\begin{align}
\dfrac{x-4}{x^2 - 4} \div \dfrac{x^2 - 3x - 4}{x^2 + 5x + 6} &= \dfrac{x-4}{x^2 - 4} \cdot \dfrac{x^2 + 5x + 6}{x^2 - 3x - 4} && \text{Fraction Law 2}
\\&= \dfrac{x-4}{(x-2)(x+2)} \cdot \dfrac{(x+2)(x+3)}{(x - 4)(x+1)} && A^2 - B^2 \text{ and new X method}
\\&= \dfrac{(x-4)(x+2)(x+3)}{(x-2)(x+2)(x-4)(x+1)} && \text{Fraction Law 1}
\\&= \dfrac{x+3}{(x-2)(x+1)} && \text{Fraction Law 5}
\end{align}
Adding (property 3 + 4)
Adding fractions required you to find the LCD.
LCD requires factors.
You can group terms together with parentheses to create a factor.
Perform the indicated operation and simplify:
$\dfrac{1}{6} + \dfrac{1}{8}$
$\dfrac{3}{x-1} + \dfrac{x}{x+2}$
Let's find the LCD. The denominator are terms, so we can group together terms into one factor.
\begin{align}
(x-1) \quad \leftarrow \text{Missing } (x+2) \text{ as a factor}
\\(x+2) \quad \leftarrow \text{Missing } (x-1) \text{ as a factor}
\end{align}
Introducing what we found to be missing:
\begin{align}
\dfrac{3}{x-1} + \dfrac{x}{x+2} &= \dfrac{(x+2)}{(x+2)}\dfrac{3}{x-1} + \dfrac{x}{x+2} \cdot \dfrac{(x-1)}{(x-1)} &&
\\&= \dfrac{3(x+2)}{(x-1)(x+2)} + \dfrac{x(x-1)}{(x+2)(x-1)} && \text{Fraction Law 1}
\\&= \dfrac{3(x+2) + x(x-1)}{(x-1)(x+2)} && \text{Fraction Law 3}
\\&= \dfrac{3x + 6 + x^2 - x}{(x-1)(x+2)} && \text{Distributive Law}
\\&= \dfrac{x^2 + 2x + 6}{(x-1)(x+2)}
\end{align}
$\dfrac{1}{x^2 - 1} - \dfrac{2}{(x + 1)^2}$
Let's find the LCD:
\begin{align}
x^2 - 1 &= (x-1)(x+1) \quad \leftarrow \text{Missing } (x+1) \text{ as a factor}
\\(x+1)^2 &= (x+1)(x+1) \quad \leftarrow \text{Missing } (x-1) \text{ as a factor}
\end{align}
Let's now introduce what we found:
\begin{align}
\dfrac{1}{x^2 - 1} - \dfrac{2}{(x + 1)^2} &= \dfrac{(x+1)}{(x+1)} \cdot \dfrac{1}{(x-1)(x+1)} - \dfrac{2}{(x+1)^2} \cdot \dfrac{(x-1)}{(x-1)} &&
\\&= \dfrac{(x+1)}{(x-1)(x+1)^2} - \dfrac{2(x-1)}{(x-1)(x+1)^2} && \text{Fraction Law 1}
\\&= \dfrac{x + 1 - 2(x-1)}{(x-1)(x+1)^2} && \text{Fraction Law 3, careful with the subtraction}
\\&= \dfrac{x + 1 - 2x + 2}{(x-1)(x+1)^2} && \text{Distributive Law}
\\&= \dfrac{-x + 3}{(x-1)(x+1)^2}
\end{align}
Compound Fractions
A compound fraction is a fraction nested within another fraction, for example \[\dfrac{\dfrac{x}{y} + 1}{1 - \dfrac{y}{x}}\] When you're asked to "simplify a compound fraction," this means getting rid of the nested fraction by removing the internal denominators.
Simplify the following:
$\dfrac{\dfrac{x}{y} + 1}{1 - \dfrac{y}{x}}$
$\dfrac{\dfrac{1}{a+h} - \dfrac{1}{a}}{h}$
The second problem will appear in calculus again! Take note of the $h$'s cancelling through fraction law #5.