Question

score: 1/5 penalty: 1 off
question
fully simplify.
$\frac{\frac{9}{x^2}-1}{\frac{x+4}{3}+\frac{1}{x}}$
answer attempt 1 out of 2

Explanation:

Step1: Simplify numerator

$\frac{9}{x^2}-1=\frac{9 - x^2}{x^2}=\frac{(3-x)(3+x)}{x^2}$

Step2: Simplify denominator

$\frac{x+4}{3}+\frac{1}{x}=\frac{x(x+4)+3}{3x}=\frac{x^2+4x+3}{3x}=\frac{(x+3)(x+1)}{3x}$

Step3: Rewrite as multiplication

$\frac{(3-x)(3+x)}{x^2}\div\frac{(x+3)(x+1)}{3x}=\frac{(3-x)(3+x)}{x^2}\times\frac{3x}{(x+3)(x+1)}$

Step4: Cancel common factors

Cancel $(3+x)$ and $x$: $\frac{3(3 - x)}{x(x + 1)}$

Answer:

$\frac{3(3 - x)}{x(x + 3)}$