Beginning and Intermediate Algebra 2nd edition doc

Order of Operations: Parenthesis Grouping Exponents Multiply and Divide Left to Right Add and Subtract Left to Right Multiply and Divide are on the same level because they are the same o

Beginning and Intermediate Algebra

An open source (CC-BY) textbook Available for free download at: http://wallace.ccfaculty.org/book/book.html

by Tyler Wallace

ISBN #978-1-4583-7768-5

Copyright 2010, Some Rights Reserved CC-BY.

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Table of Contents

Chapter 0: Pre-Algebra

0.1 Integers 7

0.2 Fractions 12

0.3 Order of Operations 18

0.4 Properties of Algebra 22

Chapter 1: Solving Linear Equations 1.1 One-Step Equations 28

1.2 Two-Step Equations 33

1.3 General Linear Equations 37

1.4 Solving with Fractions 43

1.5 Formulas 47

1.6 Absolute Value Equations 52

1.7 Variation 57

1.8 Application: Number/Geometry.64 1.9 Application: Age 72

1.10 Application: Distance 79

Chapter 2: Graphing 2.1 Points and Lines 89

2.2 Slope 95

2.3 Slope-Intercept Form 102

2.4 Point-Slope Form 107

2.5 Parallel & Perpendicular Lines.112 Chapter 3: Inequalities 3.1 Solve and Graph Inequalities 118

3.2 Compound Inequalitites 124

3.3 Absolute Value Inequalities 128

Chapter 4: Systems of Equations 4.1 Graphing 134

4.2 Substitution 139

4.3 Addition/Elimination 146

4.4 Three Variables 151

4.5 Application: Value Problems 158

4.6 Application: Mixture Problems.167 Chapter 5: Polynomials 5.1 Exponent Properties 177

5.2 Negative Exponents 183

5.3 Scientific Notation 188

5.4 Introduction to Polynomials 192

5.5 Multiply Polynomials 196

5.6 Multiply Special Products 201

5.7 Divide Polynomials 205

Chapter 6: Factoring

6.1 Greatest Common Factor 212

6.2 Grouping 216

6.3 Trinomials where a = 1 221

6.4 Trinomials where a1 226

6.5 Factoring Special Products 229

6.6 Factoring Strategy 234

6.7 Solve by Factoring 237

Chapter 7: Rational Expressions 7.1 Reduce Rational Expressions 243

7.2 Multiply and Divide 248

7.3 Least Common Denominator 253

7.4 Add and Subtract 257

7.5 Complex Fractions 262

7.6 Proportions 268

7.7 Solving Rational Equations 274

7.8 Application: Dimensional Analysis 279

Chapter 8: Radicals 8.1 Square Roots 288

8.2 Higher Roots 292

8.3 Adding Radicals 295

8.4 Multiply and Divide Radicals 298

8.5 Rationalize Denominators 303

8.6 Rational Exponents 310

8.7 Radicals of Mixed Index 314

8.8 Complex Numbers 318

Chapter 9: Quadratics 9.1 Solving with Radicals 326

9.2 Solving with Exponents 332

9.3 Complete the Square 337

9.4 Quadratic Formula 343

9.5 Build Quadratics From Roots 348

9.6 Quadratic in Form 352

9.7 Application: Rectangles 357

9.8 Application: Teamwork 364

9.9 Simultaneous Products 370

9.10 Application: Revenue and Distance.373 9.11 Graphs of Quadratics 380

Chapter 10: Functions 10.1 Function Notation 386

10.2 Operations on Functions 393

10.3 Inverse Functions 401

10.4 Exponential Functions 406

10.5 Logarithmic Functions 410

10.6 Application: Compound Interest.414 10.7 Trigonometric Functions 420

10.8 Inverse Trigonometric Functions.428 Answers 438

Chapter 0 : Pre-Algebra

0.1 Integers 7

0.2 Fractions 12

0.3 Order of Operations 18

0.4 Properties of Algebra 22

plying and dividing of integers Integers are all the positive whole numbers, zero,

and their opposites (negatives) As this is intended to be a review of integers, thedescriptions and examples will not be as detailed as a normal lesson

World View Note: The first set of rules for working with negative numbers waswritten out by the Indian mathematician Brahmagupa

When adding integers we have two cases to consider The first is if the signsmatch, both positive or both negative If the signs match we will add the num-bers together and keep the sign This is illustrated in the following examples

Example 7.

