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2 UNIT 1 ALGEBRA BASICS In medieval times, a barber called himself an algebrista. UNIT 1 Algebra Basics Copyright © 2008, K12 Inc. All rights reserved. This material may not be reproduced in whole or in part, including illustrations, without the express prior written consent of K12 Inc.

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Page 1: UNIT 1 Algebra Basics - Kitabook12.kitaboo.com/k12/ebookpdf/maths02/chapter1.pdf · 10 UNIT 1 ALGEBRA BASICS Identifying Terms Variable expressions consist of terms. Terms are the

2 UNIT 1 ALGEBRA BASICS

In medieval times, a barber called himself an algebrista.

UNIT 1 Algebra Basics

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The English word algebra and the Spanish word algebrista both come from the Arabic word al-jabr, which means “restoration.” A barber in medieval times often called himself an algebrista. The algebrista also was a bonesetter who restored or fixed bones. Mathematicians today use algebra to solve problems. Algebra can help you find solutions and “fix” certain problems that you encounter.

Big Ideas► Expressions, equations, and inequalities express relationships between

different entities.

► If you can create a mathematical model for a situation, you can use the model to solve other problems that you might not be able to solve otherwise. Algebraic equations can capture key relationships among quantities in the world.

Unit Topics ► Expressions

► Variables

► Translating Words into Variable Expressions

► Equations

► Translating Words into Equations

► Replacement Sets

► Problem Solving

ALGEBRA BAISCS 3

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EXPRESSIONS 5

Expressions

A numerical expression consists of numbers and one or more operations. Some examples of numerical expressions are shown below.

18 − 2 7 + 2 · 3 4 + 2[3 − 4 ÷ (12 − 8)] (5 + 3 ) 2

____________ 10 − 3(4 − 2)

To find the value of a numerical expression, you need to simplify the expression or evaluate the expression. To simplify a numerical expression, perform the indicated operation(s).

Consider the expression 7 + 2 · 3. If you add and then multiply, you get the value 27. But if you multiply and then add, you get the value 13. Because an expression can have only one value, mathematicians have agreed on a process to follow so that everyone simplifies expressions the same way. This process is the order of operations.

8 2 An exponent indicates repeated multiplication, so 8 2 = 8 · 8.

NOTATION

You can use “Please Excuse My Dear Aunt Sally” to remember the order of operations.

P: ParenthesesE: ExponentsM/D: Multiply and DivideA/S: Add and Subtract

TIP

So, the correct way to simplify 7 + 2 · 3 is to multiply and then add: 7 + 2 · 3 = 7 + 6 = 13.

The most common grouping symbols are parentheses ( ). Grouping symbols can affect the value of an expression. Nested grouping symbols (grouping symbols within grouping symbols) contain brackets [ ] and sometimes braces { }.

Simplifying Expressions with and without Grouping SymbolsExample 1 Simplify.A. 5 · 2 + 7Solution

5 · 2 + 7 = 10 + 7 Multiply. = 17 Add. ■B. 5 · (2 + 7)Solution

5 · (2 + 7) = 5 · (9) Add first because addition appears inside parentheses. = 45 Multiply. ■

The expression 5 · (2 + 7) is also written as 5(2 + 7).

TIP

Step 1 Perform operations within grouping symbols. For nested grouping symbols, simplify in the innermost group fi rst.

Step 2 Evaluate powers (as indicated by exponents).Step 3 Multiply and divide from left to right.Step 4 Add and subtract from left to right.

ORDER OF OPERATIONS

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6 UNIT 1 ALGEBRA BASICS

Simplifying Expressions with Several OperationsExample 2 Simplify.A. 3 · 7 − 10 ÷ 2 · 3Solution

3 · 7 − 10 ÷ 2 · 3 = 21 − 10 ÷ 2 · 3 Multiply and divide in order from left to right.

= 21 − 5 · 3 = 21 − 15 = 6 Subtract. ■B. 4 + 2[3 − 4 ÷ (12 − 8)]Solution

4 + 2[3 − 4 ÷ (12 − 8)] = 4 + 2[3 − 4 ÷ 4] Parentheses are nested within brackets. Subtract.

= 4 + 2[3 − 1] Divide inside the brackets. = 4 + 2[2] Subtract inside

the brackets. = 4 + 4 Multiply. = 8 Add. ■

Simplifying an Expression with a Fraction BarA fraction bar indicates division. It is also a grouping symbol, separating the numerator from the denominator.

Example 3 Simplify. (5 + 3 ) 2

__________ 1 + 2 5 − 23

Solution Treat the fraction bar as a grouping symbol. Simplify the numerator and denominator, and then divide.

(5 +�3 ) 2

__________ 1 + 2 5 − 23 = (8 ) 2 _________ 1 + 2 5 − 23 Perform the operation inside the parentheses.

= 64 ___________ 1 + 32 − 23 Evaluate powers: 8 2 = 8 · 8 = 64 and 2 5 = 2 · 2 · 2 · 2 · 2 = 32.

