Rational Functions: Addition and Subtraction

4.6 Rational Functions: Addition and Subtraction

Learning Objectives

Add and subtract rational functions.
Simplify complex rational expressions.

Adding and Subtracting Rational Functions

Adding and subtracting rational expressions is similar to adding and subtracting fractions. Recall that if the denominators are the same, we can add or subtract the numerators and write the result over the common denominator. When working with rational expressions, the common denominator will be a polynomial. In general, given polynomials P, Q, and R, where $Q \neq 0$ , we have the following:

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

The set of restrictions to the domain of a sum or difference of rational expressions consists of the restrictions to the domains of each expression.

Example 1

Subtract: $\frac{4 x}{x^{2} - 64} - \frac{3 x + 8}{x^{2} - 64} .$

Solution:

The denominators are the same. Hence we can subtract the numerators and write the result over the common denominator. Take care to distribute the negative 1.

$\begin{matrix} \frac{4 x}{x^{2} - 64} - \frac{3 x + 8}{x^{2} - 64} & = & \frac{4 x - (3 x + 8)}{x^{2} - 64} S u b t r a c t t h e n u m e r a t o r s . \\ = & \frac{4 x - 3 x - 8}{x^{2} - 64} S i m p l i f y . \\ = & \frac{\overset{1}{x - 8}}{(x + 8) (x - 8)} C a n c e l . \\ = & \frac{1}{x + 8} R e s t r i c t i o n s x \neq \pm 8 \end{matrix}$

Answer: $\frac{1}{x + 8}$ , where $x \neq \pm 8$

To add rational expressions with unlike denominators, first find equivalent expressions with common denominators. Do this just as you have with fractions. If the denominators of fractions are relatively prime, then the least common denominator (LCD) is their product. For example,

$\frac{1}{x} + \frac{1}{y} \Rightarrow LCD = x \cdot y = x y$

Multiply each fraction by the appropriate form of 1 to obtain equivalent fractions with a common denominator.

$\begin{array}{l} \frac{1}{x} + \frac{1}{y} & = & \frac{1 \cdot y}{x \cdot y} + \frac{1 \cdot x}{y \cdot x} \\ = & \frac{y}{x y} + \frac{x}{x y} & E q u i v a l e n t f r a c t i o n s w i t h a c o m m o n d e n o m i n a t o r \\ = & \frac{y + x}{x y} \end{array}$

In general, given polynomials P, Q, R, and S, where $Q \neq 0$ and $S \neq 0$ , we have the following:

$\frac{P}{Q} \pm \frac{R}{S} = \frac{P S \pm Q R}{Q S}$

Example 2

Given $f (x) = \frac{5 x}{3 x + 1}$ and $g (x) = \frac{2}{x + 1}$ , find $f + g$ and state the restrictions.

Solution:

Here the LCD is the product of the denominators $(3 x + 1) (x + 1) .$ Multiply by the appropriate factors to obtain rational expressions with a common denominator before adding.

$\begin{matrix} (f + g) (x) & = & f (x) + g (x) \\ = & \frac{5 x}{3 x + 1} + \frac{2}{x + 1} \\ = & \frac{5 x}{(3 x + 1)} \cdot \frac{(x + 1)}{(x + 1)} + \frac{2}{(x + 1)} \cdot \frac{(3 x + 1)}{(3 x + 1)} \\ = & \frac{5 x (x + 1)}{(3 x + 1) (x + 1)} + \frac{2 (3 x + 1)}{(x + 1) (3 x + 1)} \\ = & \frac{5 x (x + 1) + 2 (3 x + 1)}{(3 x + 1) (x + 1)} \\ = & \frac{5 x^{2} + 5 x + 6 x + 2}{(3 x + 1) (x + 1)} \\ = & \frac{5 x^{2} + 11 x + 2}{(3 x + 1) (x + 1)} \\ = & \frac{(5 x + 1) (x + 2)}{(3 x + 1) (x + 1)} \end{matrix}$

The domain of f consists all real numbers except $- \frac{1}{3}$ , and the domain of g consists of all real numbers except −1. Therefore, the domain of f + g consists of all real numbers except −1 and $- \frac{1}{3} .$

Answer: $(f + g) (x) = \frac{(5 x + 1) (x + 2)}{(3 x + 1) (x + 1)}$ , where $x \neq - 1, - \frac{1}{3}$

It is not always the case that the LCD is the product of the given denominators. Typically, the denominators are not relatively prime; thus determining the LCD requires some thought. Begin by factoring all denominators. The LCD is the product of all factors with the highest power.

Example 3

Given $f (x) = \frac{3 x}{3 x - 1}$ and $g (x) = \frac{4 - 14 x}{3 x^{2} - 4 x + 1}$ , find $f - g$ and state the restrictions to the domain.

Solution:

To determine the LCD, factor the denominator of $g .$

$\begin{matrix} (f - g) (x) & = & f (x) - g (x) \\ = & \frac{3 x}{3 x - 1} - \frac{4 - 14 x}{3 x^{2} - 4 x + 1} \\ = & \frac{3 x}{(3 x - 1)} - \frac{4 - 14 x}{(3 x - 1) (x - 1)} \end{matrix}$

In this case the $LCD = (3 x - 1) (x - 1) .$ Multiply $f$ by 1 in the form of $\frac{(x - 1)}{(x - 1)}$ to obtain equivalent algebraic fractions with a common denominator and then subtract.

