Numbers, Variables, and Expressions

Numbers, Variables, and Expressions
1.1 Numbers, Variables, and Expressions Natural Numbers and Whole Numbers Prime Numbers and Composite Numbers Variables, Algebraic Expressions, and Equations Translating Words to Expressions Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Natural Numbers and Whole Numbers
The set of natural numbers are also known as the counting numbers. 1, 2, 3, 4, 5, 6,… Because there are infinitely many natural numbers, three dots are used to show that the list continues in the same pattern without end. The whole numbers can be expressed as 0, 1, 2, 3, 4, 5, … Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Prime Numbers and Composite Numbers
When two natural numbers are multiplied, the result is another natural number. The product of 6 and 7 is  7 = 42 The numbers 6 and 7 are factors of 42. A prime number has only itself and 1 as factors. A natural number greater than 1 that is not prime is a composite number. Any composite number can be written as a product of prime numbers. Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Prime Factorization The prime factorization of 120.
Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Solution EXAMPLE Classifying numbers as prime or composite
Classify each number as prime or composite. If a number is composite, write it as a product of prime numbers. a b. 3 c d. 300 Solution a. 37 The only factors of 37 are 1 and itself. The number is prime. b. 3 The only factors of 3 are 1 and itself. The number is prime. c. 45 Composite because 9 and 5 are factors. Prime factorization: 32  5 Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Solution EXAMPLE Classifying numbers as prime or composite
Classify each number as prime or composite. If a number is composite, write it as a product of prime numbers. a b. 3 c d. 300 Solution d. 300 Prime factorization Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Variables, Algebraic Expressions, and Equations
Variables are often used in mathematics when tables of numbers are inadequate. A variable is a symbol, typically an italic letter used to represent an unknown quantity. An algebraic expression consists of numbers, variables, operation symbols, such as +, , , and , and grouping symbols, such as parentheses. An equation is a mathematical statement that two algebraic expressions are equal. A formula is a special type of equation that expresses a relationship between two or more quantities. Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Solution EXAMPLE Evaluating algebraic expressions with one variable
Evaluate each algebraic expression for x = 6. a. x + 4 b. 4x c. 20 – x d. Solution a. x + 4 b. 4x 6 + 4 = 10 4(6) = 24 c. 20 – x d. 20 – 6 = 14 Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Solution EXAMPLE Evaluating algebraic expressions with two variables
Evaluate each algebraic expression for y = 3 and z = 9 a. 5yz b. z – y c. Solution a. 5yz b. z – y 5(3)(9) = 135 9 – 3 = 6 c. Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Solution EXAMPLE Evaluating formulas
Find the value of y for x = 20 and z = 5. a. y = x b. y = 9xz Solution a. y = x + 4 b. y = 9xz y = y = 9(20)(5) = 24 = 900 Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Translating Words to Expressions

Solution EXAMPLE Translating words to expressions
Translate each phrase to an algebraic expression. a. Twice the cost of a book b. Ten less than a number c. The product of 8 and a number Solution a. Twice the cost of a book b. Ten less than a number c. The product of 8 and a number 2c where c is the cost of the book n – 10 where n is the number 8n where n is the number Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Solution EXAMPLE Finding the area of a rectangle
The area A of a rectangle equals its length L times its width W. a. Write a formula that shows the relationship between these three quantities. b. Find the area of a yard that is 100 feet long and 75 feet wide. Solution a. The word times indicates the length and width should be multiplied. The formula is A = LW. b. A = LW = (100)(75) = 7500 square feet Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

1.2 Fractions Basic Concepts Simplifying Fractions to Lowest Terms
Multiplication and Division of Fractions Addition and Subtraction of Fractions Applications Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Basic Concepts The parts of a fraction are named as follows. Numerator
Fraction bar Denominator Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Solution EXAMPLE Identifying numerators and denominators
Give the numerator and denominator of each fraction. a. b. c. Solution a. The numerator is 8 and the denominator is 19. b. The numerator is mn, and the denominator is p. c. The numerator is c + d, and the denominator is f – 7. Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Simplifying Fractions to Lowest Terms
When simplifying fractions, we usually factor out the greatest common factor (GCF) for the numerator and the denominator. The greatest common factor is the largest factor common to both the numerator and the denominator. Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 19

Solution EXAMPLE Finding the greatest common factor
Find the greatest common factor (GCF) for each pair of numbers. a. 14, 21 b. 42, 90 Solution a. Because 14 = 7 ∙ 2 and 21 = 7 ∙ 3, the number 7 is the largest factor that is common to both 14 and 21. Thus the GCF of 14 and 21 is 7. Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

