Chapter 9.7, Problem 67SR

To determine

To simplify: The expression $\frac{r^2+2r-8}{r^2+4r+3}÷\frac{r-2}{3r+3}$ .

Answer to Problem 67SR

The solution of the expression is $\frac{3\left(r+4\right)}{\left(r+3\right)}$ .

Explanation of Solution

Given:

The expression is $\frac{r^2+2r-8}{r^2+4r+3}÷\frac{r-2}{3r+3}$ .

Calculation:

Simplify the expression as follows.

  $\begin{array}{c}\frac{r^2+2r-8}{r^2+4r+3}÷\frac{r-2}{3r+3}=\frac{r^2+2r-8}{r^2+4r+3}\times \frac{r-2}{3r+3}\\ =\frac{r^2+4r-2r-8}{r^2+3r+r+3}\times \frac{3\left(r+1\right)}{r-2}\\ =\frac{\left(r+4\right)\left(r-2\right)}{\left(r+3\right)\left(r+1\right)}\times \frac{3\left(r+1\right)}{r-2}\\ =\frac{3\left(r+4\right)}{\left(r+3\right)}\end{array}$

Therefore, the solution of the expression is $\frac{3\left(r+4\right)}{\left(r+3\right)}$ .