Chapter 5.5, Problem 10WE

To determine

To simplify the given expression.

Answer to Problem 10WE

  $\frac{x^2{y}^2}{2}$

Explanation of Solution

Given:

Expression: $\frac{7x^2}{9y}÷\frac{4x}{15y^2}\cdot {\left(\frac{35}{6xy}\right)}^{-1}$

Concept used:

To divide two fractions, simply flip the second fraction into a reciprocal and change the division sign to a multiplication sign and then multiply the two fractions by multiplying the numerators by each other and denominators by each other. As illustrated below:

  $\frac{a}{b}÷\frac{c}{d}\to \frac{a}{b}\times \frac{d}{c}\to \frac{a\times d}{b\times c}$

Calculation:

Consider the given expression $\frac{7x^2}{9y}÷\frac{4x}{15y^2}\cdot {\left(\frac{35}{6xy}\right)}^{-1}$ ,

  $\frac{7x^2}{9y}÷\frac{4x}{15y^2}\cdot {\left(\frac{35}{6xy}\right)}^{-1}$

  $=\frac{7x^2}{9y}\times \frac{15y^2}{4x}\times \frac{6xy}{35}$  [Flipping second and third fraction]

  $=\frac{7x^2}{9\cancel{y}}\times \frac{15y^2}{4\cancel{x}}\times \frac{6\cancel{x}\cancel{y}}{35}$  [Reducing the fractions]

  $=\frac{7x^2}{9}\times \frac{15y^2}{4}\times \frac{6}{35}$

  $=\frac{x^2}{3}\times \frac{15y^2}{4}\times \frac{2}{5}$  [Reducing the fractions]

  $=\frac{x^2}{1}\times \frac{y^2}{2}\times \frac{1}{1}$  [Reducing the fractions]

  $\begin{array}{l}=\frac{x^2\times y^2\times 1}{1\times 2\times 1}\\ =\frac{x^2{y}^2}{2}\end{array}$

Conclusion:

Therefore, after simplification, the answer is $\frac{x^2{y}^2}{2}$ .