Question

The numerator of a fraction is 3 more than its denominator. If its reciprocal is subtracted from it, the difference is 33/28. Find the fraction.

Answer 1

 Let the fraction be x/y

Then according to the question ;

x = 3 + y

\(\frac{x}{y}-\frac{y}{x}=\frac{33}{28}\) (given)

Substitute x = 3 + y

\(\frac{3+y}{y}-\frac{y}{3+y}=\frac{33}{28}\)

\(\frac{(3+4)^2-y^2}{y(3+y)}=\frac{33}{28}\)

\(\frac{9+y^2+6y-y^2}{3y+y^2}=\frac{33}{28}\)

\(\frac{9+6y}{3y+y^ 2}=\frac{33}{28}\)

252 + 168y = 99y + 33y²

33y² - 69y - 252 = 0

11y² - 23y - 84 = 0

\(y=\frac{-b\pm \sqrt {b^ 2-4ac}}{2a}\)

\(=\frac{23 \pm \sqrt {23^ 2+4\times 11 \times 84}}{2\times 11}\)

\(=\frac{23 \pm \sqrt {529 + 3696}}{22}=\frac{23 \pm \sqrt {4225}}{22}\)

y \(=\frac{23 +65}{22}\) ⇒ \(\frac{88}{22} = 4\)

∴ x = 3 + y ⇒ 3 + 4 = 7

∴ The fraction is 7/4