= 64 ___ 10 Add and subtract from left to right to simplify the denominator.

= 32 ___ 5

= 6.4 Reduce the fraction or divide to simplify. ■

In Example 3, the expression can be written as (5 + 3 ) 2 ÷ (1 + 2 5 − 23).

THINK ABOUT IT

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EXPRESSIONS 7

Placing Grouping Symbols to Get a Specified ValueYou can get different values for an expression by changing the placement of grouping symbols.Example 4 Place grouping symbols in the expression 2 · 8 + 2 3 · 10 to get expressions that have these values: 20,000 and 80,000.Solution The method is trial-and-error. Some possible placements of group-ing symbols are 2 · (8 + 2 ) 3 · 10, [2 · (8 + 2) ] 3 · 10, and (2 · 8 + 2 ) 3 · 10. Evaluate these expressions, and try other placements if necessary. The two correct expressions are shown below. 2 · (8 + 2 ) 3 · 10 = 2 · (10 ) 3 · 10 Perform the operation inside

the parentheses. = 2 · 1000 · 10 Evaluate the power:

1 0 3 = 10 · 10 · 10 = 1000. = 2000 · 10 Multiply from left to right. = 20,000 Multiply.[2 · (8 + 2) ] 3 · 10 = [2 · (10) ] 3 · 10 Perform the operation inside

the parentheses. = [20 ] 3 · 10 Perform the operation inside

the brackets. = 8000 · 10 Evaluate the power:

2 0 3 = 20 · 20 · 20 = 8000. = 80,000 Multiply. ■

Problem Set

Simplify the expression.

1. 6 · 3 + 10

2. 2 · 10 − 7

3. 4 2 + 5

4. 3 · (8 − 3)

5. (4 + 8) · 4

6. 4 + 9 · 2 ÷ 3 − 1

7. 5 · 3 − 8 ÷ 2 · 3

8. 5 + 4[3 + 4 ÷ (11 − 7)]

9. 7 · [5 + 2 · (16 ÷ 2)] − 1

10. (7 − 2 ) 2

__________ 4 + 3 4 − 10

11. 2 · 7 2 − 10 _________ (9 + 1 ) 3

12. 10 ÷ 2 − 3

13. 7 − 6 _____ 2 + 4

14. 16 − 4[3 + 2 ÷ (9 − 7)]

15. 10 ÷ 1 5 − 3

16. 100 ÷ (3 + 22)

17. 2 · 9 − 14 ÷ 7 · 4

18. 5 + [2 · (14 − 2) ÷ 8] · 3

19. ( 3 2 + 1) · 2

20. 12 − (27 ÷ 3)

21. 8 · 2 ______ 19 − 3

22. 10 ÷ 2 + 5 − 8 _____________ 6(2 + 5) − 3

23. 12 − 5 · 2 + 6 ÷ 6

24. 30 ÷ (9 + 6)

25. 2 + 8 · 3 ÷ 4 _________________ 30 − 3 · (2 + 4) + 2

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8 UNIT 1 ALGEBRA BASICS

Solve.

*31. Challenge 2 3 7 3 18 9 Place one of each of the four basic math operations (+ − · ÷) in the blanks to get an expression with a value of 347.

*32. Challenge Simplify. 312 − (4 + 8) ÷ 3 · 5

__________________ [(1 + 4 2 ) · 9] − 7

Place grouping symbols in each expression to get expressions that have the given value.

26. 3 · 4 + 3 2 · 8

A. 312

B. 168

27. 8 + 9 2 − 7 · 10

A. 820

B. 748

28. 6 2 · 2 + 5 · 20

A. 1540

B. 5040

29. 720 ÷ 2 + 7 · 2 3

A. 10

B. 2936

30. 159 − 81 ÷ 3 3 + 12

A. 168

B. 2

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Variables

A variable is a symbol that represents a value.

Variables are usually represented by lowercase letters in italics. Most books use x, y, a, and n more than other letters. A variable expression is a combina-tion of variables, numbers, and operations. Examples of variable expressions are shown below.

x − 2 5n + 7 c + 1 _____ d− 2 2y − 21z + 6

Unlike numerical expressions, to evaluate variable expressions, substi-tute, or replace, numbers for the variables. So, the value of a variable expres-sion depends on the numbers chosen for the variable(s).

Evaluating Variable ExpressionsTo evaluate a variable expression, replace all the variables in the expression with numbers and simplify the resulting numerical expression.Example 1 Evaluate.A. 5n + 7 when n = 8Solution

5n + 7 = 5 · 8 + 7 Substitute 8 for n. = 40 + 7 Multiply. = 47 Add. ■

B. c + 1 _____ d − 2 when c = 3 and d = 4

Solution

c + 1 _____ d − 2 = 3 + 1 _____ 4 − 2 Substitute 3 for c and 4 for d.

= 4 __ 2 Simplify the numerator, then the denominator.

