Answer
The simplified form of a complex rational expression $\frac{\frac{3}{x-7}}{\frac{4x+3}{x+1}}$ is $\frac{3\left( x+1 \right)}{\left( x-7 \right)\left( 4x+3 \right)}$.
Work Step by Step
$\frac{\frac{3}{x-7}}{\frac{4x+3}{x+1}}$
Divide the numerator by the denominator,
$\frac{\frac{3}{x-7}}{\frac{4x+3}{x+1}}=\frac{3}{x-7}\div \frac{4x+3}{x+1}$
$\begin{align}
& \frac{A}{B}\div \frac{C}{D}=\frac{A}{B}\cdot \frac{D}{C} \\
& =\frac{AD}{BC}
\end{align}$
where $\frac{A}{B},\left( B\ne 0 \right)$, $\frac{C}{D},\left( D\ne 0 \right)$ are rational expressions with $\frac{C}{D}\ne 0$
The reciprocal of $\frac{4x+3}{x+1}$ is $\frac{x+1}{4x+3}$
So, multiply the reciprocal of the divisor,
$\frac{\frac{3}{x-7}}{\frac{4x+3}{x+1}}=\frac{3}{x-7}\cdot \frac{x+1}{4x+3}$
Simplify the terms,
$\begin{align}
& \frac{\frac{3}{x-7}}{\frac{4x+3}{x+1}}=\frac{3\cdot \left( x+1 \right)}{\left( x-7 \right)\cdot \left( 4x+3 \right)} \\
& =\frac{3\left( x+1 \right)}{\left( x-7 \right)\left( 4x+3 \right)}
\end{align}$
