Chapter 8.4, Problem 49PPSE

Solution

To simplify: The complex fraction $\frac{\frac{x^2-1}{x^2-9}}{\frac{x^2+3x-4}{x^2+8x+15}}$ using the result $\frac{\frac{a}{b}}{\frac{c}{d}}=\frac{a}{b}÷\frac{c}{d}$ and state the restrictions on the variables.

The simplified form of complex fraction is written as $\frac{\frac{x^2-1}{x^2-9}}{\frac{x^2+3x-4}{x^2+8x+15}}=\frac{\left(x+1\right)\left(x+5\right)}{\left(x-3\right)\left(x+4\right)}$ for $x\ne \left\{-4,-3,1,3\right\}$ .

Given:

The complex fraction to evaluate is $\frac{\frac{x^2-1}{x^2-9}}{\frac{x^2+3x-4}{x^2+8x+15}}$ .

Concept used:

Simplify the complex fraction using the result $\frac{\frac{a}{b}}{\frac{c}{d}}=\frac{a}{b}÷\frac{c}{d}$ and then simplify the rational expression using the result of restriction on variables.

Calculation:

With the help of the concept stated above and using the result $\frac{\frac{a}{b}}{\frac{c}{d}}=\frac{a}{b}÷\frac{c}{d}$ , the complex fraction $\frac{\frac{x^2-1}{x^2-9}}{\frac{x^2+3x-4}{x^2+8x+15}}$ is simplified as:

  $\begin{array}{c}\frac{\frac{x^2-1}{x^2-9}}{\frac{x^2+3x-4}{x^2+8x+15}}=\frac{x^2-1}{x^2-9}÷\frac{x^2+3x-4}{x^2+8x+15}\mathrm{....}\left(\text{From, }\frac{\frac{a}{b}}{\frac{c}{d}}=\frac{a}{b}÷\frac{c}{d}\right)\\ =\frac{x^2-1}{x^2-9}\cdot \frac{x^2+8x+15}{x^2+3x-4}\\ =\frac{\left(x-1\right)\left(x+1\right)}{\left(x-3\right)\left(x+3\right)}\cdot \frac{\left(x+3\right)\left(x+5\right)}{\left(x+4\right)\left(x-1\right)}\mathrm{....}\left(\text{Factorize}\right)\\ =\frac{\left(x+1\right)\left(x+5\right)}{\left(x-3\right)\left(x+4\right)}\mathrm{....}\left(\begin{array}{l}\text{ Cancel the common terms},\\ \text{ where }x\ne \left\{-4,-3,1,3\right\}\end{array}\right)\\ =\frac{\left(x+1\right)\left(x+5\right)}{\left(x-3\right)\left(x+4\right)}\mathrm{....}\left(\text{Restricted }x\ne \left\{-4,-3,1,3\right\}\right)\end{array}$

Conclusion:

The simplified form of complex fraction is written as $\frac{\frac{x^2-1}{x^2-9}}{\frac{x^2+3x-4}{x^2+8x+15}}=\frac{\left(x+1\right)\left(x+5\right)}{\left(x-3\right)\left(x+4\right)}$ for $x\ne \left\{-4,-3,1,3\right\}$ .