Question

which expression is equivalent to (\frac{c^2 - 4}{c + 3} div \frac{c + 2}{3(c^2 - 9)})?
options (shown in boxes):

1. (\frac{c + 3}{c^2 - 4} div \frac{c + 2}{3(c^2 - 9)})
2. (\frac{c^2 - 4}{c + 3} cdot \frac{3(c^2 - 9)}{c + 2})
3. (\frac{c^2 - 4}{c + 3} div \frac{3(c^2 - 9)}{c + 2})
4. (\frac{c + 3}{c^2 - 4} cdot \frac{c + 2}{3(c^2 - 9)})

Explanation:

Step1: Recall division of fractions rule

To divide two fractions, we multiply the first fraction by the reciprocal of the second fraction. The rule is \(\frac{a}{b} \div \frac{c}{d}=\frac{a}{b}\times\frac{d}{c}\) (where \(b
eq0\), \(c
eq0\), \(d
eq0\)).

Given the expression \(\frac{c^{2}-4}{c + 3}\div\frac{c + 2}{3(c^{2}-9)}\), let \(a = c^{2}-4\), \(b=c + 3\), \(c=c + 2\), \(d = 3(c^{2}-9)\).

Step2: Apply the division rule

Using the rule \(\frac{a}{b}\div\frac{c}{d}=\frac{a}{b}\times\frac{d}{c}\), we get:

\(\frac{c^{2}-4}{c + 3}\div\frac{c + 2}{3(c^{2}-9)}=\frac{c^{2}-4}{c + 3}\times\frac{3(c^{2}-9)}{c + 2}\)

Now, let's analyze the options:

• First option: \(\frac{c + 3}{c^{2}-4}\div\frac{c + 2}{3(c^{2}-9)}\) is not equivalent as the first fraction is reciprocal of what we have.
• Second option: \(\frac{c^{2}-4}{c + 3}\times\frac{3(c^{2}-9)}{c + 2}\) matches our derived expression.
• Third option: \(\frac{c^{2}-4}{c + 3}\div\frac{3(c^{2}-9)}{c + 2}\) is dividing by the reciprocal of the second fraction in the original, so not equivalent.
• Fourth option: \(\frac{c + 3}{c^{2}-4}\times\frac{c + 2}{3(c^{2}-9)}\) has reciprocal of the first fraction and wrong multiplication, so not equivalent.

Answer:

The second option (the one with \(\frac{c^{2}-4}{c + 3}\cdot\frac{3(c^{2}-9)}{c + 2}\))