Chapter 5.5, Problem 20WE

To determine

To simplify the given expression.

Answer to Problem 20WE

  $x\left(x-3a\right)$

Explanation of Solution

Given:

Expression: $\frac{x^2+3ax}{3a-x}\cdot \frac{x^2-4ax+3a^2}{a^2-x^2}÷\frac{x+3a}{x+a}$

Concept used:

To multiply two fractions, simply multiply numerators by each other and denominators by each other and then reduce the fraction if possible.

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

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,

  $\Rightarrow \frac{x^2+3ax}{3a-x}\cdot \frac{x^2-4ax+3a^2}{a^2-x^2}÷\frac{x+3a}{x+a}$

  $\Rightarrow \frac{x^2+3ax}{3a-x}\cdot \frac{x^2-4ax+3a^2}{a^2-x^2}\cdot \frac{x+a}{x+3a}$  [Flipping third fraction]

  $\Rightarrow \frac{x\left(x+3a\right)}{3a-x}\times \frac{x^2-3ax-ax+3a^2}{\left(a-x\right)\left(a+x\right)}\times \frac{x+a}{x+3a}$  [Factorizing the fractions]

  $\Rightarrow \frac{x\left(x+3a\right)}{3a-x}\times \frac{x\left(x-3a\right)-a\left(x-3a\right)}{\left(a-x\right)\left(x+a\right)}\times \frac{x+a}{x+3a}$

  $\Rightarrow \frac{x\left(x+3a\right)}{\left(-1\right)\left(x-3a\right)}\times \frac{\left(x-3a\right)\left(x-a\right)}{\left(-1\right)\left(x-a\right)\left(x+a\right)}\times \frac{x+a}{x+3a}$

  $\Rightarrow \frac{x}{\left(-1\right)}\times \frac{\left(x-3a\right)}{\left(-1\right)}\times \frac{1}{1}$  [Reducing the fractions]

  $\begin{array}{l}\Rightarrow \frac{x\times \left(x-3a\right)\times 1}{\left(-1\right)\times \left(-1\right)\times 1}\\ \Rightarrow x\left(x-3a\right)\end{array}$

Conclusion:

Therefore, after simplification, the answer is $x\left(x-3a\right)$ .