ORCCA Rational Functions and Equations Chapter Review

To find the number of zombies after $2$ days, we locate the point $(2, 3637.27) .$ Since we can only have a whole number of zombies, we round to $3,637$ zombies.
To find the number of days it will take for the zombie population reach $20,000,$ we locate the point $(19.999, 20000)$ so it will take about 20 days.
When we look far to the right on the graph using technology we can see that the population will level off at about $40,000$ zombies.

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Subsection 12.6.2 Multiplication and Division of Rational Expressions

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In Section 2 we covered how to simplify rational expressions. It is very important to list any domain restrictions from factors that are canceled. We also multiplied and divided rational expressions.

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Example 12.6.3. Simplifying Rational Expressions.

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Simplify the expression

\frac{8 t + 4 t^{2} - 12 t^{3}}{1 - t} .

Explanation.

To begin simplifying this expression, we will rewrite each polynomial in descending order. Then we’ll factor out the GCF, including the constant

- 1

from both the numerator and denominator because their leading terms are negative.

\begin{aligned} \frac{8 t + 4 t^{2} - 12 t^{3}}{1 - t} & = \frac{- 12 t^{3} + 4 t^{2} + 8 t}{- t + 1} \\ = \frac{- 4 t (3 t^{2} - t - 2)}{- (t - 1)} \\ = \frac{- 4 t (3 t + 2) (t - 1)}{- (t - 1)} \\ = \frac{- 4 t (3 t + 2) (t - 1)}{- (t - 1)} \\ = \frac{- 4 t (3 t + 2)}{- 1}, for t \neq 1 \\ = 4 t (3 t + 2), for t \neq 1 \end{aligned}

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Example 12.6.4. Multiplication of Rational Functions and Expressions.

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Multiply the rational expressions:

\frac{r^{3} s}{4 t} \cdot \frac{2 t^{2}}{r^{2} s^{3}} .

Explanation.

Note that we won’t need to factor anything in this problem, and can simply multiply across and then simplify. With multivariable expressions, this textbook ignores domain restrictions.

\begin{aligned} \frac{r^{3} s}{4 t} \cdot \frac{2 t^{2}}{r^{2} s^{3}} & = \frac{r^{3} s \cdot 2 t^{2}}{4 t \cdot r^{2} s^{3}} \\ = \frac{2 r^{3} s t^{2}}{4 r^{2} s^{3} t} \\ = \frac{r t}{2 s^{2}} \end{aligned}

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Example 12.6.5. Division of Rational Functions and Expressions.

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Divide the rational expressions:

\frac{2 x^{2} + 8 x y}{x^{2} - 4 x + 3} \div \frac{x^{3} + 4 x^{2} y}{x^{2} + 4 x - 5} .

Explanation.

To divide rational expressions, we multiply by the reciprocal of the second fraction. Then we will factor and cancel any common factors. With multivariable expressions, this textbook ignores domain restrictions.

\begin{aligned} \frac{2 x^{2} + 8 x y}{x^{2} - 4 x + 3} \div \frac{x^{3} + 4 x^{2} y}{x^{2} + 4 x - 5} & = \frac{2 x^{2} + 8 x y}{x^{2} - 4 x + 3} \cdot \frac{x^{2} + 4 x - 5}{x^{3} + 4 x^{2} y} \\ = \frac{2 x (x + 4 y)}{(x - 1) (x - 3)} \cdot \frac{(x - 1) (x + 5)}{x^{2} (x + 4 y)} \\ = \frac{2 x}{x - 3} \cdot \frac{x + 5}{x^{2}} \\ = \frac{2 (x + 5)}{x (x - 3)} \end{aligned}

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Subsection 12.6.3 Addition and Subtraction of Rational Expressions

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In Section 3 we covered how to add and subtract rational expressions.

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Example 12.6.6. Addition and Subtraction of Rational Expressions with the Same Denominator.

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Add the rational expressions:

\frac{5 x}{x + 5} + \frac{25}{x + 5} .

Explanation.

These expressions already have a common denominator:

\begin{aligned} \frac{5 x}{x + 5} + \frac{25}{x + 5} & = \frac{5 x + 25}{x + 5} \\ = \frac{5 (x + 5)}{x + 5} \\ = \frac{5}{1}, for x \neq - 5 \\ = 5, for x \neq - 5 \end{aligned}

Note that we didn’t stop at

\frac{5 x + 25}{x + 5} .

If possible, we must simplify the numerator and denominator.

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Example 12.6.7. Addition and Subtraction of Rational Expressions with Different Denominators.

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Add and subtract the rational expressions:

\frac{6 y}{y + 2} + \frac{y}{y - 2} - 7

Explanation.

