Chapter 1.4, Problem 43E

To determine

To calculate:Thesimplified form of the expression $\frac{1}{x+1}-\frac{1}{x+2}$

Answer to Problem 43E

The simplified form of the expression is $\underset{\_}{\frac{1}{\left(x+1\right)\left(x+2\right)}}$

Explanation of Solution

Given information:

The expression is $\frac{1}{x+1}-\frac{1}{x+2}$

Formula used:

To add or subtract rational expression, first find the common denominator and then use the following property of fractions:

  $\frac{P}{R}+\frac{Q}{R}=\frac{P+Q}{R}$

If denominators are not same then find least common denominator. The least common denominator is found factoring each denominator and taking the product of distinct factors, using the highest power that appears in any of the factors.

Calculation:

Consider the provided expression. $\frac{1}{x+1}-\frac{1}{x+2}$

Recall that,

To add or subtract rational expression, first find the common denominator and then use the following property of fractions:

  $\frac{P}{R}+\frac{Q}{R}=\frac{P+Q}{R}$

If denominators are not same then find least common denominator. The least common denominator is found factoring each denominator and taking the product of distinct factors, using the highest power that appears in any of the factors.

Since denominators are not the same then find the least common denominator.

The least common denominator of $\left(x+1\right)$ and $\left(x+2\right)$ is $\left(x+1\right)\left(x+2\right)$ .

  $\frac{1}{x+1}-\frac{1}{x+2}=\frac{1\times \left(x+2\right)}{\left(x+1\right)\left(x+2\right)}-\frac{1\times \left(x+1\right)}{\left(x+1\right)\left(x+2\right)}$

Add fractions,

  $\begin{array}{c}\frac{1}{x+1}-\frac{1}{x+2}=\frac{1\times \left(x+2\right)-1\times \left(x+1\right)}{(x+1)\left(x+2\right)}\\ =\frac{x+2-x-1}{\left(x+1\right)\left(x+2\right)}\end{array}$

Combine terms in numerator,

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

Thus,the simplified form of the expression is $\frac{1}{\left(x+1\right)\left(x+2\right)}$