Chapter 6.5, Problem 57WE

To determine

Find the simplified form of the given expression.

Answer to Problem 57WE

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

Explanation of Solution

Given:

The given expression is $\frac{b+1}{{\left(b-1\right)}^2}+\frac{2-2b}{{\left(b-1\right)}^3}+\frac{1}{\left(b-1\right)}$ .

Calculation:

Rewrite the given expression.

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

Since the denominators are different, rewrite the fractions using their least common denominator, ${\left(b-1\right)}^2$ .

  $=\frac{b+1}{{\left(b-1\right)}^2}-\frac{2}{{\left(b-1\right)}^2}+\frac{\left(b-1\right)}{{\left(b-1\right)}^2}$

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.

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

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