.9\left(\frac{x^2+y^2}{xy}\right)=2.9x^2+y^2=2.9xy(2y+1{)}^2+y^2=2.9(2y+1)y4y^2+4y+1+y^2=2.9(2y^2+y)4y^2+4y+1+y^2=(\frac{29}{10}\left)\right(2y^2+y)10\times (5y^2+4y+1)=58y^2+29y50y^2+40y+10=58y^2+29y8y^2-11y-10=08y^2-16y+5y-10=08y(y-2)+5(y-2)=0(y-2\left)\right(8y+5)=0y=2andy=(\frac{-5}{8})\therefore 2y+1=2\times 2+1=5\therefore number(\frac{5}{2})$