8 − 3 Add the opposite of 3

8 + ( − 3) Different signs, subtract 8 − 3, use sign from bigger number, positive

5 Our Solution

Example 8.

− 4 − 6 Add the opposite of 6

− 4 + ( − 6) Same sign, add 4 + 6, keep the negative

− 10 Our Solution

Example 9.

9 − ( − 4) Add the opposite of − 4

9 + 4 Same sign, add 9 + 4, keep the positive

13 Our Solution

Example 10.

− 6 − ( − 2) Add the opposite of − 2

− 6 + 2 Different sign, subtract 6 − 2, use sign from bigger number, negative

− 4 Our Solution

Multiplication and division of integers both work in a very similar pattern Theshort description of the process is we multiply and divide like normal, if the signsmatch (both positive or both negative) the answer is positive If the signs don’tmatch (one positive and one negative) then the answer is negative This is shown

in the following examples

10, but if the signs match on multiplication, the answer is positive, ( − 3)( − 7) =21

10) ( − 4) + ( − 1)12) ( − 1) + ( − 6)14) ( − 8) + ( − 1)16) ( − 1) − 818) 5 − 720) ( − 5) + 722) 1 + ( − 1)24) 8 − ( − 1)26) ( − 3) + ( − 1)28) 7 − 3

− 2

57) − 16259) 60

− 10

Pre-Algebra - Fractions

Objective: Reduce, add, subtract, multiply, and divide with fractions.

Working with fractions is a very important foundation to algebra Here we willbriefly review reducing, multiplying, dividing, adding, and subtracting fractions

As this is a review, concepts will not be explained in detail as other lessons are

World View Note: The earliest known use of fraction comes from the MiddleKingdom of Egypt around 2000 BC!

We always like our final answers when working with fractions to be reduced.Reducing fractions is simply done by dividing both the numerator and denomi-nator by the same number This is shown in the following example

numer-by multiplying straight across, multiplying numerators together and denominatorstogether.

denomi-Example 19.

Find the LCD of 8 and 12 Test multiples of 12

12?12

8 Can′tdivide 12 by 824?24

8 = 3 Yes! We can divide 24 by 8!

24 Our Soultion

Adding and subtracting fractions is identical in process If both fractions alreadyhave a common denominator we just add or subtract the numerators and keep thedenominator

If the denominators do not match we will first have to identify the LCD and build

up each fraction by multiplying the numerators and denominators by the samenumber so the denominator is built up to the LCD

Find each product.

Find each quotient.

Evaluate each expression.

80) ( −12) − ( − 35)82) 97− ( −53)

Pre-Algebra - Order of Operations

Objective: Evaluate expressions using the order of operations, including the use of absolute value.

When simplifying expressions it is important that we simplify them in the correctorder Consider the following problem done two different ways:

or exponents When we want to do something out of order and make it come first

we will put it in parenthesis (or grouping symbols) This list then is our order ofoperations we will use to simplify expressions

Order of Operations:

Parenthesis (Grouping)

Exponents Multiply and Divide (Left to Right) Add and Subtract (Left to Right)

Multiply and Divide are on the same level because they are the same operation(division is just multiplying by the reciprocal) This means they must be done left

to right, so some problems we will divide first, others we will multiply first Thesame is true for adding and subtracting (subtracting is just adding the opposite).Often students use the word PEMDAS to remember the order of operations, asthe first letter of each operation creates the word PEMDAS However, it is theauthor’s suggestion to think about PEMDAS as a vertical word written as:

