Question

For the following exercises, perform the indicated operations.

$\frac{6q+3}{9q^2-9q}÷\frac{q^2+14q+33}{q^2+4q-5}.\frac{4q^2+12q}{12q+6}$

Short answer

The solution is$\frac{2\left(q+5\right)}{9\left(q+11\right)}$

Step-by-step solution

Step 1. Given Information

$$\begin{gathered} \frac{6q+3}{9q^2-9q}÷\frac{q^2+14q+33}{q^2+4q-5}.\frac{4q^2+12q}{12q+6} \\ =\frac{6q+3}{9q^2-9q}.\frac{q^2+4q-5}{q^2+14q+33}.\frac{4q^2+12q}{12q+6} \end{gathered}$$

Step 2. Factors the numerator and denominator

$$\begin{gathered} \frac{6q+3}{9q^2-9q}.\frac{q^2+4q-5}{q^2+14q+33}.\frac{4q^2+12q}{12q+6} \\ =\frac{3\left(2q+1\right)}{9q\left(q-1\right)}.\frac{\left(q+5\right)\left(q-1\right)}{\left(q+11\right)\left(q+3\right)}.\frac{4q\left(q+3\right)}{6\left(2q+1\right)} \end{gathered}$$

Step 3. Multiply the numerators and denominators

$\frac{3\times 4q\left(2q+1\right)\left(q+5\right)\left(q-1\right)\left(q+3\right)}{9q\times 6\left(2q+1\right)\left(q+11\right)\left(q+3\right)\left(q-1\right)}.$

Step 4. Simplify by dividing out common factors

$\frac{2\left(q+5\right)}{9\left(q+11\right)}$