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Question
simplify.\\(\dfrac{1 - \dfrac{4}{x + 8}}{x + \dfrac{16}{x + 8}}\\)
Step1: Simplify numerator
Rewrite 1 as $\frac{x+8}{x+8}$, then subtract.
$$1 - \frac{4}{x+8} = \frac{x+8}{x+8} - \frac{4}{x+8} = \frac{x+8-4}{x+8} = \frac{x+4}{x+8}$$
Step2: Simplify denominator
Rewrite $x$ as $\frac{x(x+8)}{x+8}$, then add.
$$x + \frac{16}{x+8} = \frac{x(x+8)}{x+8} + \frac{16}{x+8} = \frac{x^2+8x+16}{x+8} = \frac{(x+4)^2}{x+8}$$
Step3: Divide numerator by denominator
Multiply by reciprocal of denominator.
$$\frac{\frac{x+4}{x+8}}{\frac{(x+4)^2}{x+8}} = \frac{x+4}{x+8} \times \frac{x+8}{(x+4)^2}$$
Step4: Cancel common factors
Eliminate $(x+8)$ and one $(x+4)$.
$$\frac{x+4}{x+8} \times \frac{x+8}{(x+4)^2} = \frac{1}{x+4}$$
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$\frac{1}{x+4}$ (where $x
eq -8, -4$)