PEMDAS

so we don’t forget that multiplication and division are done left to right (samewith addition and subtraction) Another way students remember the order ofoperations is to think of a phrase such as “Please Excuse My Dear Aunt Sally”where each word starts with the same letters as the order of operations start with

World View Note: The first use of grouping symbols are found in 1646 in theDutch mathematician, Franciscus van Schooten’s text, Vieta He used a bar over

the expression that is to be evaluated first So problems like 2(3 + 5) were written

2{82− 7[32− 4(10)]( − 1)} Multiply inside inner most parenthesis

2{82− 7[32 − 40]( − 1)} Subtract inside those parenthesis

2{82

˜ − 7[ − 8]( − 1)} Exponents next2{64− 7[ − 8]( − 1)} Multiply left to right, sign with the number2{64 + 56( − 1)} Finish multiplying

2{64 − 56} Subtract inside parenthesis2{8} Multiply

16 Our Solution

As the above example illustrates, it can take several steps to complete a problem.The key to successfully solve order of operations problems is to take the time toshow your work and do one step at a time This will reduce the chance of making

a mistake along the way

There are several types of grouping symbols that can be used besides parenthesis.One type is a fraction bar If we have a fraction, the entire numerator and theentire denominator must be evaluated before we reduce the fraction In thesecases we can simplify in both the numerator and denominator at the same time.

Example 29.

1 + 3| − 42

˜ − ( − 8)| + 2|3 + ( − 5)2| Evaluate absolute values first, exponents

1 + 3|− 16 − ( − 8)| + 2|3 + 25| Add inside absolute values

1 + 3| − 8| + 2|28| Evaluate absolute values

1 + 3(8) + 2(28) Multiply left to right

Expo-is also squared giving us a positive solution, 25

0.3 Practice - Order of Operation

)

14) − 3 − {3 − [ − 3(2 + 4) − ( − 2)]}16) − 4 − [2 + 4( − 6) − 4 −

22− 5 · 2

]18) 2 · ( − 3) + 3 − 6[ − 2 − ( − 1 − 3)]20) − 52+ ( − 5)2

Pre-Algebra - Properties of Algebra

Objective: Simplify algebraic expressions by substituting given values, distributing, and combining like terms

In algebra we will often need to simplify an expression to make it easier to use.There are three basic forms of simplifying which we will review here

World View Note: The term “Algebra” comes from the Arabic word al-jabrwhich means “reunion” It was first used in Iraq in 830 AD by Mohammad ibn-Musa al-Khwarizmi

The first form of simplifying expressions is used when we know what number eachvariable in the expression represents If we know what they represent we canreplace each variable with the equivalent number and simplify what remains usingorder of operations

− 6 + ( − 2)( − 6)(5)( − 2) Multiply left to right

− 6 + 12(5)( − 2) Multiply left to right

− 6 + 60( − 2) Multiply

− 6 − 120 Subtract

− 126 Our Solution

It will be more common in our study of algebra that we do not know the value ofthe variables In this case, we will have to simplify what we can and leave thevariables in our final solution One way we can simplify expressions is to combine

like terms Like terms are terms where the variables match exactly (exponents

included) Examples of like terms would be 3xy and − 7xy or 3a2band 8a2b or −

3 and 5 If we have like terms we are allowed to add (or subtract) the numbers infront of the variables, then keep the variables the same This is shown in the fol-lowing examples

fol-A final method to simplify is known as distributing Often as we work with lems there will be a set of parenthesis that make solving a problem difficult, if notimpossible To get rid of these unwanted parenthesis we have the distributiveproperty Using this property we multiply the number in front of the parenthesis

prob-by each term inside of the parenthesis

Distributive Property: a(b + c) = ab + ac

Several examples of using the distributive property are given below

It is possible to distribute just a negative through parenthesis If we have a tive in front of parenthesis we can think of it like a − 1 in front and distribute the

nega-− 1 through This is shown in the following example

Example 36.