The denominators can’t be factored, so we’ll find the least common denominator and build each expression to that denominator. Then we will be able to combine the numerators and simplify the expression.

\begin{aligned} \frac{6 y}{y + 2} + \frac{y}{y - 2} - 7 & = \frac{6 y}{y + 2} \cdot \frac{y - 2}{y - 2} + \frac{y}{y - 2} \cdot \frac{y + 2}{y + 2} - 7 \cdot \frac{(y - 2) (y + 2)}{(y - 2) (y + 2)} \\ = \frac{6 y (y - 2)}{(y - 2) (y + 2)} + \frac{y (y + 2)}{(y - 2) (y + 2)} - \frac{7 (y - 2) (y + 2)}{(y - 2) (y + 2)} \\ = \frac{6 y^{2} - 12 y + y^{2} + 2 y - \overset{↓}{(} 7 (y^{2} - 4) \overset{↓}{)}}{(y - 2) (y + 2)} \\ = \frac{6 y^{2} - 12 y + y^{2} + 2 y - 7 y^{2} + 28}{(y - 2) (y + 2)} \\ = \frac{- 10 y + 28}{(y - 2) (y + 2)} \\ = \frac{- 2 (5 y - 14)}{(y - 2) (y + 2)} \end{aligned}

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Subsection 12.6.4 Complex Fractions

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In Section 4 we covered how to simplify a rational expression that has fractions in the numerator and/or denominator.

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Example 12.6.8. Simplifying Complex Fractions.

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Simplify the complex fraction

\frac{\frac{2 t}{t^{2} - 9} + 3}{\frac{6}{t + 3} + \frac{1}{t - 3}} .

Explanation.

First, we check all quadratic polynomials to see if they can be factored and factor them:

\frac{\frac{2 t}{t^{2} - 9} + 3}{\frac{6}{t + 3} + \frac{1}{t - 3}} = \frac{\frac{2 t}{(t - 3) (t + 3)} + 3}{\frac{6}{t + 3} + \frac{1}{t - 3}}

Next, we identify the common denominator of the three fractions, which is

(t + 3) (t - 3) .

We then multiply the main numerator and denominator by that expression:

\begin{aligned} \frac{\frac{2 t}{(t - 3) (t + 3)} + 3}{\frac{6}{t + 3} + \frac{1}{t - 3}} & = \frac{\frac{2 t}{(t - 3) (t + 3)} + 3}{\frac{6}{t + 3} + \frac{1}{t - 3}} \cdot \frac{(t - 3) (t + 3)}{(t - 3) (t + 3)} \\ = \frac{\frac{2 t}{(t - 3) (t + 3)} (t - 3) (t + 3) + 3 (t - 3) (t + 3)}{\frac{6}{t + 3} (t - 3) (t + 3) + \frac{1}{t - 3} (t - 3) (t + 3)} \\ = \frac{2 t + 3 (t - 3) (t + 3)}{6 (t - 3) + 1 (t + 3)} for t \neq - 3, t \neq 3 \\ = \frac{2 t + 3 (t^{2} - 9)}{6 t - 18 + t + 3} for t \neq - 3, t \neq 3 \\ = \frac{2 t + 3 t^{2} - 27}{7 t - 15} for t \neq - 3, t \neq 3 \\ = \frac{3 t^{2} + 2 t - 27}{7 t - 15} for t \neq - 3, t \neq 3 \end{aligned}

Both the numerator and denominator are prime polynomials so this expression can neither factor nor simplify any further.

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Subsection 12.6.5 Solving Rational Equations

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In Section 5 we covered how to solve rational equations. We looked at rate problems, solved for a specified variable and used technology to solve rational equations.

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Example 12.6.9. Solving Rational Equations.

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Two pipes are being used to fill a large tank. Pipe B can fill the tank twice as fast as Pipe A can. When both pipes are turned on, it takes 12 hours to fill the tank. Write and solve a rational equation to answer the following questions:

If only Pipe A is turned on, how many hours would it take to fill the tank?
If only Pipe B is turned on, how many hours would it take to fill the tank?

Explanation.

Since both pipes can fill the tank in

12

hours, they fill

\frac{1}{12}

of the tank together each hour. We will let

a

represent the number of hours it takes pipe A to fill the tank alone, so pipe A will fill

\frac{1}{a}

of the tank each hour. Pipe B can fill the tank twice as fast so it fills

2 \cdot \frac{1}{a}

of the tank each hour or

\frac{2}{a} .

When they are both turned on, they fill

\frac{1}{a} + \frac{2}{a}

of the tank each hour.

