Chapter 8.4, Problem 20PPSE

Solution

To find: The division of $\frac{7x}{4y^3}÷\frac{21x^3}{8y}$ in simplified form and state the restrictions on the variables.

The simplified form of the division is $\frac{2}{3x^2{y}^2}$ for $x\ne \left\{0\right\}\text{ and }y\ne \left\{0\right\}$ where the restriction $x\ne \left\{0\right\}\text{ and }y\ne \left\{0\right\}$ is evident from the obtained simplified form.

Given:

The division to evaluate is $\frac{7x}{4y^3}÷\frac{21x^3}{8y}$ .

Concept used:

A rational expression $R\left(x\right)=\frac{P\left(x\right)}{Q\left(x\right)}$ where $P\left(x\right)\text{ and }Q\left(x\right)$ are polynomial and $Q\left(x\right)\ne 0$ , is said to be in its simplest form when its numerator and denominator have no common divisors in it, and in the simplified form of rational expression, the restriction is on the denominator at which the denominator is equal to zero. Further, if the rational expression has common divisors in it, then equate the common divisor equal to zero which will give another restriction that is not evident, but it is needed to prevent the denominator of the original expression from being reduced to zero.

Calculation:

With the help of the concept stated above, the simplified form of the division and the restrictions on the variables is evaluated as:

Therefore, the rational expression $\frac{7x}{4y^3}÷\frac{21x^3}{8y}$ is multiplied as:

  $\begin{array}{c}\frac{7x}{4y^3}÷\frac{21x^3}{8y}=\frac{7x}{4y^3}\cdot \frac{8y}{21x^3}\\ =\frac{7x}{4\cdot y^2\cdot y}\cdot \frac{4\cdot 2\cdot y}{3\cdot 7\cdot x\cdot x^2}\mathrm{....}\left(\begin{array}{l}\text{Factorize the numerator}\\ \text{ and denominator}\end{array}\right)\\ =\frac{1}{y^2}\cdot \frac{2}{3\cdot x^2}\mathrm{....}\left(\begin{array}{l}\text{ Cancel the common terms where,}\\ \text{the restriction is }x\ne \left\{0\right\}\text{ and }y\ne \left\{0\right\}\end{array}\right)\\ =\frac{2}{3x^2{y}^2}\end{array}$

Conclusion:

The simplified form of the division is $\frac{2}{3x^2{y}^2}$ for $x\ne \left\{0\right\}\text{ and }y\ne \left\{0\right\}$ where the restriction $x\ne \left\{0\right\}\text{ and }y\ne \left\{0\right\}$ is evident from the obtained simplified form.