− (4x − 5y + 6) Negative can be thought of as − 1

− 1(4x − 5y + 6) Multiply each term by − 1

− 4x + 5y − 6 Our SolutionDistributing through parenthesis and combining like terms can be combined intoone problem Order of operations tells us to multiply (distribute) first then add orsubtract last (combine like terms) Thus we do each problem in two steps, dis-tribute then combine

Example 37.

5 + 3(2x − 4) Distribute 3, multipling each term

5 + 6x − 12 Combine like terms 5 − 12

we now have a positive answer Following are more involved examples of tributing and combining like terms

0.4 Practice - Properties of Algebra

Evaluate each using the values given.

Combine Like Terms

68) − 7(4x − 6) + 2(10x − 10)70) − 3(4 + a) + 6a(9a + 10)72) − 7(4x + 3) − 10(10x + 10)74) (7x2− 3) − (5x2+ 6x)76) (3x2− x) − (7 − 8x)78) (2b − 8) + (b − 7b2)80) (7a2+ 7a) − (6a2+ 4a)82) (3 − 7n2) + (6n2+ 3)

Chapter 1 : Solving Linear Equations

1.1 One-Step Equations 281.2 Two-Step Equations 331.3 General Linear Equations 371.4 Solving with Fractions 431.5 Formulas 471.6 Absolute Value Equations 521.7 Variation 571.8 Application: Number and Geometry 641.9 Application: Age 721.10 Application: Distance, Rate and Time 79

Solving Linear Equations - One Step Equations

Objective: Solve one step linear equations by balancing using inverse operations

Solving linear equations is an important and fundamental skill in algebra Inalgebra, we are often presented with a problem where the answer is known, butpart of the problem is missing The missing part of the problem is what we seek

to find An example of such a problem is shown below

Example 41.

4x + 16 = − 4Notice the above problem has a missing part, or unknown, that is marked by x If

we are given that the solution to this equation is − 5, it could be plugged into theequation, replacing the x with − 5 This is shown in Example 2

to solving equations Here we will focus on what are called “one-step equations” orequations that only require one step to solve While these equations often seemvery fundamental, it is important to master the pattern for solving these problems

so we can solve more complex problems

Addition Problems

To solve equations, the general rule is to do the opposite For example, considerthe following example

Example 44.

x+ 7 = − 5 The 7 is added to the x

− 7 − 7 Subtract 7 from both sides to get rid of it

− 3 = x

5 = − 8 + x+ 8 + 8

Then we get our solution x = 5

With multiplication problems it is very important that care is taken with signs If

x is multiplied by a negative then we will divide by a negative This is shown inexample 9

5= − 3(5) Multiply both sides by 5

x= 40

x

− 4= 9( − 4)− 4x = 9( − 4)

x= − 36

Table 4. Division Examples

The process described above is fundamental to solving equations once this cess is mastered, the problems we will see have several more steps These prob-lems may seem more complex, but the process and patterns used will remain thesame

pro-World View Note: The study of algebra originally was called the “Cossic Art”from the Latin, the study of “things” (which we now call variables)

1.1 Practice - One Step Equations

Solve each equation.

20) 13k = − 1622) 21 = x + 524) m − 4 = − 1326) 3n = 2428) − 17 = 12x30) n + 8 = 1032) v − 16 = − 3034) − 15 = x − 1636) − 8k = 12038) − 15 = x940) − 19 = 20n

Linear Equations - Two-Step Equations

Objective: Solve two-step equations by balancing and using inverse opperations.

After mastering the technique for solving equations that are simple one-step tions, we are ready to consider two-step equations As we solve two-step equa-tions, the important thing to remember is that everything works backwards!When working with one-step equations, we learned that in order to clear a “plusfive” in the equation, we would subtract five from both sides We learned that toclear “divided by seven” we multiply by seven on both sides The same patternapplies to the order of operations When solving for our variable x, we use order

equa-of operations backwards as well This means we will add or subtract first, thenmultiply or divide second (then exponents, and finally any parentheses orgrouping symbols, but that’s another lesson) So to solve the equation in the firstexample,

Example 53.

