Question

divide the rational expressions. write your answer in simplest form. $\frac{7x}{x + 3}div\frac{7}{x^{2}-9}$
$\frac{7x}{x + 3}cdot\frac{x^{2}-9}{7}=x(x - 3)$
$\frac{7x}{x + 3}cdot\frac{7}{x^{2}-9}=\frac{49x}{(x + 3)(x^{2}-9)}$
$\frac{x + 3}{7x}cdot\frac{7}{x^{2}-9}=\frac{1}{x(x - 3)}$
$\frac{x + 3}{7x}cdot\frac{x^{2}-9}{7}=\frac{(x + 3)(x^{2}-9)}{49x}$
steps:

1. multiply by reciprocal
2. factor all terms
3. reduce common terms
4. write in simplest factored form

Explanation:

Step1: Change division to multiplication

Dividing by a fraction is the same as multiplying by its reciprocal. So, $\frac{7x}{x + 3}\div\frac{7}{x^{2}-9}=\frac{7x}{x + 3}\times\frac{x^{2}-9}{7}$.

Step2: Factor the difference - of - squares

We know that $x^{2}-9=(x + 3)(x - 3)$ (using the formula $a^{2}-b^{2}=(a + b)(a - b)$ where $a=x$ and $b = 3$). So the expression becomes $\frac{7x}{x + 3}\times\frac{(x + 3)(x - 3)}{7}$.

Step3: Cancel out common terms

Cancel out the common factors. We can cancel out the 7 in the numerator and denominator, and also the $(x + 3)$ in the numerator and denominator. After canceling, we get $x(x - 3)=x^{2}-3x$.

Answer:

$x^{2}-3x$