# Topic 3 Algebra of Rational Expressions

## 3.1 Is There Enough Time

Matt is kayaking upstream on a river with his best effort. After 30 minutes, he received an emergency call and has to return in 20 minute. The speed of the current of the river is 1 mph. Under normal condition, a paddler’s average paddling speed is between 2 and 5 mph. Do you think Matt can return on time? Why?

A construction team is building a house. After half of the work was done, to expedite the construction process, the another team joins in the construction. The first team normally takes 7-10 days to build a hours. The second team normally takes 2 extra days to build a house. How many days it takes to build the house?

## 3.2 Rational Expressions

Let \(p\) and \(q\) be polynomial functions of \(x\) and \(p\) is not a constant function. We call the function \(r(x)=\frac{p(x)}{q(x)}\) a ** rational function**. The domain of \(r\) is \(\{x\mid Q(x)\neq 0\}\).
The expression \(\frac{p(x)}{q(x)}\) is called a

**, the polynomial \(q(x)\) is called the**

*rational expression***, and the polynomial \(q(x)\) is called the**

*numerator***. A rational expression is**

*denominator***if the numerator and the denominator have no common factor other than \(1\).**

*simplified*Let \(p(x)\), \(q(x)\) be polynomials with \(q(x)\neq 0\) and \(c(x)\) be a nonzero expression. Then \[ \dfrac{~p(x)\cdot c(x)~}{~q(x)\cdot c(x)~}=\dfrac{~p(x)~}{~q(x)~}. \]

**Example 3.1 **
Simplify \(\dfrac{x^2+4x+3}{x^2+3x+2}\).

*Solution. *

Factor both the top and the bottom. \[ \dfrac{x^2+4x+3}{x^2+3x+2}=\dfrac{(x+1)(x+3)}{(x+1)(x+2)}. \]

Divide out common factors. \[ \dfrac{(x+1)(x+3)}{(x+1)(x+2)}=\dfrac{x+3}{x+2}. \]

**Example 3.2 **
Simplify \(\dfrac{2x^2-x-3}{2x^2-3x-5}\).

*Solution. *

Factor both the top and the bottom. \[\dfrac{2x^2-x-3}{2x^2-3x-5}=\dfrac{(x+1)(2x-3)}{(x+1)(2x-5)}.\]

Divide out common factors. \[\dfrac{(x+1)(2x-3)}{(x+1)(2x-5)}=\dfrac{2x-3}{2x-5}.\]

## 3.3 Multiplying Rational Expressions

If \(p\), \(q\), \(s\), \(t\) are polynomials with \(q\neq 0\) and \(t\neq 0\), then \[ \dfrac{~p~}{~q~}\cdot\dfrac{~s~}{~t~}=\dfrac{~ps~}{~qt~}. \]

**Example 3.3 **
Multiply and then simplify.
\[\dfrac{3x^2}{x^2+x-6}\cdot\dfrac{x^2-4}{6x}.\]

*Solution. *

Factor numerators and denominators. \[ \dfrac{3x^2}{x^2+x-6}\cdot\dfrac{x^2-4}{6x}=\dfrac{3\cdot x\cdot x}{(x-2)(x+3)}\cdot\dfrac{(x-2)(x+2)}{2\cdot3\cdot x} \]

Multiply and simplify. \[ \dfrac{\cancel{3}\cdot \cancel{x}\cdot x\cdot\cancel{(x-2)}(x+2)}{\cancel{(x-2)}(x+3)\cdot 2\cdot\cancel{3}\cdot \cancel{x}}=\dfrac{x(x+2)}{2(x+3)} \]

**Example 3.4 **
Multiply and then simplify.
\[
\dfrac{3x^2-8x-3}{x^2+8x+16}\cdot\dfrac{x^2-16}{5x^2-14x-3}.
\]

*Solution. *

\[ \dfrac{3x^2-8x-3}{x^2+8x+16}\cdot\dfrac{x^2-16}{5x^2-14x-3} =\dfrac{(3x+1)\cancel{(x-3)}\cancel{(x+4)}(x-4)}{\cancel{(x+4)}(x+4)(5x+1)\cancel{(x-3)}} =\dfrac{(3x+1)(x-4)}{(x+4)(5x+1)} \]

## 3.4 Dividing Rational Expressions

If \(p\), \(q\), \(s\), \(t\) are polynomials where \(q\neq 0\), \(s\neq 0\) and \(t\neq 0\), then \[ \dfrac{~p~}{~q~}\div\dfrac{~s~}{~t~}=\dfrac{~p~}{~q~}\cdot\dfrac{~t~}{~s~}=\dfrac{~pt~}{~qs~}. \]