4x − 20 = − 8

We have two numbers on the same side as the x We need to move the 4 and the

20 to the other side We know to move the four we need to divide, and to movethe twenty we will add twenty to both sides If order of operations is done back-wards, we will add or subtract first Therefore we will add 20 to both sides first.Once we are done with that, we will divide both sides by 4 The steps are shownbelow

4x − 20 = − 8 Start by focusing on the subtract 20

+ 20 + 20 Add 20 to both sides4x = 12 Now we focus on the 4 multiplied by x

4 4 Divide both sides by 4

The same process is used to solve any two-step equations Add or subtract first,then multiply or divide Consider our next example and notice how the same pro-cess is applied.

Example 54.

5x + 7 = 7 Start by focusing on the plus 7

− 7 − 7 Subtract 7 from both sides5x = 0 Now focus on the multiplication by 5

5 5 Divide both sides by 5

x= 0 Our Solution!

Notice the seven subtracted out completely! Many students get stuck on thispoint, do not forget that we have a number for “nothing left” and that number iszero With this in mind the process is almost identical to our first example

A common error students make with two-step equations is with negative signs.Remember the sign always stays with the number Consider the followingexample

Example 55.

4 − 2x = 10 Start by focusing on the positive 4

− 4 − 4 Subtract 4 from both sides

− 2x = 6 Negative (subtraction) stays on the 2x

8 − x = 2 Start by focusing on the positive 8

− 8 − 8 Subtract 8 from both sides

− x = − 6 Negative (subtraction) stays on the x

− 1x = − 6 Remember, no number in front of variable means 1

− 1 − 1 Divide both sides by − 1

x= 6 Our Solution!

Solving two-step equations is a very important skill to master, as we studyalgebra The first step is to add or subtract, the second is to multiply or divide.This pattern is seen in each of the following examples

− 3x = 0

− 3 − 3

x= 0

− 3 =x5− 4+ 4 + 4(5)(1) =x5(5)

5 = x

Table 5. Two-Step Equation Examples

As problems in algebra become more complex the process covered here willremain the same In fact, as we solve problems like those in the next example,each one of them will have several steps to solve, but the last two steps are a two-step equation like we are solving here This is why it is very important to mastertwo-step equations now!

alge-1.2 Practice - Two-Step Problems

Solve each equation.

Solving Linear Equations - General Equations

Objective: Solve general linear equations with variables on both sides.

Often as we are solving linear equations we will need to do some work to set them

up into a form we are familiar with solving This section will focus on lating an equation we are asked to solve in such a way that we can use our pat-tern for solving two-step equations to ultimately arrive at the solution

manipu-One such issue that needs to be addressed is parenthesis Often the parenthesiscan get in the way of solving an otherwise easy problem As you might expect wecan get rid of the unwanted parenthesis by using the distributive property This isshown in the following example Notice the first step is distributing, then it issolved like any other two-step equation

equa-Example 60.

3(2x − 4) + 9 = 15 Distribute the 3 through the parenthesis

6x − 12 + 9 = 15 Combine like terms, − 12 + 96x − 3 = 15 Focus on the subtraction first+ 3 + 3 Add 3 to both sides

6x = 18 Now focus on multiply by 6

6 6 Divide both sides by 6

x= 3 Our Solution

A second type of problem that becomes a two-step equation after a bit of work isone where we see the variable on both sides This is shown in the followingexample

Example 61.

