You do it the same way we do with numbers: find a common denominator and then just add the numerators.
For example, if you needed to add
$$\frac{1}{2}+\frac{1}{3}$$
you need to find a "common denominator" (a number which is a multiple of both $2$ and $3$), convert the two fractions to the common denominator, and then add them. The simplest common denominator is just the product of the two denominators, we one way to do this is:
$$\begin{align*}
\frac{1}{2}+\frac{1}{3} &= \frac{1}{2}\cdot\frac{3}{3} + \frac{1}{3}\cdot\frac{2}{2}\\
&= \frac{3}{6} + \frac{2}{6}\\
&= \frac{3+2}{6}\\
&= \frac{5}{6}.\end{align*}$$
So do the same thing here:
$$\begin{align*}
\frac{4x+4h}{x+h+1} - \frac{4x}{x+1} &= \frac{4x+4h}{x+h+1}\cdot\frac{x+1}{x+1} + \frac{4x}{x+1}\cdot\frac{x+h+1}{x+h+1} \\
&= \frac{(4x+4h)(x+1)}{(x+h+1)(x+1)} - \frac{4x(x+h+1)}{(x+h+1)(x+1)}.\end{align*}$$
Now you can just add them, since they have the same denominator, by keeping the denominator and adding numerators. Do the algebra in the numerator, and simplify as needed.
Note: If you are more comfortable with the algebra, there are things you can do to simplify your life; when adding fractions, sometimes the product of the denominators is not the best common denominator: if you can spot a simpler one, use it. For instance, if we had $\frac{1}{6}+\frac{1}{10}$, then we could use $30$ as the common denominator, instead of $6\times 10 = 60$, which leads to smaller numbers, easier arithmetic; likewise, if we had
$$\frac{1}{x^2-1} + \frac{2}{x^2+2x+1}$$
then you could use $(x^2-1)(x^2+2x+1)$ as a common denominator, but if you could spot that $x^2-1 = (x+1)(x-1)$ and $x^2+2x+1 = (x+1)^2$, then you might realize that you can use $(x+1)^2(x-1)$ instead, which is smaller (degree 3 instead of degree 4).
In your situation, on possible simplification is to note that both fractions are multiples of $4$. So you can factor out $4$ and deal with simpler fractions if you want. But if you are unsure about these kinds of simplifying steps, then don't do them. They are not strictly necessary, though they are useful when dealing with more complicated expressions.
Best Answer
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.