EXAMPLE continued b. When working with larger numbers, one way to determine the greatest common factor is to find the prime factorization of each number. 42 = 6 ∙ 7 = 2 ∙ 3 ∙ 7 and 90 = 6 ∙ 15 = 2 ∙ 3 ∙ 3 ∙ 5 The prime factorizations have one 2 and one 3 in common. Thus the GCF for 42 and 90 is 6. Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 21

Solution EXAMPLE Simplifying fractions to lowest terms
Simplify each fraction to lowest terms. a. b. c. Solution a. The GCF of 9 and 15 is 3. b. The GCF of 20 and 28 is 4. c. The GCF of 45 and 135 is 45. Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Multiplication of Fractions
Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 23

Solution EXAMPLE Multiplying fractions
Multiply each expression and simplify the result when appropriate. a. b. c. Solution a. b. c. Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 24

Solution EXAMPLE Finding fractional parts Find each fractional part.
One-third of one-fourth One half of three-fourths Solution a. b. Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 25

Division of Fractions Slide 26

Solution EXAMPLE Dividing fractions Divide each expression. a. b. c.

Fractions with Like Denominators

EXAMPLE Adding and subtracting fractions with common denominators Add or subtract as indicated. Simplify your answer to lowest terms when appropriate. a. b. Solution a. b. Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 30

Fractions with Unlike Denominators

Solution EXAMPLE Rewriting fractions with the LCD
Rewrite each set of fractions using the LCD. a. b. Solution a. The LCD is 24 b. The LCD is 40. Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 32

EXAMPLE Adding and subtracting fractions with unlike denominators Add or subtract as indicated. Simplify your answer to lowest terms when appropriate. a. b. Solution a. b. Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 33

Solution EXAMPLE Applying fractions to carpentry
A pipe measures inches long and needs to be cut into three equal pieces. Find the length of each piece. Solution Begin by writing as the improper fraction Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 34

Exponents and Order of Operations
1.3 Exponents and Order of Operations Natural Number Exponents Order of Operations Translating Words to Expressions Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Natural Number Exponents
The area of a square equals the length of one of its sides times itself. If the square is 5 inches on a side, then its area is 5  5 = 52 = 25 square inches The expression 52 is an exponential expression with base 5 and exponent 2. Exponent Base Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Writing products in exponential notation
EXAMPLE Writing products in exponential notation Write each product as an exponential expression. a. b. c. Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Evaluating exponential notation
EXAMPLE Evaluating exponential notation Evaluate each expression. a. b. Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Solution EXAMPLE Evaluating arithmetic expressions
Evaluate each expression by hand. a. 12 – 6 – 2 b. 12 – (6 – 2) c. Solution a. 12 – 6 – 2 b. 12 – (6 – 2) c. 6 – 2 12 – 4 4 8 Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Solution EXAMPLE Evaluating arithmetic expressions
Evaluate each expression. a b. c. Solution a. b. c. 15 – 6 9 Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Solution EXAMPLE Writing and evaluating expressions
Write each expression and then evaluate it. a. Two to the fifth power plus three b. Twenty-four less two times four Solution a. Two to the fifth power plus three b. Twenty-four less two times four Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Real Numbers and the Number Line
1.4 Real Numbers and the Number Line Signed Numbers Integers and Rational Numbers Square Roots Real and Irrational Numbers The Number Line Absolute Value Inequality Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Signed Numbers The opposite, or additive inverse, of a number a is −a.

Solution EXAMPLE Finding opposites (or additive inverses)
Find the opposite of each expression. a. 29 b. c. d. −(−13) Solution a. The opposite of 29 is −29. b. The opposite of is c. d. −(−13) = 13, so the opposite of −(−13) is −13. Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Solution EXAMPLE Finding an additive inverse (or opposite)
Find the additive inverse of –x, if x = Solution The additive inverse of −x is x = because −(−x) = x by the double negative rule. Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 49

Integers and Rational Numbers
The integers include the natural numbers, zero, and the opposite of the natural numbers. …,−2, −1, 0, 1, 2,… A rational number is any number that can be expressed as the ratio of two integers, where q ≠ 0. Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 50

Solution EXAMPLE Classifying numbers
Classify each number as one or more of the following: natural number, whole number, integer, or rational number. a. b. −9 c. Solution Because , the number is a natural number, whole number, integer, and rational number. b. The number −9 is an integer and rational number, but not a natural number or a whole number. c. The fraction is a rational number because it is the ratio of two integers. However it is not a natural number, a whole number, or an integer. Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 51

Square Roots Square roots are frequently used in algebra. The number b is a square root of a number a if b ∙ b = a. Every positive number has one positive square root and one negative square root. Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 52

Solution EXAMPLE Calculating principal square roots
Evaluate each square root. Approximate to three decimal places when appropriate. a. b. c. Solution because 8 ∙ 8 = 64 and 8 is nonnegative. b because 13 ∙ 13 = 169 and 13 is nonnegative. c is a number between 4 and 5. We can estimate the value of with a calculator. Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 53

