Chapter 6.5, Problem 55WE

To determine

Find the simplified form of the given expression.

Answer to Problem 55WE

The simplified form of the expression is $\frac{1}{a-2}$ .

Explanation of Solution

Given:

The given expression is $\frac{a+2}{a^2+5a+6}-\frac{2+a}{4-a^2}+\frac{2-a}{a^2+a-6}$ .

Calculation:

Rewrite the given expression.

  $\begin{array}{l}\frac{a+2}{a^2+5a+6}-\frac{2+a}{4-a^2}+\frac{2-a}{a^2+a-6}\\ =\frac{a+2}{a^2+3a+2a+6}-\frac{2+a}{2^2-a^2}+\frac{2-a}{a^2+3a-2a-6}\\ =\frac{a+2}{a\left(a+3\right)+2\left(a+3\right)}-\frac{2+a}{\left(2+a\right)\left(2-a\right)}+\frac{2-a}{a\left(a+3\right)-2\left(a+3\right)}\\ =\frac{a+2}{\left(a+2\right)\left(a+3\right)}-\frac{2+a}{\left(2+a\right)\left(2-a\right)}+\frac{2-a}{\left(a+3\right)\left(a-2\right)}\\ =\frac{1}{\left(a+3\right)}-\frac{1}{\left(2-a\right)}-\frac{2-a}{\left(a+3\right)\left(2-a\right)}\end{array}$

Since the denominators are different, rewrite the fractions using their least common denominator, $\left(a+3\right)\left(2-a\right)$ .

  $=\frac{\left(2-a\right)}{\left(a+3\right)\left(2-a\right)}-\frac{\left(a+3\right)}{\left(2-a\right)\left(a+3\right)}-\frac{2-a}{\left(a+3\right)\left(2-a\right)}$

Use addition/subtraction rule of fraction.

  $\frac{a}{c}\pm \frac{b}{c}=\frac{a\pm b}{c}$

To add/subtract fractions with the same denominator, add/subtract their numerators and write the result over the common denominator.

  $\begin{array}{l}=\frac{2-a-a-3-2+a}{\left(a+3\right)\left(2-a\right)}\\ =\frac{-a-3}{\left(a+3\right)\left(2-a\right)}\\ =-\frac{\left(a+3\right)}{\left(a+3\right)\left(2-a\right)}\\ =-\frac{1}{2-a}\\ =\frac{1}{a-2}\end{array}$

Hence the simplified form of the expression is $\frac{1}{a-2}$ .