$\begin{matrix} = & \frac{3 x}{(3 x - 1)} \cdot \frac{(x - 1)}{(x - 1)} - \frac{4 - 14 x}{(3 x - 1) (x - 1)} \\ = & \frac{3 x (x - 1) - 4 + 14 x}{(3 x - 1) (x - 1)} \\ = & \frac{3 x^{2} + 11 x - 4}{(3 x - 1) (x - 1)} \\ = & \frac{(3 x - 1) (x + 4)}{(3 x - 1) (x - 1)} \\ = & \frac{(x + 4)}{(x - 1)} \end{matrix}$

The domain of f consists of all real numbers except $\frac{1}{3}$ , and the domain of g consists of all real numbers except 1 and $\frac{1}{3} .$ Therefore, the domain of f − g consists of all real numbers except 1 and $\frac{1}{3} .$

Answer: $(f - g) (x) = \frac{x + 4}{x - 1}$ , where $x \neq \frac{1}{3}, 1$

Example 4

Simplify and state the restrictions: $\frac{- 2 x}{x + 6} - \frac{3 x}{6 - x} - \frac{18 (x - 2)}{x^{2} - 36} .$

Solution:

Begin by applying the opposite binomial property $6 - x = - (x - 6) .$

$\begin{matrix} \frac{- 2 x}{x + 6} - \frac{3 x}{6 - x} - \frac{18 (x - 2)}{x^{2} - 36} \\ = \frac{- 2 x}{(x + 6)} - \frac{3 x}{- 1 \cdot (x - 6)} - \frac{18 (x - 2)}{(x + 6) (x - 6)} \\ = \frac{- 2 x}{(x + 6)} + \frac{3 x}{(x - 6)} - \frac{18 (x - 2)}{(x + 6) (x - 6)} \end{matrix}$

Next, find equivalent fractions with the $L C D = (x + 6) (x - 6)$ and then simplify.

$\begin{matrix} = & \frac{- 2 x}{(x + 6)} \cdot \frac{(x - 6)}{(x - 6)} + \frac{3 x}{(x - 6)} \cdot \frac{(x + 6)}{(x + 6)} - \frac{18 (x - 2)}{(x + 6) (x - 6)} \\ = & \frac{- 2 x (x - 6) + 3 x (x + 6) - 18 (x - 2)}{(x + 6) (x - 6)} \\ = & \frac{- 2 x^{2} + 12 x + 3 x^{2} + 18 x - 18 x + 36}{(x + 6) (x - 6)} \\ = & \frac{x^{2} + 12 x + 36}{(x + 6) (x - 6)} \\ = & \frac{(x + 6) (x + 6)}{(x + 6) (x - 6)} \\ = & \frac{x + 6}{x - 6} \end{matrix}$

Answer: $\frac{x + 6}{x - 6}$ , where $x \neq \pm 6$

Try this! Simplify and state the restrictions: $\frac{x + 1}{{(x - 1)}^{2}} - \frac{2}{x^{2} - 1} - \frac{4}{(x + 1) {(x - 1)}^{2}}$

Answer: $\frac{1}{x - 1}$ , where $x \neq \pm 1$

(click to see video)

Rational expressions are sometimes expressed using negative exponents. In this case, apply the rules for negative exponents before simplifying the expression.

Example 5

Simplify and state the restrictions: $5 a^{- 2} + {(2 a + 5)}^{- 1} .$

Solution:

Recall that $x^{- n} = \frac{1}{x^{n}}$ . Begin by rewriting the rational expressions with negative exponents as fractions.

$5 a^{- 2} + {(2 a + 5)}^{- 1} = \frac{5}{a^{2}} + \frac{1}{{(2 a + 5)}^{1}}$

Then find the LCD and add.

$\begin{array}{l} \frac{5}{a^{2}} + \frac{1}{{(2 a + 5)}^{1}} & = & \frac{5}{a^{2}} \cdot \frac{(2 a + 5)}{(2 a + 5)} + \frac{1}{(2 a + 5)} \cdot \frac{a^{2}}{a^{2}} \\ = & \frac{5 (2 a + 5)}{a^{2} (2 a + 5)} + \frac{a^{2}}{a^{2} (2 a + 5)} & E q u i v a l e n t e x p r e s s i o n s w i t h a c o m m o n d e n o m i n a t o r \\ = & \frac{10 a + 25 + a^{2}}{a^{2} (2 a + 5)} & A d d . \\ = & \frac{a^{2} + 10 a + 25}{a^{2} (2 a + 5)} & S i m p l f i y . \\ = & \frac{(a + 5) (a + 5)}{a^{2} (2 a + 5)} \end{array}$

Answer: $\frac{{(a + 5)}^{2}}{a^{2} (2 a + 5)}$ , where $a \neq - \frac{5}{2}, 0$

Simplifying Complex Rational Expressions

A complex rational expressionA rational expression that contains one or more rational expressions in the numerator or denominator or both. is defined as a rational expression that contains one or more rational expressions in the numerator or denominator or both. For example, $\frac{4 - \frac{12}{x} + \frac{9}{x^{2}}}{2 - \frac{5}{x} + \frac{3}{x^{2}}}$ is a complex rational expression. We simplify a complex rational expression by finding an equivalent fraction where the numerator and denominator are polynomials. There are two methods for simplifying complex rational expressions, and we will outline the steps for both methods. For the sake of clarity, assume that variable expressions used as denominators are nonzero.

