Question

simplify.
\\(\dfrac{7 + \dfrac{21}{x + 4}}{\dfrac{18}{x + 4} + 6}\\)

Explanation:

Step1: Combine terms in numerator

First, rewrite 7 as a fraction with denominator $x+4$ to combine with $\frac{21}{x+4}$.
$$7 + \frac{21}{x+4} = \frac{7(x+4)}{x+4} + \frac{21}{x+4} = \frac{7x + 28 + 21}{x+4} = \frac{7x + 49}{x+4}$$

Step2: Combine terms in denominator

Rewrite 6 as a fraction with denominator $x+4$ to combine with $\frac{18}{x+4}$.
$$\frac{18}{x+4} + 6 = \frac{18}{x+4} + \frac{6(x+4)}{x+4} = \frac{18 + 6x + 24}{x+4} = \frac{6x + 42}{x+4}$$

Step3: Rewrite as fraction division

The original expression becomes the numerator fraction divided by the denominator fraction.
$$\frac{\frac{7x + 49}{x+4}}{\frac{6x + 42}{x+4}} = \frac{7x + 49}{x+4} \times \frac{x+4}{6x + 42}$$

Step4: Cancel common factors

Cancel $(x+4)$ from numerator and denominator, then factor the remaining terms.
$$\frac{7(x + 7)}{x+4} \times \frac{x+4}{6(x + 7)} = \frac{7}{6}$$
(Note: $x
eq -4, -7$ to avoid division by zero)

Answer:

$\frac{7}{6}$