= 2 Divide. ■

A variable expression can also be called an algebraic expression.

THINK ABOUT IT

VARIABLES 9

5n When a number and variable are written together, the operation is multiplication, so 5n = 5 · n.

NOTATION

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10 UNIT 1 ALGEBRA BASICS

Identifying TermsVariable expressions consist of terms. Terms are the parts of an expression that are added or subtracted. A term that has no variables is a constant.

3 terms

constant

2y – 21z + 6

constant

2 terms

n––2

– 3mn

4 terms

2a + 3b + 4c + 7

Example 2 Identify the terms of the expression 3x + y + 23.Solution Each term is separated by a “+” sign. The terms are 3x, y, and 23. The number 23 is a constant since there is no variable. ■

Identifying Coefficients and FactorsTerms can consist of two or more factors. The numerical part of a term is the coefficient. When the coefficient of a term is 1, it is usually not written.

3 factors

coefficient

–3mn

Example 3 Identify the coefficient and factors of each term. A. −mnSolution You can write the term −mn as −1 · m · n. So, the coefficient is −1. The factors are −1, m, and n. ■

B. n __ 2

Solution You can write the term n __ 2 as 1 __ 2 · n. So, the coefficient is 1 __ 2 . The factors are 1 __ 2 and n. ■

Applications: Measurement and Distance TraveledYou can use expressions to solve many types of problems, including those involving measurement. Example 4

A. Natalie jogged once around the perimeter of a football field, which has a length of 100 yards and a width of 60 yards. Find the total number of yards she jogged.

Solution Use the expression 2l + 2w, where l is the length of the rectangle and w is the width of the rectangle, to find the perimeter. Evaluate 2l + 2w when l = 100 and w = 60.2l + 2w = 2 · 100 + 2 · 60 Substitute 100 for l and 60 for w. = 200 + 120 Multiply. = 320 Add.Natalie jogged 320 yards. ■

A constant is a term.

REMEMBER

In 2 · 3 = 6, 2 and 3 are called factors and 6 is called the product.

REMEMBER

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B. Find the area of the triangle shown below.

30 feet

12 feet

Solution Use the expression 1 __ 2 bh, where b is the length of the base and h is the height of the triangle, to find the area. Evaluate 1 __ 2 bh when b = 30 and h = 12.

1 __ 2 bh = 1 __ 2 · 30 · 12 Substitute 30 for b and 12 for h.

= 15 · 12 Multiply from left to right. = 180 Multiply.The area of the triangle is 180 square feet. ■C. Find the number of miles the Perez family traveled if they drove

nonstop for 2.5 hours at a rate of 60 miles per hour. Solution Use the expression rt, where r is the rate of travel in miles per hour and t is the number of hours traveled, to find the number of miles the vehicle traveled. Evaluate rt when r = 60 and t = 2.5.rt = 60 · 2.5 Substitute 60 for r and 2.5 for t. = 150 Multiply.The Perez family traveled 150 miles. ■

Area is measured in square units.

REMEMBER

Problem Set

Evaluate the expression.

1. 3x − 9 when x = 5

2. 5 + 4z when z = 7

3. m + 3 ______ n + 4 when m = 2 and n = 6

4. 5 − a _____ b + 2 when a = 1 and b = 10

5. 5a + 3b − 6c when a = 2, b = 8, and c = 3

6. 2a − 7 ______ 6b + 3 when a = 5 and b = 9

7. 2x − 2y + 2 ___________ 36 − 3x when x = 7 and y = 2

8. 14 + 5x when x = 13

9. 75m + 11n when m = 3 and n = 8

10. 20 + 3c + 17d when c = 8 and d = 2

11. 9a ___ c when a = 3 and c = 14

Identify the terms of the expression.

12. a + 3b − 18 13. 5 x 2 − 3y

VARIABLES 11

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12 UNIT 1 ALGEBRA BASICS

Solve.For problems 24 and 25: The expression for the perimeter of a rectangle is 2l + 2w. The expression for the area of a rectangle is lw.

24.

20 ft

14 ft

A. Find the perimeter of the rectangle shown.

B. Find the area of the rectangle.

25.

21cm

15 cm

A. Find the perimeter of the rectangle shown.

B. Find the area of the rectangle.

19. z 3 − 6x + 4

20. 2m ___ 4 − n 2 + 3p

___ 2

21. 3m ___ 2n − 12

22. 3x − y + 5z

23. 47 ___ 3x + 2y − 19 ___ 51

For problems 26 and 27: The expression for the area of a triangle is 1 __ 2 bh.

27. Jaya’s garden is in the shape of this triangle. In order to buy the correct amount of fertilizer, she needs to know how many square feet the garden measures. Find the area of the garden.

18 ft

17 ft

For problems 28 through 30: The expression for distance traveled is rt, where r represents rate and t represents time.