Now we can write this equation:

\frac{1}{a} + \frac{2}{a} = \frac{1}{12}

To clear away denominators, we multiply both sides of the equation by the common denominator of

12

and

a,

which is

12 a :

\begin{aligned} \frac{1}{a} + \frac{2}{a} & = \frac{1}{12} \\ 12 a \cdot (\frac{1}{a} + \frac{2}{a}) & = 12 a \cdot \frac{1}{12} \\ 12 a \cdot \frac{1}{a} + 12 a \cdot \frac{2}{a} & = 12 a \cdot \frac{1}{12} \\ 12 + 24 & = a \\ 36 & = a \\ a & = 36 \end{aligned}

The possible solution

a = 36

should be checked

\begin{aligned} \frac{1}{36} + \frac{2}{36} & \overset{?}{=} \frac{1}{12} \\ \frac{3}{36} & \overset{✓}{=} \frac{1}{12} \end{aligned}

So it is a solution.

If only Pipe A is turned on, it would take $36$ hours to fill the tank.
Since Pipe B can fill the tank twice as fast, it would take half the time, or $18$ hours to fill the tank.

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Example 12.6.10. Solving Rational Equations for a Specific Variable.

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Solve the rational equation

y = \frac{2 x + 5}{3 x - 1}

for

x .

Explanation.

To get the

x

out of the denominator, our first step will be to multiply each side by the LCD, which is

3 x - 1 .

Then we’ll isolate all terms containing

x,

factor out

x,

and then finish solving for that variable.

\begin{aligned} y & = \frac{2 x + 5}{3 x - 1} \\ y \cdot (3 x - 1) & = \frac{2 x + 5}{3 x - 1} \cdot (3 x - 1) \\ 3 x y - y & = 2 x + 5 \\ 3 x y & = 2 x + 5 + y \\ 3 x y - 2 x & = y + 5 \\ x (3 y - 2) & = y + 5 \\ \frac{x (3 y - 2)}{3 y - 2} & = \frac{y + 5}{3 y - 2} \\ x & = \frac{y + 5}{3 y - 2} \end{aligned}

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Example 12.6.11. Solving Rational Equations Using Technology.

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Solve the equation

\frac{1}{x + 2} + 1 = \frac{10 x}{x^{2} + 5}

using graphing technology.

Explanation.

We will define

f (x) = \frac{1}{x + 2} + 1

and

g (x) = \frac{10 x}{x^{2} + 5},

and then find a window where we can see all of the points of intersection.

Since the two functions intersect at approximately

(- 2.309, - 2.235),

(0.76, 1.362)

and

(8.549, 1.095),

the solutions to

\frac{1}{x + 2} + 1 = \frac{10 x}{x^{2} + 5}

are approximately

- 2.309,

0.76

and

8.549 .

The solution set is approximately

{- 2.309 \dots, 0.76 \dots, 8.549 \dots} .

Figure 12.6.12. Graph of

f (x) = \frac{1}{x + 2} + 1

and

g (x) = \frac{10 x}{x^{2} + 5}

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A function is graphed.

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This function has domain and range .

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2.

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A function is graphed.

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This function has domain and range .

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3.

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The population of deer in a forest can be modeled by

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P (x) = \frac{2700 x + 2450}{6 x + 7}

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where

x

is the number of years since the year 1900.

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How many deer lived in this forest in the year 1900?
How many deer live in this forest $20$ years after 1900?
How many years since 1900 was it (or will it be) when the deer population was (or will be) $441 ?$
Use a calculator or graphing calculator to answer this question: As time goes on, the population levels off at about how many deer?

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4.

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The population of deer in a forest can be modeled by

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P (x) = \frac{2030 x + 1560}{7 x + 4}

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where

x

is the number of years since the year 1900.

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How many deer lived in this forest in the year 1900?
How many deer live in this forest $23$ years after 1900?
How many years since 1900 was it (or will it be) when the deer population was (or will be) $297 ?$
Use a calculator or graphing calculator to answer this question: As time goes on, the population levels off at about how many deer?

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5.

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In a certain store, cashiers can serve

60

customers per hour on average. If

x

customers arrive at the store in a given hour, then the average number of customers

C

waiting in line can be modeled by the function

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C (x) = \frac{x^{2}}{3600 - 60 x}

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where

x < 60 .

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Answer the following questions with a graphing calculator. Round your answers to integers.

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If $48$ customers arrived in the store in the past hour, there are approximately customers waiting in line.
If there are $10$ customers waiting in line, approximately customers arrived in the past hour.

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6.

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In a certain store, cashiers can serve

60

customers per hour on average. If

x

customers arrive at the store in a given hour, then the average number of customers

C

waiting in line can be modeled by the function

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C (x) = \frac{x^{2}}{3600 - 60 x}

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where

x < 60 .

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Answer the following questions with a graphing calculator. Round your answers to integers.

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If $44$ customers arrived in the store in the past hour, there are approximately customers waiting in line.
If there are $10$ customers waiting in line, approximately customers arrived in the past hour.

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7.