Real and Irrational Numbers

Solution EXAMPLE Classifying numbers
Identify the natural numbers, whole numbers, integers, rational numbers, and irrational numbers in the following list. Solution Natural numbers: Whole numbers: Integers: Rational numbers: Irrational numbers: Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 55

Solution EXAMPLE Plotting numbers on a number line
Plot each real number on a number line. a. b. c. Solution Plot a dot halfway between −2 and −3. b Plot a dot between 2 and 3. c Plot a dot halfway between 3 and 4. Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 56

Absolute Value The absolute value of a real number equals its distance on the number line from the origin. Because distance is never negative, the absolute value of a real number is never negative. Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 57

Solution EXAMPLE Finding the absolute value of a real number
Write the expression without the absolute value sign. a. b. c. d. Solution a because the distance between the origin and −9 is 9. b because the distance is 0 between the origin and 0. c. d. Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 58

Solution EXAMPLE Ordering real numbers
List the following numbers from least to greatest. Then plot these numbers on a number line. Solution Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 59

Addition and Subtraction of Real Numbers
1.5 Addition and Subtraction of Real Numbers Addition of Real Numbers Subtraction of Real Numbers Applications Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

There are four arithmetic operations: addition, subtraction, multiplication, and division.

Addition of Real Numbers
In an addition problem the two numbers added are called addends, and the answer is called the sum. 5 + 8 = 13 5 and 8 are the addends 13 is the sum The opposite (or additive inverse) of a real number a is a. Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

EXAMPLE Adding Opposites
Find the opposite of each number and calculate the sum of the number and its opposite. a. b. Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Addition of Real Numbers

Evaluate each expression. a.
EXAMPLE Adding real numbers Evaluate each expression. a. b. The numbers are both negative, add the absolute values. The sign would be negative as well. Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Subtraction of Real Numbers

Solution EXAMPLE Subtracting real numbers
Find each difference by hand. a. 12 – 16 b. –6 – 2 Solution a. 12 – 16 b. –6 – 2 12 + (–16) –6 + (–2) 4 8 Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Solution EXAMPLE Adding and subtracting real numbers
Evaluate each expression. a b. Solution a. b. Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Solution EXAMPLE Balancing a checking account
The initial balance in a checking account is $326. Find the final balance if the following represents a list of withdrawals and deposits: $20, $15, $200, and $150 Solution Find the sum of the five numbers. The final balance is $341. Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Multiplication and Division of Real Numbers
1.6 Multiplication and Division of Real Numbers Multiplication of Real Numbers Division of Real Numbers Applications Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Multiplication of Real Numbers

Solution EXAMPLE Multiplying real numbers Find each product by hand.
a. −4 ∙ 8 b c d. Solution a. The resulting product is negative because the factors have unlike signs. Thus −4 ∙ 8 = −32. b. The product is positive because both factors are positive. c. Since both factors are negative, the product is positive. d. Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Solution EXAMPLE Evaluating real numbers with exponents
Evaluate each expression by hand. a. (−6)2 b. −62 Solution a. Because the exponent is outside of parentheses, the base of the exponential expression is −6. The expression is evaluated as (−6)2 = (−6)(−6) = 36. b. This is the negation of an exponential expression with base 6. Evaluating the exponent before negative results in −62 = −(6)(6) = −36. Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 76

Division of Real Numbers

Solution EXAMPLE Dividing real numbers
Evaluate each expression by hand. a. b c d. Solution a. b. c. d. 4 ÷ 0 is undefined. The number 0 has no reciprocal. Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 78

Solution EXAMPLE Converting fractions to decimals
Convert the measurement to a decimal number. Solution a. Begin by dividing 5 by 16. 0.3 1 2 5 Thus the mixed number is equivalent to − 48 20 − 16 40 − 32 80 −80 Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 80

Solution EXAMPLE Converting decimals to fractions
Convert each decimal number to a fraction in lowest terms. a b Solution a. The decimal 0.32 equals thirty-two hundredths. b. The decimal equals eight hundred seventy-five thousandths. Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 81

Solution EXAMPLE Application
After surveying 125 pediatricians, 92 stated that they had admitted a patient to the children’s hospital in the last month for pneumonia. Write the fraction as a decimal. Solution One method for writing the fraction as a decimal is to divide 92 by 125 using long division. An alternative method is to multiply the fractions by so the denominator becomes Then, write the numerator in the thousandths place in the decimal. Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 82