28. Find the distance Amita bicycled if she rode for 2 hours at a rate of 19 miles per hour.

29. Find the distance traveled if a plane fl ew for 3.5 hours at 550 miles per hour.

30. Challenge Julio walked his dog around his neighborhood 3 times. He walked for 0.75 hour at a rate of 4 miles per hour. How many miles would Julio have walked if he walked his dog around his neighborhood once?

*

Identify the coefficient and factors of the term.

14. 3bc

15. 2x ___ 3

16. bcd

17. 3xy

___ 7

18. −xyz

Identify the terms and constant of the expression.

26. Find the area of the triangle shown.

9�in.

12�in.

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Translating Words into Variable ExpressionsIt is sometimes necessary to write a variable expression from words or phrases.

The table below shows common words and phrases for each of the four basic operations.

Operation Words and Phrases

Addition plus, more than, increased by, sum

Subtraction minus, less than, decreased by, difference

Multiplication times, product

Division divided by, quotient

The phrase “the quantity” indicates that grouping symbols should appear in the expression. For example, the expression 4x + 8 represents the phrase “four times x plus eight,” but the expression 4(x + 8) represents the phrase “four times the quantity x plus eight.”

Translating Word Phrases into Variable ExpressionsExample 1 Translate each word phrase into a variable expression. Use n to represent “a number” in each expression.A. twenty-five times a number B. six less than twice a number Solution 25n ■ Solution 2n − 6 ■

C. twice the sum of nine and a D. the quotient of three and a number number

Solution 2(9 + n) ■ Solution 3 __ n ■

The phrases “y less than x” and “x is decreased by y” are represented by the same expression: x − y.

TIP

Unless stated otherwise, you can choose any letter for the variable(s).

REMEMBER

TRANSLATING WORDS INTO VARIABLE EXPRESSIONS 13

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14 UNIT 1 ALGEBRA BASICS

Using Only One Variable to Write a Variable ExpressionSometimes it is useful to write different quantities in terms of the same variable. For example, suppose you know the length of a rectangle is three times its width. Instead of using two variables, l and w, to represent the length and width, you can use w for width and 3w for length.Example 2 There are twelve more girls than boys in a classroom. Write expressions for the number of boys and the number of girls using the same variable b.Solution If b represents the number of boys, you can also represent the number of girls in terms of b. Number of boys: b

Number of girls: b + 12You can check your work by substituting a number for b. So, if there are 8 boys, then the number of girls is 8 + 12, or 20. ■

Application: Measurement You can write a variable expression to help you convert from one unit of measure to another.Example 3 Find the number of feet in 2 yards, 5 yards, and y yards. Solution There are 3 feet in 1 yard, so the number of feet is equal to the number of yards times 3.

Number of Yards Number of Feet

2 3 · 2 = 6

5 3 · 5 = 15

y 3 · y = 3y

There are 6 feet in 2 yards, 15 feet in 5 yards, and 3y feet in y yards. ■

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TRANSLATING WORDS INTO VARIABLE EXPRESSIONS 15

Application: GeometryGeometric formulas are basically expressions that describe how to figure out values such as perimeter or area.Example 4

A. Write and simplify an expression for the perimeter of the quadrilateral.

12

x

97

B. Use the simplified expression to find the perimeter when x = 5.Solution

A. x + 7 + 9 + 12 Sum of all four sides = x + 28 Add. The simplified expression for the perimeter is x + 28. ■B. Evaluate x + 28 when x = 5. x + 28 = 5 + 28 Substitute 5 for x. = 33 Add. When x = 5, the perimeter is 33 units. ■

Problem Set

Translate each word phrase into a variable expression. 1. twelve times a number

2. fi ve more than three times a number

3. four times the sum of fi ve and a number

4. three less than the product of a number and six

5. a number decreased by seven

6. one-half the sum of a number and three

7. the quotient of a number and ten

*8. Challenge two more than the sum of a number and four divided by nine

For each situation, write variable expressions using only one variable.

9. Armen owns twelve fewer DVDs than CDs. Express the number of DVDs and the number of CDs Armen owns in terms of the number of CDs c.

10. Fatima is three times the age of her younger sister Melody. Express both ages in terms of Melody’s age m.

11. There are a total of one hundred nine cars and trucks in a parking lot. Express the number of each automobile in terms of the number of trucks t.

12. Ciara’s algebra book cost half as much as her biology book. Express the price of each book in terms of the price of the biology book b.

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16 UNIT 1 ALGEBRA BASICS

Solve.

Solve. Write a variable expression when needed.

17. Find the number of inches in each.

A. 2 feet

B. 6 feet

C. x feet

18. Find the number of quarts in each.

A. 3 gallons

B. 5 gallons

C. n gallons

19. Find the number of minutes in each.

A. 5 hours

B. 7 hours

C. 2y hours

20. How many hours are in 3x days?

21. How many meters are in 5 __ n kilometers?

13. A band traveled all day to reach a state competi-tion. In the morning, they traveled twenty-fi ve miles less than twice the distance they traveled in the afternoon. Express each distance traveled in terms of the distance traveled in the afternoon a.