Method 1: Simplify Using Division

We begin our discussion on simplifying complex rational expressions using division. Before we can multiply by the reciprocal of the denominator, we must simplify the numerator and denominator separately. The goal is to first obtain single algebraic fractions in the numerator and the denominator. The steps for simplifying a complex algebraic fraction are illustrated in the following example.

Example 6

Simplify: $\frac{4 - \frac{12}{x} + \frac{9}{x^{2}}}{2 - \frac{5}{x} + \frac{3}{x^{2}}}$ .

Solution:

Step 1: Simplify the numerator and denominator to obtain a single algebraic fraction divided by another single algebraic fraction. In this example, find equivalent terms with a common denominator in both the numerator and denominator before adding and subtracting.

$\begin{array}{l} \frac{4 - \frac{12}{x} + \frac{9}{x^{2}}}{2 - \frac{5}{x} + \frac{3}{x^{2}}} & = & \frac{\frac{4}{1} \cdot \frac{x^{2}}{x^{2}} - \frac{12}{x} \cdot \frac{x}{x} + \frac{9}{x^{2}}}{\frac{2}{1} \cdot \frac{x^{2}}{x^{2}} - \frac{5}{x} \cdot \frac{x}{x} + \frac{3}{x^{2}}} \\ = & \frac{\frac{4 x^{2}}{x^{2}} - \frac{12 x}{x^{2}} + \frac{9}{x^{2}}}{\frac{2 x^{2}}{x^{2}} - \frac{5 x}{x^{2}} + \frac{3}{x^{2}}} & E q u i v a l e n t f r a c t i o n s w i t h c o m m o n d e n o m i n a t o r s . \\ = & \frac{\frac{4 x^{2} - 12 x + 9}{x^{2}}}{\frac{2 x^{2} - 5 x + 3}{x^{2}}} & A d d t h e f r a c t i o n s i n t h e n u m e r a t o r a n d d e n o m i n a t o r . \end{array}$

At this point we have a single algebraic fraction divided by another single algebraic fraction.

Step 2: Multiply the numerator by the reciprocal of the denominator.

$\frac{\frac{4 x^{2} - 12 x + 9}{x^{2}}}{\frac{2 x^{2} - 5 x + 3}{x^{2}}} = \frac{4 x^{2} - 12 x + 9}{x^{2}} \cdot \frac{x^{2}}{2 x^{2} - 5 x + 3}$

Step 3: Factor all numerators and denominators completely.

$= \frac{(2 x - 3) (2 x - 3)}{x^{2}} \cdot \frac{x^{2}}{(2 x - 3) (x - 1)}$

Step 4: Cancel all common factors.

$\begin{matrix} = & \frac{(2 x - 3) (2 x - 3)}{x^{2}} \cdot \frac{x^{2}}{(2 x - 3) (x - 1)} \\ = & \frac{2 x - 3}{x - 1} \end{matrix}$

Answer: $\frac{2 x - 3}{x - 1}$

Example 7

Simplify: $\frac{\frac{2 x}{x - 1} + \frac{7}{x + 3}}{\frac{2 x}{x - 1} - \frac{5}{x - 3}}$ .

Solution:

Obtain a single algebraic fraction in the numerator and in the denominator.

$\begin{matrix} \frac{\frac{2 x}{x - 1} + \frac{7}{x + 3}}{\frac{2 x}{x - 1} - \frac{5}{x - 3}} & = & \frac{\frac{2 x}{x - 1} \cdot \frac{(x + 3)}{(x + 3)} + \frac{7}{x + 3} \cdot \frac{(x - 1)}{(x - 1)}}{\frac{2 x}{x - 1} \cdot \frac{(x - 3)}{(x - 3)} - \frac{5}{x - 3} \cdot \frac{(x - 1)}{(x - 1)}} \\ = & \frac{\frac{2 x (x + 3) + 7 (x - 1)}{(x - 1) (x + 3)}}{\frac{2 x (x - 3) - 5 (x - 1)}{(x - 1) (x - 3)}} \\ = & \frac{\frac{2 x^{2} + 6 x + 7 x - 7}{(x - 1) (x + 3)}}{\frac{2 x^{2} - 6 x - 5 x + 5}{(x - 1) (x - 3)}} \\ = & \frac{\frac{2 x^{2} + 13 x - 7}{(x - 1) (x + 3)}}{\frac{2 x^{2} - 11 x + 5}{(x - 1) (x - 3)}} \end{matrix}$

Next, multiply the numerator by the reciprocal of the denominator, factor, and then cancel.

