What is the difference between numerator and denominator? The sum of the numerator and denominator of a positive fraction is 11. If 2 is added to
both numerator and denominator, the fraction is increased by 1/24. What is the difference
between the numerator and denominator of the fraction?
Do we have any shortcuts for questions like these?
 A: The quickest solution that I’ve found is to write the fraction as $\frac{a}b$ and note that
$$\frac1{24}=\frac{a+2}{b+2}-\frac{a}b=\frac{2(b-a)}{b(b+2)}\;,$$
so
$$\frac{b-a}{b(b+2)}=\frac1{48}\;.$$
$6\cdot8=48$, so try $b=6$; then $a=5$, which works. 
Replacing $a$ by $11-b$ to get
$$\frac{13-b}{b+2}=\frac{11-b}b+\frac1{24}=\frac{264-23b}{24b}$$
and solving the resulting quadratic $24b(13-b)=(b+2)(264-23b)$ for $b$ confirms that this is the only solution. However, it’s hardly a general technique, unlike the solution of the quadratic.
A: If the numerator and denominator are supposed to be positive integers that sum to $11$, there are only $10$ possibilities, and it's pretty easy to check them one by one.
A: $\newcommand{\angles}[1]{\left\langle #1 \right\rangle}%
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With the fraction $\ds{p \over q}$ we get
$\ds{47q - 48 p = q\pars{q + 1}}$. Since $q\pars{q + 1}$ and $48p$ are even, $q$ must be even. We only have to check $q = 2, 4, 6, 8 , 10$ such that
$$
p = {q\pars{46 - q} \over 48} = {s\pars{23 - s} \over 12}
\quad
\mbox{is an integer}\,,
\quad
q= 2s\,,\quad s = 1,2,3,4,5
$$ 
That yields the allowed $s = 3\quad\imp\quad q = 6,\quad p = 5\imp\quad$
$\ds{\large{p \over q} = {5 \over 6}}$ 
A: $\dfrac{x}{y}=\dfrac{x+a}{y+a}+b=\dfrac{x+a+by+ba}{y+a}$
$xa=ya+by^2+bay$
$x=y+by\dfrac{y+a}{a}$
$x+y=n$
$n=y(b\dfrac{y+a}{a}+2)=\dfrac{b}ay^2+(b
+2)y$,
so
$y=\dfrac{-(b+2)\pm \sqrt{(b+2)^2+4bn/a}}{2b/a}$
Yours is the particular case $a=2,b=-1/24,n=11$. This gives two solutions, but one can be cast out as it would require one of $x,y$ to be negative, which would make the fraction negative as well.