14. A deluxe salad costs seven times more than one-fourth of the price of a garden salad. Express the price of each salad in terms of the price of the garden salad g.

15. There are four more women than men at a golf club. Express the number of men and the number of women in terms of the number of men m.

*16. Challenge A clothing store has twice as many blue shirts as red shirts, and one-tenth as many yellow shirts as blue shirts.

A. Express the number of each shirt in terms of the number of red shirts r.

B. If the store has twenty red shirts, how many red, blue, and yellow shirts does the store have total?

22.

x cm

20 cm

A. Write an expression for the perimeter of the rectangle.

B. Find the perimeter when x = 42.

C. Write an expression for the area of the rectangle.

D. Find the area when x = 42.

23.

4n

n

A. Write an expression for the perimeter of the rectangle.

B. Find the perimeter when n = 3.

C. Write an expression for the area of the rectangle.

D. Find the area when n = 3.

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24.

y in.

8 in.10 in.

A. Write an expression for the perimeter of the triangle.

B. Find the perimeter when y = 7.

25.

b cm

14 cm

A. Write an expression for the area of the triangle.

B. Find the area when b = 9.

26.

5

a

2

4

12

A. Write and simplify an expression for the perimeter of the fi gure.

B. Find the perimeter when a = 8.

27.

nn

10

7

A. Write and simplify an expression for the perimeter of the quadrilateral.

B. Find the perimeter when n = 6.

TRANSLATING WORDS INTO VARIABLE EXPRESSIONS 17

Use the expression rt to answer the following questions.

28. A. Write an expression for the distance traveled for t hours at 50 miles per hour.

B. Find the distance when t = 2.5.

29. A. Write an expression for the distance traveled for t hours at 35 miles per hour.

B. Find the distance when t = 8.

30. A. Write an expression for the distance traveled for 3.6 hours at r miles per hour.

B. Find the distance when r = 48.

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18 UNIT 1 ALGEBRA BASICS

Equations

When you join two expressions with an equals sign, you create an equation.

The expressions 2 + 5 and 1 + 6 each simplify to the same value, 7. Because they have the same value, you can join them together with an equals sign to form an equation: 2 + 5 = 1 + 6.

An equation is a number sentence that indicates that two expressions have the same value.

DEFINITION

One way to distinguish between an equation and an expression is to look for an equals sign. Equations have an equals sign and expressions do not. Also, equations correspond to sentences, and expressions correspond to phrases.

Expression Equation

7 + 3 7 + 3 = 10

Phrase: the sum of seven and three

Sentence: The sum of seven and three is ten.

Determining if Two Expressions Form an EquationWhen two expressions are not equal, they do not form an equation. Instead of joining them with an equals sign, join them with the “not equals” sign (≠). Example 1 Write = or ≠.A. 5 · 2 + 7 5 · (2 + 7)Solution Use the order of operations to simplify the expressions.5 · 2 + 7 = 10 + 7 5 · (2 + 7) = 5 · (9) = 17 = 45The expression on the left has a different value than the expression on the right. So, the expressions do not form an equation: 5 · 2 + 7 ≠ 5 · (2 + 7). ■B. 3[5(2 + 1)] 100 − 5 · 11Solution [5(2 + 1)] = 3[5(3)] 100 − 5 · 11 = 100 − 55 = 3[15] = 45 = 45Both expressions have the same value. So, the expressions form an equation: 3[5(2 + 1)] = 100 − 5 · 11. ■

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EQUATIONS 19

Determining if a Given Value Makes an Open Sentence TrueAn equation that contains a variable is an open sentence. This means that you do not know if the sentence is true or false until a value is substituted for the variable. Example 2 Determine if the given value makes the open sentence true or false.A. 5a − 4 = 2; a = 2Solution Substitute the given value into each occurrence of the variable to see if the left and right sides of the equation are the same value. 5a − 4 = 2 5 · 2 − 4 � 2 10 − 4 � 2 6 ≠ 2 The sentence is false when a = 2. ■B. 10 = 0.5x; x = 40Solution 10 = 0.5x 10 � 0.5 · 40 10 ≠ 20 The sentence is false when x = 40. ■C. 5n = 7n − 6; n = 3Solution 5n = 7n − 6 5 · 3 � 7 · 3 − 6 15 � 21 − 6 15 = 15 � The sentence is true when n = 3. ■

Application: Phone PlanExample 3 A phone company offers two monthly plans. The first plan costs $25 per month plus $0.10 per minute. The second plan costs $10 per month plus $0.25 per minute. Determine if the monthly cost is the same when 100 minutes are used.Solution To determine if the cost is the same when 100 minutes are used, evaluate 25 + 0.10x = 10 + 0.25x when x = 100. 25 + 0.10x = 10 + 0.25x25 + 0.10 · 100 � 10 + 0.25 · 100 25 + 10 � 10 + 25 35 = 35 �Yes, the monthly costs are the same when 100 minutes are used. ■

Not all open sentences are equations. For instance, 3 + y ≠ 5 is an open sentence.

