# How do you find the quotient of (21b^3)/(4c^2)div7/(6c^2)?

Jan 29, 2017

See the entire solution process below:

#### Explanation:

We can rewrite this problem as:

$\frac{\frac{21 {b}^{3}}{4 {c}^{2}}}{\frac{7}{6 {c}^{2}}}$

We can now use this rule for dividing fractions to find the quotient:

$\frac{\frac{\textcolor{red}{a}}{\textcolor{b l u e}{b}}}{\frac{\textcolor{\mathmr{and} a n \ge}{c}}{\textcolor{p u r p \le}{d}}} = \frac{\textcolor{red}{a} \times \textcolor{p u r p \le}{d}}{\textcolor{b l u e}{b} \times \textcolor{\mathmr{and} a n \ge}{c}}$

$\frac{\frac{\textcolor{red}{21 {b}^{3}}}{\textcolor{b l u e}{4 {c}^{2}}}}{\frac{\textcolor{\mathmr{and} a n \ge}{7}}{\textcolor{p u r p \le}{6 {c}^{2}}}} = \frac{\textcolor{red}{21 {b}^{3}} \times \textcolor{p u r p \le}{6 {c}^{2}}}{\textcolor{b l u e}{4 {c}^{2}} \times \textcolor{\mathmr{and} a n \ge}{7}} = \frac{126 {b}^{3} {c}^{2}}{28 {c}^{2}} = \frac{126 {b}^{3} \textcolor{red}{\cancel{\textcolor{b l a c k}{{c}^{2}}}}}{28 \textcolor{red}{\cancel{\textcolor{b l a c k}{{c}^{2}}}}} = \frac{\left(14 \times 9\right) {b}^{3}}{14 \times 2} =$

$\frac{\left(\textcolor{red}{\cancel{\textcolor{b l a c k}{14}}} \times 9\right) {b}^{3}}{\textcolor{red}{\cancel{\textcolor{b l a c k}{14}}} \times 2} = \frac{9 {b}^{3}}{2}$ or $4.5 {b}^{3}$