Question

The numerator of a fraction is 3 less than the denominator. If 2 is added to both the numerator and the denominator, then the sum of the new fraction and the original fraction is \[\frac{29}{20}\]. Find the original fraction.

Answer 1

Let the denominator of the original fraction be x then the numerator be x − 3.

Then according to question,

\[\frac{x - 3}{x} + \frac{x - 3 + 2}{x + 2} = \frac{29}{20}\]

\[ \Rightarrow \frac{x - 3}{x} + \frac{x - 1}{x + 2} = \frac{29}{20}\]

\[ \Rightarrow \frac{\left( x - 3 \right)\left( x + 2 \right) + \left( x - 1 \right)x}{x\left( x + 2 \right)} = \frac{29}{20}\]

\[ \Rightarrow \frac{x^2 - 3x + 2x - 6 + x^2 - x}{x^2 + 2x} = \frac{29}{20}\]

\[ \Rightarrow 20(2 x^2 - 2x - 6) = 29( x^2 + 2x)\]

\[ \Rightarrow 40 x^2 - 40x - 120 - 29 x^2 - 58x = 0\]

\[ \Rightarrow 11 x^2 - 98x - 120 = 0\]

\[ \Rightarrow 11 x^2 - 110x + 12x - 120 = 0\]

\[ \Rightarrow 11x(x - 10) + 12(x - 10) = 0\]

\[ \Rightarrow (11x + 12)(x - 10) = 0\]

\[ \Rightarrow 11x + 12 = 0 \text { or } x - 10 = 0\]

\[ \Rightarrow x = - \frac{12}{11} \text { or }x = 10\]

Since, x being an integer,

Therefore, x = 10.

Then the numerator will be \[x - 3 = 10 - 3 = 7\].

Thus, the original fraction is \[\frac{7}{10}\].