THINK ABOUT IT

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20 UNIT 1 ALGEBRA BASICS

Problem Set

Write = or ≠ to make a true statement.

1. 3 · 6 − 4 3 · (6 − 4)

2. 10 · 7 − 3 10 · (7 − 3)

3. 4[6(2 + 3)] 125 − 5 · 1

4. 4[3(5 + 2)] 95 − 4 · 3

5. 2 · 10 − 4 2 · (10 − 4)

6. 4 · 7 − 2 · 10 4(7 − 2) · 10

7. 5 · 11 + 12 · 3 19 + 3(8 · 3)

8. 2[5(7 − 1)] 15 + 3(5)

9. 12[1 + (2 · 2)] 36 − 4(2 + 4)

10. 4 + 3 ·�6 ________ 3 · 2 + 5 12 − 2 ·�3 _________ 2 · 1 +�1

11. 16 ÷ [10 + (7 − 1)] 16 ÷ (10 + 7 − 1)

Determine if the given value makes the open sentence true or false.

12. 6a + 3 = 15; a = 1

13. 16 = 0.4x; x = 40

14. 3z = 2z + 5; z = 5

15. 4b − 2 = 10; b = 3

16. 8 − 4x = 4x; x = 1

17. 0.1x = 12; x = 120

18. 3 + 0.25x = x; x = 16

19. 8b = 48; b = 7

20. 15 = 3d − 3; d = 6

21. 9 = 0.5y; y = 45

22. 6g = 2g + 28; g = 7

Solve.

23. FitGym offers two monthly membership plans. Suri chose the plan that is $35 per month so she can attend all offered classes for free. Arni chose the plan that is $10 per month and he must pay $5 to attend each offered class. Determine if Suri and Arni were charged the same in the month of July if they each attended fi ve classes.

24. Khali has $300 and is saving $50 every month. Tara has $650 and is spending $75 every month. Will the girls have an equal amount of money after three months?

25. Speed-ee car rental charges $22 per day to rent a car, and $0.15 per mile driven over the limit. Zip-ee car rental charges $18 per day to rent a car, and $0.20 per mile driven over the limit. Determine if a car will cost the same at both agencies if it is driven 30 miles over the limit.

26. A local salad bar charges $3 per pound or $7.50 per container. Bianca chose to pay per pound and her salad weighed 2.5 pounds. Simon fi lled one container and chose to pay per container. Determine if Bianca and Simon spent the same amount.

27. At an amusement park, you can choose to pay $60 for a wrist band and ride unlimited rides or pay $10 to enter and $3 per ride. Determine if the two ways of paying result in the same cost if you ride 20 rides.

*28. Challenge Window Fixers charges a fee of $55 and $20 per window for installation. We Fix Panes charges an $85 fee and $10 per window for installation.

A. Is the cost of having four windows installed the same at both companies?

B. Which company will charge less to install eight windows?

*29. Challenge Arturo and Andi took a test that had 30 multiple choice questions worth 2 points each and 8 open-ended questions worth 5 points each. Arturo answered all of the multiple choice questions correctly and 6 of open-ended questions correctly. Andi answered 25 of the multiple choice questions correctly and all of the open-ended questions correctly. Determine if Arturo and Andi received the same score.

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Translating Words into EquationsUse the same words and phrases that allow you to translate word phrases into variable expressions to translate word sentences into equations.

Verbal sentences often use the word “is” and the phrase “is equal to.” Represent each by an equals sign in an equation.

Translating Sentences into EquationsExample 1 Translate each sentence into an equation.A. Two less than five times a number is equal to seven.Solution Two less than five times a number is equal to seven.

5n − 2 = 7The equation is 5n − 2 = 7. ■B. The quotient of a number and four is the sum of six and the number.

TRANSLATING WORDS INTO EQUATIONS 21

Solution The quotient of a number and four is the sum of six and the number.

n __ 4 = 6 + n

The equation is n __ 4 = 6 + n. ■

Finding True StatementsYou may be asked to find the value of a variable that would make a verbal sentence true. If so, start by translating the sentence into an equation.Example 2 Find a value of the variable that makes each statement true.A. The product of six and a number is forty-two.Solution The product of six and a number is forty-two.

