Chapter 8.5, Problem 4LC

Solution

To simplify: The difference of $\frac{b-4}{b^2+2b-8}-\frac{b+2}{b^2-16}$ in simplified form and state the restrictions on the variables.

The difference of the expression is written as $\frac{b-4}{b^2+2b-8}-\frac{b+2}{b^2-16}=-\frac{8}{\left(b-2\right)\left(b+4\right)}$ where the restriction on variable is $b\ne \left\{-4,-2,2\right\}$ .

Given:

The expression to evaluate is $\frac{b-4}{b^2+2b-8}-\frac{b+2}{b^2-16}$ .

Concept used:

Simplify the difference of rational expression by simplifying the expression using the least common denominator and apply the restriction on variables for which the denominator is zero.

Calculation:

With the help of the concept stated above, the difference of the expression $\frac{b-4}{b^2+2b-8}-\frac{b+2}{b^2-16}$ is evaluated as:

  $\begin{array}{c}\frac{b-4}{b^2+2b-8}-\frac{b+2}{b^2-16}=\frac{b-4}{\left(b+4\right)\left(b-2\right)}-\frac{b+2}{b^2-4^2}\mathrm{....}\left(\text{Factorize}\right)\\ =\frac{b-4}{\left(b+4\right)\left(b-2\right)}-\frac{b+2}{\left(b-2\right)\left(b+2\right)}\mathrm{....}\left(\because a^2-b^2=\left(a-b\right)\left(a+b\right)\right)\\ =\frac{\left(b-4\right)\cdot \left(b+2\right)-\left(b+2\right)\cdot \left(b+4\right)}{\left(b-2\right)\left(b+2\right)\left(b+4\right)}\mathrm{....}\left(\begin{array}{l}\text{ Simplify by least}\\ \text{common denominator}\end{array}\right)\\ =\frac{\left(b^2-2b-8\right)-\left(b^2+6b+8\right)}{\left(b-2\right)\left(b+2\right)\left(b+4\right)}\\ =\frac{-8b-16}{\left(b-2\right)\left(b+2\right)\left(b+4\right)}\\ =-\frac{8\left(b+2\right)}{\left(b-2\right)\left(b+2\right)\left(b+4\right)}\\ =-\frac{8}{\left(b-2\right)\left(b+4\right)}\mathrm{.....}\left(\begin{array}{l}\text{Cancel the common terms}\\ \text{ where }b\ne \left\{-4,-2,2\right\}\end{array}\right)\end{array}$

Conclusion:

The difference of the expression is written as $\frac{b-4}{b^2+2b-8}-\frac{b+2}{b^2-16}=-\frac{8}{\left(b-2\right)\left(b+4\right)}$ where the restriction on variable is $b\ne \left\{-4,-2,2\right\}$ .