Chapter 11.2, Problem 52PPE

To determine

Divide the expressions: $\frac{c^2+3c+2}{c^2-4c+3}÷\frac{c+2}{c-3}$

Answer to Problem 52PPE

  $\frac{c^2+3c+2}{c^2-4c+3}÷\frac{c+2}{c-3}=\frac{c+1}{c-1}$

Explanation of Solution

Given:

Divide the expressions: $\frac{c^2+3c+2}{c^2-4c+3}÷\frac{c+2}{c-3}$

Concept Used:

Find the factors of numerator and denominator, and then cancel the common factor and simplify.

For division of rational expression, we always flip the 2^nd expression and put a multiplication sign.

Rule: $\frac{a}{b}÷\frac{c}{d}=\frac{a}{b}·\frac{d}{c}$

Calculation:

  $\begin{array}{l}\frac{c^2+3c+2}{c^2-4c+3}÷\frac{c+2}{c-3}\\ \\ \frac{c^2+3c+2}{c^2-4c+3}·\frac{c-3}{c+2}=\frac{(c+2)(c+1)}{(c-3)(c-1)}·\frac{c-3}{c+2}\\ \\ \frac{c+2}{c+2}·\frac{c-3}{c-3}·\frac{c+1}{c-1}=\frac{c+1}{c-1}\end{array}$

Thus, $\frac{c^2+3c+2}{c^2-4c+3}÷\frac{c+2}{c-3}=\frac{c+1}{c-1}$