6n = 42 The equation is 6n = 42. Find the value of n by answering the question: What number times 6 equals 42? The answer is 7. ■

(continued)

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22 UNIT 1 ALGEBRA BASICS

B. Twelve less than a number is equal to nine.Solution Twelve less than a number is equal to nine.

n − 12 = 9 The equation is n − 12 = 9. Find the value of n by answering the question: What number reduced by 12 equals 9? The answer is 21. ■

Applications: Distance and GeometryEquations can represent many real-world situations. These applications include problems involving distance traveled, as well as dimensions of common geometric shapes.Example 3

Carly rode her bicycle at a constant speed of 4 miles per hour to her friend’s house. Write an equation to represent the distance Carly rode.Solution The formula for the distance traveled is d = rt, where d is the distance, r is the rate (or speed), and t is the time traveled. Since Carly’s speed is 4 miles per hour, r = 4. Substitute 4 for r to obtain the equation d = 4t. ■Example 4

A. A rectangle is 2 meters long, but the width is unknown. Write equations for the perimeter and area of the rectangle.

Solution The formula for the perimeter of a rectangle is P = 2l + 2w, and the formula for the area of a rectangle is A = lw. Substitute the known information into each formula and simplify as needed. P = 2l + 2w A = lw P = 2 · 2 + 2w A = 2w P = 4 + 2wThe equations are P = 4 + 2w and A = 2w. ■B. Write an equation for the perimeter of the triangle.

3b

12

6.5

Solution The perimeter of a figure is the sum of the side lengths of the figure. P = 12 + 6.5 + 3b P = 18.5 + 3bThe equation is P = 18.5 + 3b. ■

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Problem Set

Translate each sentence into an equation.

1. Three more than four times a number is equal to twelve.

2. The sum of a number and two is twice the number.

3. The quotient of a number and twelve is the sum of eight and the number.

4. Five more than three times a number is equal to eight.

5. The sum of a number and six is seven times the number.

6. The quotient of a number and six is equal to nine.

7. Seven less than a number is two.

8. Twenty-seven is the product of three and a number.

9. Twelve less than six times a number is equal to three times the number.

10. The product of two and the quotient of a number and four is equal to three less than ten times the number.

Find a value of the variable that makes each statement true.

11. The product of fi ve and a number is equal to forty-fi ve.

12. Fifteen is equal to ten less than a number.

13. The quotient of nine and a number is one.

14. Seventeen is equal to eleven less than a number.

15. The quotient of a number and fi ve is eight.

16. The sum of a number and six is equal to the sum of two times the number and one.

17. The product of twelve and a number is equal to seventy-two.

18. The quotient of twenty and a number is four.

19. The product of eight and the quotient of a number and three is equal to three.

20. Forty less than a number is equal to fi ve.

TRANSLATING WORDS INTO EQUATIONS 23

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24 UNIT 1 ALGEBRA BASICS

Solve.

21. Tran drove his car at a constant speed of 55 miles per hour. Write an equation for the distance Tran drove.

22.

3x3x

12

12

A. Write an equation for the perimeter of the rectangle shown.

B. Write an equation for the area of the rectangle.

23. The formula for the area of a triangle is

A = 1 __ 2 bh, where b is the length of the base of

the triangle and h is the height of the triangle. Write an equation to fi nd the area of a triangle with a base length of 6 cm and an unknown height.

24. Hooke’s law states that F = −kx, where F is the force by the spring, k is the spring constant, and x is the displacement (distance). Write an equation for the force by a spring with a spring constant of 30 N/m and an unknown displacement.

25. The weight of an object is found by the formula W = mg, where m is the mass of the object and g is the acceleration due to gravity. Write an equation for the weight of an object with an unknown mass, with an acceleration due to gravity of 9.8 m/ s 2 .

26. Lorrae swims laps at a constant speed. If Lorrae swims for 2 hours, write a formula for the distance she swims.

27. Write an equation for the area of the square.

s

*28. Challenge Projectile motion can be modeled

using the equation y = v 0 t − 1 __ 2 g t 2 , where y is

the height the object reaches, g is the acceleration due to gravity, t is the time, and v 0 is the initial velocity.

A. Write an equation for the height of a baseball thrown into the air with an initial velocity of 16 m/s, and an acceleration due to gravity of 9.8 m/ s 2 .

B. What is the height of the baseball after 3 seconds?

*29. Challenge The formula for the area of a

trapezoid is A = 1 __ 2 ( b 1 + b 2 )h.

h

12

10

A. Write an equation for the area of the trapezoid shown.

B. Find the area if h = 5.

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One method you can use to solve an open sentence involves replacement sets.

Because 10 − 3x = 4 is a true statement when x = 2, the number 2 is a solution of the equation. The process of finding a value that makes an open sentence true is called solving the equation or inequality.

A set is a collection of objects, and each member of the set is an element of the set. The elements of a set are enclosed in braces, so the set containing the elements 10, 15, and 20 is written as {10, 15, 20}. To show that a value is an element of a set, you can use the symbol ∈.

∈ is an element of

For instance, 2 ∈ {2, 4, 6, 8}.

NOTATION

Replacement Sets

A set of values that are allowable as a solution to an open sentence is called the replacement set, or domain. The elements of the replacement set that make an open sentence true become the solution set.