4x − 6 = 2x + 10

Notice here the x is on both the left and right sides of the equation This can

make it difficult to decide which side to work with We fix this by moving one ofthe terms with x to the other side, much like we moved a constant term Itdoesn’t matter which term gets moved, 4x or 2x, however, it would be theauthor’s suggestion to move the smaller term (to avoid negative coefficients) Forthis reason we begin this problem by clearing the positive 2x by subtracting 2xfrom both sides

4x − 6 = 2x + 10 Notice the variable on both sides

− 2x − 2x Subtract 2x from both sides2x − 6 = 10 Focus on the subtraction first+ 6 + 6 Add 6 to both sides

2x = 16 Focus on the multiplication by 2

2 2 Divide both sides by 2

x= 8 Our Solution!

The previous example shows the check on this solution Here the solution is

plugged into the x on both the left and right sides before simplifying.

Example 62.

4(8) − 6 = 2(8) + 10 Multiply 4(8) and 2(8) first

32 − 6 = 16 + 10 Add and Subtract

26 = 26 True!

The next example illustrates the same process with negative coefficients Noticefirst the smaller term with the variable is moved to the other side, this time byadding because the coefficient is negative

Example 63.

− 3x + 9 = 6x − 27 Notice the variable on both sides, − 3x is smaller

+ 3x + 3x Add 3x to both sides

9 = 9x − 27 Focus on the subtraction by 27

+ 27 + 27 Add 27 to both sides

36 = 9x Focus on the mutiplication by 9

9 9 Divide both sides by 9

4 = x Our Solution

Linear equations can become particularly intersting when the two processes arecombined In the following problems we have parenthesis and the variable on bothsides Notice in each of the following examples we distribute, then combine liketerms, then move the variable to one side of the equation

Example 64.

2(x − 5) + 3x = x + 18 Distribute the 2 through parenthesis

2x − 10 + 3x = x + 18 Combine like terms 2x + 3x

5x − 10 = x + 18 Notice the variable is on both sides

− x − x Subtract x from both sides4x − 10 = 18 Focus on the subtraction of 10

+ 10 + 10 Add 10 to both sides4x = 28 Focus on multiplication by 4

4 4 Divide both sides by 4

x= 7 Our Solution

Sometimes we may have to distribute more than once to clear several parenthesis.Remember to combine like terms after you distribute!

Example 65.

3(4x − 5) − 4(2x + 1) = 5 Distribute 3 and − 4 through parenthesis

12x − 15 − 8x − 4 = 5 Combine like terms 12x − 8x and − 15 − 4

4x − 19 = 5 Focus on subtraction of 19

+ 19 + 19 Add 19 to both sides4x = 24 Focus on multiplication by 4

4 4 Divide both sides by 4

x= 6 Our Solution

This leads to a 5-step process to solve any linear equation While all five stepsaren’t always needed, this can serve as a guide to solving equations

1 Distribute through any parentheses

2 Combine like terms on each side of the equation

3 Get the variables on one side by adding or subtracting

4 Solve the remaining 2-step equation (add or subtract then multiply ordivide)

5 Check your answer by plugging it back in for x to find a true statement.

The order of these steps is very important

World View Note: The Chinese developed a method for solving equations thatinvolved finding each digit one at a time about 2000 years ago!

We can see each of the above five steps worked through our next example

Example 66.

4(2x − 6) + 9 = 3(x − 7) + 8x Distribute 4 and 3 through parenthesis

8x − 24 + 9 = 3x − 21 + 8x Combine like terms − 24 + 9 and 3x + 8x8x − 15 = 11x − 21 Notice the variable is on both sides

− 8x − 8x Subtract 8x from both sides

Pre -Algebra - Properties of Algebra< /b>

Objective: Simplify algebraic expressions by substituting given values, distributing, and combining like terms

In algebra. .. (exponents

included) Examples of like terms would be 3xy and − 7xy or 3a2band 8a2b or −

3 and If we have like terms we are allowed to add (or subtract)... balancing using inverse operations

Solving linear equations is an important and fundamental skill in algebra Inalgebra, we are often presented with a problem where the answer is known, butpart