Chapter 11.4, Problem 30PPE

To determine

Add or subtract $\frac{9}{m+2}+\frac{8}{m-7}$

Answer to Problem 30PPE

  $\frac{17m-47}{\left(m+2\right)\left(m-7\right)}$

Explanation of Solution

Given:

  $\frac{9}{m+2}+\frac{8}{m-7}$ ,

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

  $\frac{9}{m+2}+\frac{8}{m-7}$

Here observe that the denominators are not same, so first we will make both denominators equal to LCD. The LCD of $\left(m+2\right)\text{and}\left(m-7\right)$ is $\left(m+2\right)\left(m-7\right)$ .

So in order to subtract the fractions, we make both the denominators equal to LCD by multiplying the first fraction by $\left(m-7\right)$ up and down and second fraction by $\left(m+2\right)$ . And then when denominators are equal, we directly subtract the numerators and put over the common denominator. So, we get

  $\begin{array}{c}\frac{9}{m+2}+\frac{8}{m-7}=\frac{9}{\left(m+2\right)}\cdot \frac{\left(m-7\right)}{\left(m-7\right)}+\frac{8}{\left(m-7\right)}\cdot \frac{\left(m+2\right)}{\left(m+2\right)}\\ =\frac{9\cdot m+9\cdot \left(-7\right)}{\left(m+2\right)\left(m-7\right)}+\frac{\left(8\cdot m+8\cdot 2\right)}{\left(m+2\right)\left(m-7\right)}\\ =\frac{9m-63+8m+16}{\left(m+2\right)\left(m-7\right)}\\ =\frac{17m-47}{\left(m+2\right)\left(m-7\right)}\end{array}$

Thus, the given expression simplifies to $\frac{17m-47}{\left(m+2\right)\left(m-7\right)}$ .