Chapter 11.2, Problem 61PPE

To determine

Find the error of the student.

Answer to Problem 61PPE

  $\frac{3a}{a+2}÷\frac{{(a+2)}^2}{a-4}=\frac{3a(a-4)}{{(a+2)}^3}$

Explanation of Solution

Given:

  

Concept Used:

For multiplication of a rational expression, 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: $\frac{a}{b}÷\frac{c}{d}=\frac{a}{b}·\frac{d}{c}$

We are not allowed to cancel the factor if the expression is in division form.

Step 1: By using the rule: $\frac{a}{b}÷\frac{c}{d}=\frac{a}{b}·\frac{d}{c}$ , we need to flip the 2^nd expression and put a multiplication sign. Factor it and we can only cancel any factor from numerator and denominator while the expressions are in multiplication form.

Calculation:

The correct way to solve the division:

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

Thus, $\frac{3a}{a+2}÷\frac{{(a+2)}^2}{a-4}=\frac{3a(a-4)}{{(a+2)}^3}$