Chapter 11.4, Problem 29PPE

To determine

Add or subtract $\frac{a}{a+3}-\frac{4}{a+5}$

Answer to Problem 29PPE

  $\frac{a^2+a-20}{\left(a+3\right)\left(a+5\right)}$

Explanation of Solution

Given:

  $\frac{a}{a+3}-\frac{4}{a+5}$ ,

Concept used:

The LCD (Least Common Denominator) is the smallest number that is divisible by all the denominators.

Calculation:

The LCD (Least Common Denominator) is the smallest number that is divisible by all the denominators.

The given expression can be written as

  $\frac{a}{a+3}-\frac{4}{a+5}$

Here observe that the denominators are not same, so first we will make both denominators equal to LCD. The LCD of $\left(a+3\right)\text{and}\left(a+5\right)$ is $\left(a+3\right)\left(a+5\right)$ .

So in order to subtract the fractions, we make both the denominators equal to LCD by multiplying the first fraction by $\left(a+5\right)$ up and down and second fraction by $\left(a+3\right)$ . And then when denominators are equal, we directly subtract the numerators and put over the common denominator. So, we get

  $\begin{array}{c}\frac{a}{a+3}-\frac{4}{a+5}=\frac{a}{\left(a+3\right)}\cdot \frac{\left(a+5\right)}{\left(a+5\right)}-\frac{4}{\left(a+5\right)}\cdot \frac{\left(a+3\right)}{\left(a+3\right)}\\ =\frac{a\cdot a+a\cdot 5}{\left(a+3\right)\left(a+5\right)}-\frac{\left(4\cdot a+4\cdot 5\right)}{\left(a+3\right)\left(a+5\right)}\\ =\frac{a^2+5a-4a-20}{\left(a+3\right)\left(a+5\right)}\\ =\frac{a^2+a-20}{\left(a+3\right)\left(a+5\right)}\end{array}$

Thus, the given expression simplifies to $\frac{a^2+a-20}{\left(a+3\right)\left(a+5\right)}$ .