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

Let the denominator be 𝑥. 

Therefore numerator = 𝑥 − 3 

After adding 2 to both the numerator and denominator, 

New numerator = 𝑥 − 3 + 2 = 𝑥 − 1 

New denominator = 𝑥 + 2

According to question, $\frac{x-1}{x+2}+\frac{x-3}{x}=\frac{29}{20}$

$\begin{array}{r}\Rightarrow \frac{x(x-1)+(x+2)(x-3)}{x(x+2)}=\frac{29}{20}\\ \Rightarrow 40x^2-40x-120=29x^2+58x\\ \Rightarrow 11x^2-98x-120=0\\ \Rightarrow 11x^2-110x+12x-120=0\\ \Rightarrow 11x^2(x-10)+12(x-10)=0\\ \Rightarrow (11x+12)(x-10)=0\\ \Rightarrow x=\frac{-12}{11}\text{ or }10\end{array}$

$\text{ If }x=\frac{-12}{11}$,

$\text{ the fraction would }be=\frac{\frac{-12}{11}-3}{\frac{-12}{11}}$

=  $\frac{35}{12}$(this does not satisfy the equation)

If x=10, the fraction would be = $\frac{10-3}{7}$

Hence, the original fraction is $\frac{10}{7}$