$\begin{matrix} = & \frac{2 x^{2} + 13 x - 7}{(x - 1) (x + 3)} \cdot \frac{(x - 1) (x - 3)}{2 x^{2} - 11 x + 5} \\ = & \frac{(2 x - 1) (x + 7)}{(x - 1) (x + 3)} \cdot \frac{(x - 1) (x - 3)}{(2 x - 1) (x - 5)} \\ = & \frac{(x + 7) (x - 3)}{(x + 3) (x - 5)} \end{matrix}$

Answer: $\frac{(x + 7) (x - 3)}{(x + 3) (x - 5)}$

Try this! Simplify using division: $\frac{\frac{1}{y^{2}} - \frac{1}{x^{2}}}{\frac{1}{y} + \frac{1}{x}}$ .

Answer: $\frac{x - y}{x y}$

(click to see video)

Sometimes complex rational expressions are expressed using negative exponents.

Example 8

Simplify: $\frac{2 y^{- 1} - x^{- 1}}{x^{- 2} - 4 y^{- 2}}$ .

Solution:

We begin by rewriting the expression without negative exponents.

$\frac{2 y^{- 1} - x^{- 1}}{x^{- 2} - 4 y^{- 2}} = \frac{\frac{2}{y} - \frac{1}{x}}{\frac{1}{x^{2}} - \frac{4}{y^{2}}}$

Obtain single algebraic fractions in the numerator and denominator and then multiply by the reciprocal of the denominator.

$\begin{matrix} \frac{\frac{2}{y} - \frac{1}{x}}{\frac{1}{x^{2}} - \frac{4}{y^{2}}} & = & \frac{\frac{2 x - y}{x y}}{\frac{y^{2} - 4 x^{2}}{x^{2} y^{2}}} \\ = & \frac{2 x - y}{x y} \cdot \frac{x^{2} y^{2}}{y^{2} - 4 x^{2}} \\ = & \frac{2 x - y}{x y} \cdot \frac{x^{2} y^{2}}{(y - 2 x) (y + 2 x)} \end{matrix}$

Apply the opposite binomial property $(y - 2 x) = - (2 x - y)$ and then cancel.

$\begin{matrix} = & \frac{(2 x - y)}{x y} \cdot \frac{\overset{x}{x^{2}} \overset{y}{y^{2}}}{- (2 x - y) (y + 2 x)} \\ = & - \frac{x y}{y + 2 x} \end{matrix}$

Answer: $- \frac{x y}{y + 2 x}$

Method 2: Simplify Using the LCD

An alternative method for simplifying complex rational expressions involves clearing the fractions by multiplying the expression by a special form of 1. In this method, multiply the numerator and denominator by the least common denominator (LCD) of all given fractions.

Example 9

Simplify: $\frac{4 - \frac{12}{x} + \frac{9}{x^{2}}}{2 - \frac{5}{x} + \frac{3}{x^{2}}}$ .

Solution:

Step 1: Determine the LCD of all the fractions in the numerator and denominator. In this case, the denominators of the given fractions are 1, $x$ , and $x^{2} .$ Therefore, the LCD is $x^{2} .$

Step 2: Multiply the numerator and denominator by the LCD. This step should clear the fractions in both the numerator and denominator.

$\begin{array}{l} \frac{4 - \frac{12}{x} + \frac{9}{x^{2}}}{2 - \frac{5}{x} + \frac{3}{x^{2}}} & = & \frac{(4 - \frac{12}{x} + \frac{9}{x^{2}}) \cdot x^{2}}{(2 - \frac{5}{x} + \frac{3}{x^{2}}) \cdot x^{2}} & M u l t i p l y n u m e r a t o r a n d d e n o m i n a t o r b y t h e L C D . \\ = & \frac{4 \cdot x^{2} - \frac{12}{x} \cdot x^{2} + \frac{9}{x^{2}} \cdot x^{2}}{2 \cdot x^{2} - \frac{5}{x} \cdot x^{2} + \frac{3}{x^{2}} \cdot x^{2}} & D i s t r i b u t e a n d t h e n c a n c e l . \\ = & \frac{4 x^{2} - 12 x + 9}{2 x^{2} - 5 x + 3} \end{array}$

This leaves us with a single algebraic fraction with a polynomial in the numerator and in the denominator.

Step 3: Factor the numerator and denominator completely.

$\begin{matrix} = & \frac{4 x^{2} - 12 x + 9}{2 x^{2} - 5 x + 3} \\ = & \frac{(2 x - 3) (2 x - 3)}{(x - 1) (2 x - 3)} \end{matrix}$

Step 4: Cancel all common factors.

$\begin{matrix} = & \frac{(2 x - 3) (2 x - 3)}{(x - 1) (2 x - 3)} \\ = & \frac{2 x - 3}{x - 1} \end{matrix}$

Note: This was the same problem presented in Example 6 and the results here are the same. It is worth taking the time to compare the steps involved using both methods on the same problem.

Answer: $\frac{2 x - 3}{x - 1}$

It is important to point out that multiplying the numerator and denominator by the same nonzero factor is equivalent to multiplying by 1 and does not change the problem.