**Example 3.5 **
Divide and then simplify.

\[
\dfrac{2x+6}{x^2-6x-7}\div \dfrac{6x-2}{2x^2-x-3}.
\]

*Solution. *

- Rewrite as a multiplication.

\[ \dfrac{2x+6}{x^2-6x-7}\div \dfrac{6x-2}{2x^2-x-3}=\dfrac{2x+6}{x^2-6x-7}\cdot \dfrac{2x^2-x-3}{6x-2} \] - Factor and simplify.

\[ \dfrac{2x+6}{x^2-6x-7}\cdot\dfrac{2x^2-x-3}{6x-2} =\dfrac{\cancel{2}(x+3)\cancel{(x+1)}(2x-3)}{\cancel{2}\cancel{(x+1)}(x-7)(3x-1)} =\dfrac{(x+3)(2x-3)}{(x-7)(3x-1)} \]

## 3.5 Adding or Subtracting Rational Expressions with the Same Denominator

If \(P\), \(Q\) and \(R\) are polynomials with \(R\neq 0\), then \[ \dfrac{~P~}{~R~}+\dfrac{~Q~}{~R~}=\dfrac{~P+Q~}{~R~}\qquad \text{and} \qquad \dfrac{~P~}{~R~}-\dfrac{~Q~}{~R~}=\dfrac{~P-Q~}{~R~}. \]

**Example 3.6 **
Add and simplify

\[
\dfrac{x^2+4}{x^2+3x+2}+\dfrac{5x}{x^2+3x+2}.
\]

*Solution. *

- Determine if the rational expressions have the same denominator. If so, the new numerator will be the sum/difference of the numerators. \[ \dfrac{x^2+4}{x^2+3x+2}+\dfrac{5x}{x^2+3x+2}=\dfrac{x^2+5x+4}{x^2+3x+2}. \]
- Simplify the resulting rational expression. \[ \dfrac{x^2+5x+4}{x^2+3x+2}=\dfrac{(x+1)(x+4)}{(x+1)(x+2)}=\dfrac{x+4}{x+2}. \]

**Example 3.7 **
Subtract and simplify \(\dfrac{2x^2}{2x^2-x-3}-\dfrac{3x+5}{2x^2-x-3}\).

*Solution. *

\[ \begin{aligned} \dfrac{2x^2}{2x^2-x-3}-\dfrac{3x+5}{2x^2-x-3}=&\dfrac{2x^2-3x-5}{2x^2-x-3}\\ =&\dfrac{(2x-5)(x+1)}{(2x-3)(x+1)}\\ =&\dfrac{2x-5}{2x-3}. \end{aligned} \]

## 3.6 Adding or Subtracting Rational Expressions with Different Denominators

To add or subtract rational expressions with different denominators, we need to rewrite the rational expressions to equivalent rational expressions with the same denominator, say the LCD.

**Equivalent Reduction.**

What if all denominators are the same? How to make denominators the same? Reducing the problem to an easier one using equivalent operations helps solve the problem.

**Example 3.8 **
Find the LCD of \(\dfrac{3}{x^2-x-6}\) and \(\dfrac{6}{x^2-4}\).

*Solution. *

Factor each denominator. \[ x^2-x-6=(x+2)(x-3)\qquad x^2-4=(x-2)(x+2) \]

List the factors of the first denominator and add unlisted factors of the second factor to get the final list.

First list \((x+2)\) \((x-3)\) Second list \((x+2)\) \((x-2)\) Final list \((x+2)\) \((x-3)\) \((x-2)\) The LCD is the product of factors in the final list.

\[(x+2)(x-3)(x-2).\]

**Example 3.9 **
Subtract and simplify

\[\dfrac{x-3}{x^2-2x-8}- \dfrac{1}{x^2-4}\]

*Solution. *

- Find the LCD.
First factor denominators.

\[x^2-2x-8=(x+2)(x-4)\] \[x^2-4=(x-2)(x+2)\]Then using the table to find the final list of factors of the LCD.

First list \((x+2)\) \((x-4)\) Second list \((x+2)\) \((x-2)\) Final list \((x+2)\) \((x-2)\) \((x-4)\)

- Rewrite each rational expression into an equivalent expression with the LCD as the new denominator.

\[ \dfrac{x-3}{x^2-2x-8}- \dfrac{1}{x^2-4}=\dfrac{(x-3)(x-2)}{(x+2)(x-2)(x-4)}-\dfrac{(x-4)}{(x+2)(x-2)(x-4)} \] - Subtract and simplify.

\[ \dfrac{(x-3)(x-2)-(x-4)}{(x+2)(x-2)(x-4)}=\dfrac{(x^2-5x+6)-(x-4)}{(x+2)(x-2)(x-4)}=\dfrac{x^2-6x+10}{(x+2)(x-2)(x-4)} \]

## 3.7 Simplifying Complex Rational Expressions

A ** complex rational expression** is a rational expression whose denominator or numerator contains a rational expression.

