why componendo and dividendo is failing in this question If a positive fraction numerator and denominator is increased by 2 the fraction increases by $\frac{1}{24}$ find the difference between the numerator and denominator, given the sum of Numerator+Denominator =11? 
GIven
$$\frac{N+2}{D+2}=\frac{N}{D}+\frac{1}{24}$$
If we solve it using c&d we get 
$$\frac{N+2}{D+2}=\frac{24N+1}{D24}$$
$$\frac{N+2+D+2}{N+2-D-2}=\frac{24N+1+D24}{24N+1-D24}$$
Which will solve up to 
$$\frac{15}{N-D}=\frac{265}{24N-D24-1}$$
$$N-D= 3/19$$
But the solution comes out to be $1$ if we solve without using C&D . 
any suggestions will be helpful 
 A: You have made a mistake (as it's consistent later on, I don't think it's a typo), as your second line should be
$$\frac{N+2}{D+2}=\frac{24N+D}{D24} \tag{1}\label{eq1}$$
Cross-multiplying gives
$$24ND + 48D = 24ND + 48N + D^2 + 2D \; \Rightarrow \; 48\left(D - N\right) = D^2 + 2D \tag{2}\label{eq2}$$
Using $N + D = 11$, use $N = 11 - D$ to get $D - N = 2D - 11$ so \eqref{eq2} becomes
$$96D - 528 = D^2 + 2D \; \Rightarrow \; D^2 - 94D + 528 = 0 \tag{3}\label{eq3}$$
This is a quadratic equation you can solve for $D$ (but make sure that the fraction $\frac{N}{D}$ is positive as the question states). Then substitute the result(s) into the RHS of \eqref{eq2} and divide by $-48$ to get $N - D$, i.e., the difference between the numerator and denominator. I trust you can finish these final calculations.
Using your "c&d" technique (i.e., if $\frac{a}{b} = \frac{c}{d}$, then $\frac{a \; + \; b}{a \; - \; b} = \frac{c \; + \; d}{c \; - \; d}$) instead gives
$$\frac{N+2+D+2}{N+2-D-2}=\frac{24N+D+D24}{24N+D-D24}$$
$$\frac{15}{N-D}=\frac{264+D}{24\left(N-D\right) + D} \tag{4}\label{eq4}$$
Cross-multiplying now gives
$$360\left(N-D\right) + 15D = 264\left(N-D\right) + D\left(N - D\right) \tag{5}\label{eq5}$$
If you substitute $N - D = 11 - 2D$, you get
$$360\left(11 - 2D\right) + 15D = 264\left(11 - 2D\right) + D\left(11 - 2D\right) \tag{6}\label{eq6}$$
Moving everything to the LHS and dividing by $2$ gives the RHS of \eqref{eq3}. As such, the "c&d" technique does work but, in this case at least, it seems to make the solution a bit longer & more complicated.