Properties of Real Numbers
1.7 Properties of Real Numbers Commutative Properties Associative Properties Distributive Properties Identity and Inverse Properties Mental Calculations Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Commutative Properties
The commutative property for addition states that two numbers, a and b, can be added in any order and the result will be the same. 6 + 8 = 8 + 6 The commutative property for multiplication states that two numbers, a and b, can be multiplied in any order and the result will be the same. 9  4 = 4  9 Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Applying the commutative properties
EXAMPLE Applying the commutative properties Use the commutative properties to rewrite each expression. a. b. c. Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Applying the associative properties
EXAMPLE Applying the associative properties Use the associative property to rewrite each expression. a. b. Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Identifying properties of real numbers
EXAMPLE Identifying properties of real numbers State the property that each equation illustrates. a. b. Associative property of multiplication because the grouping of the numbers has been changed. Commutative property for addition because the order of the numbers has changed. Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Distributive Properties
The distributive properties are used frequently in algebra to simplify expressions. 7(3 + 8) = 7   8 The 7 must be multiplied by both the 3 and the 8. Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Solution EXAMPLE Applying the distributive properties
Apply a distributive property to each expression. a. 4(x + 3) b. –8(b – 5) c. 12 – (a + 2) Solution a. 4(x + 3) c. 12 – (a + 2) = 4  x + 4  3 = 12 + (1)(a + 2) = 4x + 12 = 12 + (1)  a + (–1)  2 = 12  a – 2 b. –8(b – 5) = 10  a = 8  b  (8)  5 = 8b + 40 Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Solution Inserting parentheses using the distributive property EXAMPLE
Use the distributive property to insert parentheses in the expression and then simplify the result. a b. Solution a. b. Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Identifying properties of real numbers
EXAMPLE Identifying properties of real numbers State the property or properties illustrated by each equation. a. b. Distributive property . Commutative and associative properties for addition. Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Identity and Inverse Properties
The identity property of 0 states that if 0 is added to any real number a, the result is a. The number 0 is called the additive identity = 3 The identity property of 1 states that if any number a is multiplied by 1, the result is a. The number 1 is called the multiplicative identity. 4  1 = 4 Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Identity and Inverse Properties

Identifying identity and inverse properties
EXAMPLE Identifying identity and inverse properties State the property or properties illustrated by each equation. a. b. Identity property for 0. Additive inverse property and the identity property for 0. Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Performing calculations mentally
EXAMPLE Performing calculations mentally Use the properties of real numbers to calculate each expression mentally. a. b. Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Simplifying and Writing Algebraic Expressions
1.8 Simplifying and Writing Algebraic Expressions Terms Combining Like Terms Simplifying Expressions Writing Expressions Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Terms A term is a number, a variable, or a product of numbers and variables raised to powers. Examples of terms include 4, z, 5x, and −6xy2. The coefficient of a term is the number that appears in the term. Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Solution EXAMPLE Identifying terms
Determine whether each expression is a term. If it is a term, identify its coefficient. a. 97 b. 17x c. 4a – 6b d. 9y2 Solution a. A number is a term. The coefficient is 97. b. The product of a number and a variable is a term. The coefficient is 17. c. The difference of two terms in not a term. d. The product of a number and a variable with an exponent is a term. Its coefficient is 9. Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Solution EXAMPLE Identifying like terms
Determine whether the terms are like or unlike. a. 9x, −15x b. 16y2, c. 5a3, 5b d. 11, −8z Solution a. The variable in both terms is x, with the same power of 1, so they are like terms. b. The term 1 has no variable and the 16 has a variable of y2. They are unlike terms. c. The variables are different, so they are unlike terms. d. The term 11 has no variable and the −8 has a variable of z. They are unlike terms. Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 102

Solution EXAMPLE Combining like terms
Combine terms in each expression, if possible. a. −2y + 7y b. 4x2 – 6x Solution a. Combine terms by applying the distributive property. −2y + 7y = (−2 + 7)y = 5y b. They are unlike terms, so they can not be combined. Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 103

Solution EXAMPLE Simplifying expressions Simplify each expression.
a z – 9 + 7z b. 9x – 2(x – 5) Solution a. 13 + z – 9 + 7z b. 9x – 2(x – 5) = 13 +(– 9) + z + 7z = 9x + (– 2)x + (−2)(– 5) = 13 +(– 9) + (1+ 7)z = 9x – 2x + 10 = 7x + 10 = 4 + 8z Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 104

Solution EXAMPLE Simplifying expressions Simplify each expression.
a. 6x2 – y + 9x2 – 3y b. Solution a. 6x2 – y + 9x2 – 3y b. = 6x2 + 9x2 + (–1y) + (–3y) = (6 + 9)x2 + (–1+ (– 3))y = 15x2 –4y Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 105

Solution EXAMPLE Writing and simplifying an expression
A sidewalk has a constant width w and comprises several short sections with lengths 11, 4, and 18 feet. Write and simplify an expression that gives the number of square feet of sidewalk. Find the area of the sidewalk if its width is 3 feet. Solution 11 ft 4 ft 18 ft w 11w + 4w + 18w = ( )w = 33w b. 33w = 33 ∙ 3 = 99 square feet Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 106

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