A complex rational expression is equivalent to the quotient of its numerator by its denominator. That suggests the following strategy to simplify a complex rational expression.

**Simplify and Change the Viewpoint.** A complex rational expression is a quotient of two rational expressions. You may rewrite it as an multiplication by flipping the denominator. However, it’s better to simply the numerator and denominator or you won’t see a good looking new expression.

**Example 3.10 **
Simplify
\[
\dfrac{~\dfrac{2x-1}{x^2-1}+\dfrac{x-1}{x+1}~}{~\dfrac{x+1}{x-1}-\dfrac{1}{x^2-1}~}
\]

*Solution. *

- Simplify the numerator and the denominator. \[ \begin{aligned} \dfrac{~\dfrac{2x-1}{x^2-1}+\dfrac{x-1}{x+1}~}{~\dfrac{x+1}{x-1}-\dfrac{1}{x^2-1}~} =&\dfrac{ ~\dfrac{2x-1}{(x-1)(x+1)}+\dfrac{(x-1)(x-1)}{(x-1)(x+1)}~ }{~ \dfrac{(x+1)(x+1)}{(x-1)(x+1)}-\dfrac{1}{(x-1)(x+1)}~ }\\[5pt] =&\dfrac{ ~\dfrac{(2x-1)+(x-1)(x-1)}{(x-1)(x+1)}~ }{ ~\dfrac{(x+1)(x+1)-1}{(x-1)(x+1)}~ }\\[5pt] =&\dfrac{ ~\dfrac{(2x-1)+(x^2-2x+1)}{(x-1)(x+1)}~ }{ ~\dfrac{(x^2+2x+1)-1}{(x-1)(x+1)}~ }\\[5pt] =&\dfrac{ ~\dfrac{x^2}{(x-1)(x+1)}~ }{ ~\dfrac{x^2+2x}{(x-1)(x+1)}~ } \end{aligned} \]
- Rewrite as a product. \[ \dfrac{ ~\dfrac{x^2}{(x-1)(x+1)}~ }{ ~\dfrac{x^2+2x}{(x-1)(x+1)}~ } =\dfrac{x^2}{(x-1)(x+1)}\cdot \dfrac{(x-1)(x+1)}{x^2+2x} \]
- Multiply and simplify. \[ \begin{aligned} \dfrac{x^2}{(x-1)(x+1)}\cdot \dfrac{(x-1)(x+1)}{x^2+2x} =&\dfrac{x\cdot x}{(x-1)(x+1)}\cdot \dfrac{(x-1)(x+1)}{x(x+2)}\\[5pt] =& \dfrac{\bcancel{x}x\cancel{(x-1)(x+1)}}{\bcancel{x}(x+2)\cancel{(x-1)(x+1)}}\\[5pt] =& \dfrac{x}{x+2} \end{aligned} \]

Another way to simplify a complex rational expression is to multiply the LCD to both the denominator and numerator and then simplify.

**Example 3.11 **
Simplify
\[\dfrac{~\dfrac{x+1}{x-1}+\dfrac{x-1}{x+1}~}{~\dfrac{x+1}{x-1}-\dfrac{x-1}{x+1}~}\]

*Solution. *

- Find the LCD of all denominators.

In this case, the LCD is \((x-1)(x+1)\). - Multiply the complex rational expression by \(\dfrac{(x-1)(x+1)}{(x-1)(x+1)}\) and simplify. \[ \begin{aligned} \dfrac{~\dfrac{x+1}{x-1}+\dfrac{x-1}{x+1}~}{~\dfrac{x+1}{x-1}-\dfrac{x-1}{x+1}~} =&\dfrac{~\dfrac{x+1}{x-1}+\dfrac{x-1}{x+1}~}{~\dfrac{x+1}{x-1}-\dfrac{x-1}{x+1}~}\cdot \dfrac{(x-1)(x+1)}{(x-1)(x+1)}\\[5pt] =& \dfrac{~(x-1)(x+1)\left(\dfrac{x+1}{x-1}+\dfrac{x-1}{x+1}\right)~}{~(x-1)(x+1)\left(\dfrac{x+1}{x-1}-\dfrac{x-1}{x+1}\right)~}\\[5pt] =& \dfrac{~(x+1)^2+(x-1)^2~}{~(x+1)^2-(x-1)^2~}\\[5pt] =& \dfrac{x^2+1}{2x} \end{aligned} \]