Try this! Simplify using the LCD: $\frac{\frac{1}{y^{2}} - \frac{1}{x^{2}}}{\frac{1}{y} + \frac{1}{x}}$ .

Answer: $\frac{x - y}{x y}$

(click to see video)

Key Takeaways

Adding and subtracting rational expressions is similar to adding and subtracting fractions. A common denominator is required. If the denominators are the same, then we can add or subtract the numerators and write the result over the common denominator.
The set of restrictions to the domain of a sum or difference of rational functions consists of the restrictions to the domains of each function.
Complex rational expressions can be simplified into equivalent expressions with a polynomial numerator and polynomial denominator. They are reduced to lowest terms if the numerator and denominator are polynomials that share no common factors other than 1.
One method of simplifying a complex rational expression requires us to first write the numerator and denominator as a single algebraic fraction. Then multiply the numerator by the reciprocal of the denominator and simplify the result.
Another method for simplifying a complex rational expression requires that we multiply it by a special form of 1. Multiply the numerator and denominator by the LCD of all the denominators as a means to clear the fractions. After doing this, simplify the remaining rational expression.

Topic Exercises

Part A: Adding and Subtracting Rational Functions

State the restrictions and simplify.

$\frac{3 x}{3 x + 4} + \frac{2}{3 x + 4}$
$\frac{3 x}{2 x - 1} - \frac{2 x + 1}{2 x - 1}$
$\frac{x - 2}{2 x^{2} - 11 x - 6} + \frac{x + 3}{2 x^{2} - 11 x - 6}$
$\frac{4 x - 1}{3 x^{2} + 2 x - 5} - \frac{x - 6}{3 x^{2} + 2 x - 5}$
$\frac{1}{x} - 2 x$
$\frac{4}{x^{3}} - \frac{1}{x}$
$\frac{1}{x - 1} + 5$
$\frac{1}{x + 7} - 1$
$\frac{1}{x - 2} - \frac{1}{3 x + 4}$
$\frac{2}{5 x - 2} + \frac{x}{x + 3}$
$\frac{1}{x^{2}} + \frac{1}{x - 2}$
$\frac{2 x}{x} + \frac{2}{x - 2}$
$\frac{3 x - 7}{x (x - 7)} + \frac{1}{7 - x}$
$\frac{2}{8 - x} + \frac{3 x^{2} - 1}{x^{2} (x - 8)}$
$\frac{x - 1}{x^{2} - 25} - \frac{2}{x^{2} - 10 x + 25}$
$\frac{x + 1}{2 x^{2} + 5 x - 3} - \frac{x}{4 x^{2} - 1}$
$\frac{x}{x^{2} + 4 x} - \frac{2}{x^{2} + 8 x + 16}$
$\frac{2 x - 1}{4 x^{2} + 8 x - 5} - \frac{3}{4 x^{2} + 20 x + 25}$
$\frac{5 - x}{7 x + x^{2}} - \frac{x + 2}{49 - x^{2}}$
$\frac{2 x}{4 x^{2} + x} - \frac{x + 1}{8 x^{2} + 6 x + 1}$
$\frac{x - 1}{2 x^{2} - 7 x - 4} + \frac{2 x - 1}{x^{2} - 5 x + 4}$
$\frac{2 (x + 3)}{3 x^{2} - 5 x - 2} + \frac{4 - x}{3 x^{2} + 10 x + 3}$
$\frac{x^{2}}{4 + 2 x^{2}} - \frac{2}{x^{4} + 2 x^{2}}$
$\frac{3 x}{4 x^{4} + 6 x^{3}} - \frac{2 x^{2}}{6 x^{3} + 9 x^{2}}$
$\frac{3 x^{2} - 12}{x^{4} - 8 x^{2} + 16} - \frac{x^{2} + 2}{4 - x^{2}}$
$\frac{x^{2}}{2 x^{2} + 1} + \frac{6 x^{2} - 24}{2 x^{4} - 7 x^{2} - 4}$

Given $f$ and $g$ , simplify the sum $f + g$ and difference $f - g .$ Also, state the domain using interval notation.

$f (x) = \frac{1}{x}, g (x) = \frac{5}{x^{2}}$
$f (x) = \frac{1}{x + 2}, g (x) = \frac{2}{x - 1}$
$f (x) = \frac{x - 2}{x + 2}, g (x) = \frac{x + 2}{x - 2}$
$f (x) = \frac{x}{2 x - 1}, g (x) = \frac{2 x}{2 x + 1}$
$f (x) = \frac{6}{3 x^{2} + x}, g (x) = \frac{18}{9 x^{2} + 6 x + 1}$
$f (x) = \frac{x - 1}{x^{2} - 8 x + 16}, g (x) = \frac{x - 2}{x^{2} - 4 x}$
$f (x) = \frac{x}{x^{2} - 25}, g (x) = \frac{x - 1}{x^{2} - 4 x - 5}$
$f (x) = \frac{2 x - 3}{x^{2} - 4}, g (x) = \frac{x}{2 x^{2} + 3 x - 2}$
$f (x) = \frac{1}{3 x^{2} - x - 2}, g (x) = - \frac{1}{4 x^{2} - 3 x - 1}$
$f (x) = \frac{6}{6 x^{2} + 13 x - 5}, g (x) = - \frac{2}{2 x^{2} + x - 10}$

State the restrictions and simplify.