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The concentration of a drug in a patient’s blood stream, in milligrams per liter, can be modeled by the function

C (t) = \frac{9 t}{t^{2} + 6},

where

t

is the number of hours since the drug is injected. Answer the following question with technology.

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hours since injection, the drug’s concentration is at the maximum value of milligrams per liter.

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8.

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The concentration of a drug in a patient’s blood stream, in milligrams per liter, can be modeled by the function

C (t) = \frac{2 t}{t^{2} + 4},

where

t

is the number of hours since the drug is injected. Answer the following question with technology.

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Simplify the function formula, and if applicable, write the restricted domain.

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f (y) = \frac{y^{4} - 6 y^{3} + 9 y^{2}}{5 y^{4} - 14 y^{3} - 3 y^{2}}

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Reduced

f (y) =

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12.

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Simplify the function formula, and if applicable, write the restricted domain.

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Simplify this expression.

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\frac{x^{5}}{x^{2} r - 5 x} \div \frac{1}{x^{2} r^{2} - 11 x r + 30} =

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\frac{t^{2} - 4}{t^{2} + t} - \frac{t - 4}{t} =

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\frac{5 y x}{y^{2} - 11 y x + 24 x^{2}} - \frac{y}{y - 8 x} =

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\frac{\frac{5}{y}}{20 + 5 r} =

Use technology to solve the equation

2 x - \frac{1}{x + 4} = \frac{3}{x + 6} .

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50.

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Use technology to solve the equation

\frac{1}{x^{2} - 1} - \frac{2}{x - 4} = \frac{3}{x - 2} .

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51.

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Two pipes are being used to fill a tank. Pipe A can fill the tank

4.5

times as fast as Pipe B does. When both pipes are turned on, it takes

18

hours to fill the tank. Answer the following questions:

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If only Pipe A is turned on, it would take hours to fill the tank.

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If only Pipe B is turned on, it would take hours to fill the tank.

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52.

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Two pipes are being used to fill a tank. Pipe A can fill the tank

5.5

times as fast as Pipe B does. When both pipes are turned on, it takes

11

hours to fill the tank. Answer the following questions:

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If only Pipe A is turned on, it would take hours to fill the tank.

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If only Pipe B is turned on, it would take hours to fill the tank.

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53.

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Town A and Town B are

450

miles apart. A boat traveled from Town A to Town B, and then back to Town A. Since the river flows from Town B to Town A, the boat’s speed was

15

miles per hour faster when it traveled from Town B to Town A. The whole trip took

25

hours. Answer the following questions:

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The boat traveled from Town A to Town B at the speed of miles per hour.

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The boat traveled from Town B back to Town A at the speed of miles per hour.

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54.

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Town A and Town B are

360

miles apart. A boat traveled from Town A to Town B, and then back to Town A. Since the river flows from Town B to Town A, the boat’s speed was

15

miles per hour faster when it traveled from Town B to Town A. The whole trip took

20

hours. Answer the following questions:

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The boat traveled from Town A to Town B at the speed of miles per hour.

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The boat traveled from Town B back to Town A at the speed of miles per hour.

You have attempted 1 of 2 activities on this page.

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Section 12.6 Rational Functions and Equations Chapter Review

Subsection 12.6.1 Introduction to Rational Functions

Example 12.6.1. Graphs of Rational Functions.

Subsection 12.6.2 Multiplication and Division of Rational Expressions

Example 12.6.3. Simplifying Rational Expressions.

Example 12.6.4. Multiplication of Rational Functions and Expressions.

Example 12.6.5. Division of Rational Functions and Expressions.

Subsection 12.6.3 Addition and Subtraction of Rational Expressions

Example 12.6.6. Addition and Subtraction of Rational Expressions with the Same Denominator.

Example 12.6.7. Addition and Subtraction of Rational Expressions with Different Denominators.

Subsection 12.6.4 Complex Fractions

Example 12.6.8. Simplifying Complex Fractions.

Subsection 12.6.5 Solving Rational Equations

Example 12.6.9. Solving Rational Equations.

Example 12.6.10. Solving Rational Equations for a Specific Variable.

Example 12.6.11. Solving Rational Equations Using Technology.

Exercises 12.6.6 Exercises

Introduction to Rational Functions.

1.

2.

3.

4.

5.

6.

7.

8.

Multiplication and Division of Rational Expressions.

9.

10.

11.

12.

13.

14.

15.

16.

17.

18.

Addition and Subtraction of Rational Expressions.

19.

20.

21.

22.

23.

24.

Exercise Group.

25.

26.

27.

28.

Complex Fractions.

29.

30.

31.

32.

33.

34.

35.

36.

37.

38.

Solving Rational Equations.

39.

40.

41.

42.

43.

44.

45.

46.

Exercise Group.

47.

48.

49.

50.

51.

52.

53.

54.