## 3.8 Practice

**Problem 3.1 **
Simplify.

\(\dfrac{3x^2-x-4}{x+1}\)

\(\dfrac{2x^2-x-3}{2x^2+3x+1}\)

\(\dfrac{x^2-9}{3x^2-8x-3}\)

**Problem 3.2 **
Multiply and simplify.

\(\dfrac{x+5}{x+4}\cdot\dfrac{x^2+3x-4}{x^2-25}\)

\(\dfrac{3x^2-2x}{x+2}\cdot\dfrac{3x^2-4x-4}{9x^2-4}\)

\(\dfrac{6y-2}{3-y}\cdot\dfrac{y^2-6y+9}{3y^2-y}\)

**Problem 3.3 **
Divide and simplify.

\(\dfrac{9x^2-49}{6}\div\dfrac{3x^2-x-14}{2x+4}\)

\(\dfrac{x^2+3x-10}{2x-2}\div\dfrac{x^2-5x+6}{x^2-4x+3}\)

\(\dfrac{y-x}{xy}\div\dfrac{x^2-y^2}{y^2}\)

**Problem 3.4 **
Simplify.
\[
\frac{-x^2+11x-18}{x^2-4x+4}\div \frac{x^2-5x-36}{x^2-7x+12}\cdot \frac{2x^2+7x-4}{x^2+2x-15}
\]

**Problem 3.5 **
Add/subtract and simplify.

- \(\dfrac{x^2+2x-2}{x^2+2x-15}+\dfrac{5x+12}{x^2+2x-15}\)
- \(\dfrac{3x-10}{x^2-25}-\dfrac{2x-15}{x^2-25}\)
- \(\dfrac{4}{(x-3)(x+2)}+\dfrac{3x-2}{x^2-x-6}\)

**Problem 3.6 **
Find the LCD of rational expressions.

- \(\dfrac{2x}{2x^2-5x-3}\)and \(\dfrac{x-1}{x^2-x-6}\)
- \(\dfrac{9}{7x^2-28x}\) and \(\dfrac{2}{x^2-8x+16}\)

**Problem 3.7 **
Add and simplify.

- \(\dfrac{x}{x+1}+\dfrac{x-1}{x+2}\)
- \(\dfrac{x+2}{2x^2-x-3}+\dfrac{1}{x^2+3x+2}\)
- \(\dfrac{4}{x-3}+\dfrac{3x-2}{x^2-x-6}\)

**Problem 3.8 **
Subtract and simplify.

- \(\dfrac{3x+5}{x^2-7x+12}-\dfrac{3}{x-3}\)
- \(\dfrac{y}{y^2-5y-6}-\dfrac{7}{y^2-4y-5}\)
- \(\dfrac{2x-3}{x^2+3x-10}-\dfrac{x+2}{x^2+2x-8}\)

**Problem 3.9 **
Simplify.
\[
\frac{x+11}{7x^2-2x-5}+\frac{x-2}{x-1}-\frac{x}{7x+5}
\]

**Problem 3.10 **
Subtract and simplify.
\[
\frac{x-1}{x^2-3x}+\frac{4}{x^2-2\:x-3}-\frac{1}{x\left(x+1\right)}
\]

**Problem 3.11 **
Simplify.

- \(\dfrac{~1+\dfrac{2}{x}~}{~1-\dfrac{2}{x}~}\)
- \(\dfrac{~\dfrac{1}{x^2}-1~}{~\dfrac{1}{x^2}-\dfrac{1}{x}~}\)

**Problem 3.12 **
Simplify.

- \(\dfrac{~\dfrac{x^2-y^2}{y^2}~}{~\dfrac1x-\dfrac{1}{y}~}\)
- \(\dfrac{~\dfrac{2}{(x+1)^2}-\dfrac{1}{x+1}~}{~1-\dfrac{4}{(x+1)^2}~}\)

**Problem 3.13 **
Simplify.

- \(\dfrac{~\dfrac{5x}{x^2-x-6}~}{~\dfrac2{x+2}+\dfrac{3}{x-3}~}\)
- \(\dfrac{~\dfrac{x+1}{x-1}+\dfrac{x-1}{x+1}~}{~\dfrac{x+1}{x-1}-\dfrac{x-1}{x+1}~}\)

**Problem 3.14 **
Tim and Jim refill their cars at the same gas station twice last month. Each time Tim got $20 gas and Jim got 8 gallon. Suppose they refill their cars on same days. The price was $2.5 per gallon the first time. The price on the second time changed. Can you find out who had the better average price?