$1 + \frac{3}{x} - \frac{5 x - 1}{x^{2}}$
$4 + \frac{2}{x} - \frac{6 x - 1}{x^{2}}$
$\frac{2 x}{x - 8} - \frac{1}{3 x + 1} - \frac{2 x + 9}{3 x^{2} - 23 x - 8}$
$\frac{4 x}{x - 2} - \frac{10}{3 x + 1} - \frac{19 x + 18}{3 x^{2} - 5 x - 2}$
$\frac{1}{x - 1} + \frac{1}{{(x - 1)}^{2}} - \frac{1}{x^{2} - 1}$
$\frac{1}{x - 2} - \frac{1}{x^{2} - 4} + \frac{1}{{(x - 2)}^{2}}$
$\frac{2 x + 1}{x - 1} - \frac{3 x}{2 x^{2} - 3 x + 1} + \frac{x + 1}{x - 2 x^{2}}$
$\frac{5 x^{2}}{2 x^{2} + 2 x} - \frac{x^{2} - 4 x}{x^{2} - 2 x} + \frac{4 + 2 x^{2}}{4 + 2 x - 2 x^{2}}$
$\frac{x + 2}{2 x (3 x - 2)} + \frac{4 x}{(x - 2) (3 x - 2)} - \frac{3 x + 2}{2 x (x - 2)}$
$\frac{10 x}{x (x - 5)} - \frac{2 x^{2}}{(2 x - 5) (x - 5)} - \frac{5 x}{x (2 x - 5)}$

Simplify the given algebraic expressions. Assume all variable expressions in the denominator are nonzero.

$x^{- 2} + y^{- 2}$
$x^{- 2} + {(2 y)}^{- 2}$
$2 x^{- 1} + y^{- 2}$
$x^{- 2} - 4 y^{- 1}$
$16 x^{- 2} + y^{2}$
$x y^{- 1} - y x^{- 1}$
$3 {(x + y)}^{- 1} + x^{- 2}$
$2 {(x - y)}^{- 2} - {(x - y)}^{- 1}$
$a^{- 2} - {(a + b)}^{- 1}$
${(a - b)}^{- 1} - {(a + b)}^{- 1}$
$x^{- n} + y^{- n}$
$x y^{- n} + y x^{- n}$

Part B: Simplifying Complex Rational Expressions

Simplify. Assume all variable expressions in the denominators are nonzero.

$\frac{\frac{75 x^{2}}{{(x - 3)}^{2}}}{\frac{25 x^{3}}{x - 3}}$
$\frac{\frac{x + 5}{36 x^{3}}}{\frac{{(x + 5)}^{3}}{9 x^{2}}}$
$\frac{\frac{x^{2} - 36}{32 x^{5}}}{\frac{x - 6}{4 x^{3}}}$
$\frac{\frac{x - 8}{56 x^{2}}}{\frac{x^{2} - 64}{7 x^{3}}}$
$\frac{\frac{5 x + 1}{2 x^{2} + x - 10}}{\frac{25 x^{2} + 10 x + 1}{4 x^{2} - 25}}$
$\frac{\frac{4 x^{2} - 27 x - 7}{4 x^{2} - 1}}{\frac{x - 7}{6 x^{2} - x - 1}}$
$\frac{\frac{x^{2} - 4 x - 5}{2 x^{2} + 3 x + 1}}{\frac{x^{2} - 10 x + 25}{2 x^{2} + 7 x + 3}}$
$\frac{\frac{5 x^{2} + 9 x - 2}{x^{2} + 4 x + 4}}{\frac{10 x^{2} + 3 x - 1}{4 x^{2} + 7 x - 2}}$
$\frac{x^{2}}{\frac{1}{5} - \frac{3}{x}}$
$\frac{\frac{4}{x} - 3}{2 x^{2}}$
$\frac{\frac{1}{3} - \frac{1}{x}}{\frac{1}{9} - \frac{1}{x^{2}}}$
$\frac{\frac{2}{5} + \frac{1}{x}}{\frac{4}{25} - \frac{1}{x^{2}}}$
$\frac{\frac{1}{y^{2}} - 36}{6 - \frac{1}{y}}$
$\frac{\frac{1}{5} - \frac{1}{y}}{\frac{1}{y^{2}} - \frac{1}{25}}$
$\frac{1 - \frac{6}{x} + \frac{8}{x^{2}}}{3 - \frac{5}{x} - \frac{2}{x^{2}}}$
$\frac{2 + \frac{13}{x} - \frac{7}{x^{2}}}{3 + \frac{1}{x} - \frac{10}{x^{2}}}$
$\frac{9 - \frac{12}{x} + \frac{4}{x^{2}}}{9 - \frac{4}{x^{2}}}$
$\frac{4 - \frac{25}{x^{2}}}{4 - \frac{8}{x} - \frac{5}{x^{2}}}$
$\frac{\frac{1}{x} + \frac{5}{3 x - 1}}{\frac{2}{3 x - 1} - \frac{1}{x}}$
$\frac{\frac{2}{x - 5} - \frac{1}{x}}{\frac{1}{x} - \frac{3}{x - 5}}$
$\frac{\frac{1}{x + 1} + \frac{2}{x - 2}}{\frac{2}{x - 3} - \frac{1}{x - 2}}$
$\frac{\frac{4}{x + 5} - \frac{1}{x - 3}}{\frac{3}{x - 3} + \frac{1}{2 x - 1}}$
$\frac{\frac{x - 1}{3 x - 1} - \frac{1}{x + 1}}{\frac{x - 1}{x + 1} - \frac{2}{x + 1}}$
$\frac{\frac{x + 1}{3 x + 5} - \frac{1}{x + 3}}{\frac{2}{x + 3} - \frac{x + 1}{x + 3}}$
$\frac{\frac{2 x + 3}{2 x - 3} + \frac{2 x - 3}{2 x + 3}}{\frac{2 x + 3}{2 x - 3} - \frac{2 x - 3}{2 x + 3}}$
$\frac{\frac{x - 1}{x + 1} - \frac{x + 1}{x - 1}}{\frac{x + 1}{x - 1} - \frac{x - 1}{x + 1}}$
$\frac{\frac{1}{2 x + 5} - \frac{1}{2 x - 5} + \frac{4 x}{4 x^{2} - 25}}{\frac{1}{2 x + 5} + \frac{1}{2 x - 5} + \frac{4 x}{4 x^{2} - 25}}$
$\frac{\frac{1}{3 x - 1} + \frac{1}{3 x + 1}}{\frac{3 x}{3 x - 1} - \frac{1}{3 x + 1} - \frac{6 x}{9 x^{2} - 1}}$
$\frac{1}{1 + \frac{1}{1 + \frac{1}{x}}}$
$\frac{\frac{1}{x}}{1 - \frac{1}{1 + \frac{1}{x}}}$
$\frac{\frac{1}{y} - \frac{1}{x}}{\frac{1}{y^{2}} - \frac{1}{x^{2}}}$
$\frac{\frac{2}{y} + \frac{1}{x}}{\frac{4}{y^{2}} - \frac{1}{x^{2}}}$
$\frac{\frac{1}{25 y^{2}} - \frac{1}{x^{2}}}{\frac{1}{x} - \frac{1}{5 y}}$
$\frac{16 y^{2} - \frac{1}{x^{2}}}{\frac{1}{x} - 4 y}$
$\frac{\frac{1}{b} + \frac{1}{a}}{\frac{1}{b^{3}} + \frac{1}{a^{3}}}$
$\frac{\frac{1}{a} - \frac{1}{b}}{\frac{1}{b^{3}} - \frac{1}{a^{3}}}$
$\frac{\frac{x}{y} - \frac{y}{x}}{\frac{1}{y^{2}} - \frac{2}{x y} + \frac{1}{x^{2}}}$
$\frac{\frac{2}{y} - \frac{5}{x}}{4 x - \frac{25 y^{2}}{x}}$
$\frac{x^{- 1} + y^{- 1}}{y^{- 2} - x^{- 2}}$
$\frac{y^{- 2} - 25 x^{- 2}}{5 x^{- 1} - y^{- 1}}$
$\frac{1 - x^{- 1}}{x - x^{- 1}}$
$\frac{16 - x^{- 2}}{x^{- 1} - 4}$
$\frac{1 - 4 x^{- 1} - 21 x^{- 2}}{1 - 2 x^{- 1} - 15 x^{- 2}}$
$\frac{x^{- 1} - 4 {(3 x^{2})}^{- 1}}{3 - 8 x^{- 1} + 16 {(3 x^{2})}^{- 1}}$
$\frac{{(x - 3)}^{- 1} + 2 x^{- 1}}{x^{- 1} - 3 {(x - 3)}^{- 1}}$
$\frac{{(4 x - 5)}^{- 1} + x^{- 2}}{x^{- 2} + {(3 x - 10)}^{- 1}}$
Given $f (x) = \frac{1}{x}$ , simplify $\frac{f (b) - f (a)}{b - a} .$
Given $f (x) = \frac{1}{x - 1}$ , simplify $\frac{f (b) - f (a)}{b - a} .$
Given $f (x) = \frac{1}{x}$ , simplify the difference quotient $\frac{f (x + h) - f (x)}{h} .$
Given $f (x) = \frac{1}{x} + 1$ , simplify the difference quotient $\frac{f (x + h) - f (x)}{h} .$

Part C: Discussion Board

Explain why the domain of a sum of rational functions is the same as the domain of the difference of those functions.
Two methods for simplifying complex rational expressions have been presented in this section. Which of the two methods do you feel is more efficient, and why?

Answers

$\frac{3 x + 2}{3 x + 4}$ ; $x \neq - \frac{4}{3}$
$\frac{1}{x - 6}$ ; $x \neq - \frac{1}{2}, 6$
$\frac{1 - 2 x^{2}}{x}$ ; $x \neq 0$
$\frac{5 x - 4}{x - 1}$ ; $x \neq 1$
$\frac{2 (x + 3)}{(x - 2) (3 x + 4)}$ ; $x \neq - \frac{4}{3}, 2$
$\frac{(x - 1) (x + 2)}{x^{2} (x - 2)}$ ; $x \neq 0, 2$
$\frac{2 x - 7}{x (x - 7)}$ ; $x \neq 0, 7$
$\frac{x^{2} - 8 x - 5}{(x + 5) {(x - 5)}^{2}}$ ; $x \neq \pm 5$
$\frac{x + 2}{{(x + 4)}^{2}}$ ; $x \neq 0, - 4$
$\frac{7 (5 - 2 x)}{x (7 + x) (7 - x)}$ ; $x \neq - 7, 0, 7$
$\frac{x (5 x - 2)}{(x - 4) (x - 1) (2 x + 1)}$ ; $x \neq - \frac{1}{2}, 1, 4$
$\frac{x^{2} - 2}{2 x^{2}}$ ; $x \neq 0$
$\frac{x^{2} + 5}{(x + 2) (x - 2)}$ ; $x \neq \pm 2$
$(f + g) (x) = \frac{x + 5}{x^{2}}$ ; $(f - g) (x) = \frac{x - 5}{x^{2}}$ ; Domain: $(- \infty, 0) \cup (0, \infty)$
$(f + g) (x) = \frac{2 (x^{2} + 4)}{(x + 2) (x - 2)}$ ; $(f - g) (x) = - \frac{8 x}{(x + 2) (x - 2)}$ ; Domain: $(- \infty, - 2) \cup (- 2, 2) \cup (2, \infty)$
$(f + g) (x) = \frac{6 (6 x + 1)}{x {(3 x + 1)}^{2}}$ ; $(f - g) (x) = \frac{6}{x {(3 x + 1)}^{2}}$ ; Domain: $(- \infty, - \frac{1}{3}) \cup (- \frac{1}{3}, 0) \cup (0, \infty)$
$(f + g) (x) = \frac{2 x^{2} + 5 x - 5}{(x + 1) (x + 5) (x - 5)}$ ; $(f - g) (x) = - \frac{3 x - 5}{(x + 1) (x + 5) (x - 5)}$ ; Domain: $(- \infty, - 5) \cup (- 5, - 1) \cup (- 1, 5) \cup (5, \infty)$
$(f + g) (x) = \frac{1}{(3 x + 2) (4 x + 1)}$ ; $(f - g) (x) = - \frac{7 x + 3}{(x - 1) (3 x + 2) (4 x + 1)}$ ; Domain: $(- \infty, - \frac{2}{3}) \cup (- \frac{2}{3}, \frac{1}{4}) \cup (\frac{1}{4}, 1) \cup (1, \infty)$
$\frac{{(x - 1)}^{2}}{x^{2}}$ ; $x \neq 0$
$\frac{2 x - 1}{x - 8}$ ; $x \neq - \frac{1}{3}, 8$
$\frac{x^{2} + 1}{{(x - 1)}^{2} (x + 1)}$ ; $x \neq \pm 1$
$\frac{2 x + 1}{x}$ ; $x \neq 0, \frac{1}{2}, 1$
0; $x \neq 0, \frac{2}{3}, 2$
$\frac{y^{2} + x^{2}}{x^{2} y^{2}}$
$\frac{x + 2 y^{2}}{x y^{2}}$
$\frac{x^{2} y^{2} + 16}{x^{2}}$
$\frac{3 x^{2} + x + y}{x^{2} (x + y)}$
$\frac{a + b - a^{2}}{a^{2} (a + b)}$
$\frac{x^{n} + y^{n}}{x^{n} y^{n}}$

$\frac{3}{x (x - 3)}$
$\frac{x + 6}{8 x^{2}}$
$\frac{2 x - 5}{(x - 2) (5 x + 1)}$
$\frac{x + 3}{x - 5}$
$\frac{5 x^{3}}{x - 15}$
$\frac{3 x}{x + 3}$
$- \frac{6 y + 1}{y}$
$\frac{x - 4}{3 x + 1}$
$\frac{3 x - 2}{3 x + 2}$
$- \frac{8 x - 1}{x - 1}$
$\frac{3 x (x - 3)}{(x + 1) (x - 1)}$
$\frac{x}{3 x - 1}$
$\frac{4 x^{2} + 9}{12 x}$
$\frac{2 x - 5}{4 x}$
$\frac{x + 1}{2 x + 1}$
$\frac{x y}{x + y}$
$- \frac{x + 5 y}{5 x y}$
$\frac{a^{2} b^{2}}{a^{2} - a b + b^{2}}$
$\frac{x y (x + y)}{x - y}$
$\frac{x y}{x - y}$
$\frac{1}{x + 1}$
$\frac{x - 7}{x - 5}$
$- \frac{3 (x - 2)}{2 x + 3}$
$- \frac{1}{a b}$
$- \frac{1}{x (x + h)}$

Answer may vary