DEFINITIONS

Finding a Solution SetExample 1 Find the solution set for 5x + 3 = 18 given the replacement set {0, 1, 2, 3, 4, 5}.Solution Substitute each element of the replacement set for x to see which, if any, make the open sentence true. 5x + 3 = 18 5x + 3 = 18 5x + 3 = 185 · 0 + 3 � 18 5 · 1 + 3 � 18 5 · 2 + 3 � 18 0 + 3 � 18 5 + 3 � 18 10 + 3 � 18 3 ≠ 18 8 ≠ 18 13 ≠ 18

5x + 3 = 18 5x + 3 = 18 5x + 3 = 185 · 3 + 3 � 18 5 · 4 + 3 � 18 5 · 5 + 3 � 18 15 + 3 � 18 20 + 3 � 18 25 + 3 � 18 18 = 18 � 23 ≠ 18 28 ≠ 18The open sentence is true when x = 3. The solution set is {3}. ■

The open sentence 5x + 3 = 18 is an equation with solution x = 3.

THINK ABOUT IT

REPLACEMENT SETS 25

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26 UNIT 1 ALGEBRA BASICS

Special Cases: Empty and Null SetsIn Example 1, only one element from the replacement set became part of the solution set. It is possible for there to be two or more solutions of an open sentence. It is also possible that none of the elements in the replacement set make the open sentence true. When no elements in the replacement set make the open sentence true, the solution is the empty set, written as � or { }.Example 2 Find the solution set for each open sentence given the replacement set {1, 2, 3, 4}.A. 5x + 3 = 2(x + 3) − 3 + 3xSolution

{ } or � null or empty set

NOTATION

5x + 3 = 2(x + 3) − 3 + 3x 5x + 3 = 2(x + 3) − 3 + 3x 5 · 1 + 3 � 2(1 + 3) − 3 + 3 · 1 5 · 2 + 3 � 2(2 + 3) − 3 + 3 · 2 5 + 3 � 2 · 4 − 3 + 3 10 + 3 � 2 · 5 − 3 + 6 8 � 8 − 3 + 3 13 � 10 − 3 + 6 8 � 5 + 3 13 � 7 + 6 8 = 8 � 13 = 13 �

5x + 3 = 2(x + 3) − 3 + 3 x 5x + 3 = 2(x + 3) − 3 + 3x 5 · 3 + 3 � 2(3 + 3) − 3 + 3 · 3 5 · 4 + 3 � 2(4 + 3) − 3 + 3 · 4 15 + 3 � 2 · 6 − 3 + 9 20 + 3 � 2 · 7 − 3 + 12 18 � 12 − 3 + 9 23 � 14 − 3 + 12 18 � 9 + 9 23 � 11 + 12 18 = 18 � 23 = 23 �

The solution set is every member of the replacement set {1, 2, 3, 4}. ■B. 5x + 3 = 2(x + 3) − 6 + 3xSolution

5x + 3 = 2(x + 3) − 6 + 3x 5x + 3 = 2(x + 3) − 6 + 3x 5 · 1 + 3 � 2(1 + 3) − 6 + 3 · 1 5 · 2 + 3 � 2(2 + 3) − 6 + 3 · 2 5 + 3 � 2 · 4 − 6 + 3 10 + 3 � 2 · 5 − 6 + 6 8 � 8 − 6 + 3 13 � 10 − 6 + 6 8 � 2 + 3 13 � 4 + 6 8 ≠ 5 13 ≠ 10

5x + 3 = 2(x + 3) − 6 + 3 x 5x + 3 = 2(x + 3) − 6 + 3x 5 · 3 + 3 � 2(3 + 3) − 6 + 3 · 3 5 · 4 + 3 � 2(4 + 3) − 6 + 3 · 4 15 + 3 � 2 · 6 − 6 + 9 20 + 3 � 2 · 7 − 6 + 12 18 � 12 − 6 + 9 23 � 14 − 6 + 12 18 � 6 + 9 23 � 8 + 12 18 ≠ 15 23 ≠ 20

No member of the replacement set makes the open sentence true. The solution set is the empty set, �. ■

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Application: PerimeterReplacement sets can help you solve real-world problems that have a set list of possible answers.Example 3 Darrell is building a tree house with a rectangular floor. Two opposite sides are already set in place, each 10 feet long. He wants the perimeter of the floor to be 50 feet. Find the solution set for 50 = 2l + 20 given the replacement set {8, 10, 12, 15} to determine which size boards Darrell should buy to build the remaining sides of the floor.

l ft

l ft

10 ft10 ft

Solution Substitute each element of the replacement set for l.50 = 2l + 20 50 = 2l + 20 50 = 2l + 20 50 = 2l + 2050 � 2 · 8 + 20 50 � 2 · 10 + 20 50 � 2 · 12 + 20 50 � 2 · 15 + 2050 � 16 + 20 50 � 20 + 20 50 � 24 + 20 50 � 30 + 2050 ≠ 36 50 ≠ 40 50 ≠ 44 50 = 50 �The solution set is {15}. So, Darrell should buy 15-foot boards for the remaining sides. ■

Problem Set

Find the solution set for each open sentence with the given replacement set.

1. 2x − 3 = 5, {0, 1, 2, 3, 4, 5}

2. y + 5 = −1, {−2, −4, −6, −8}

3. 4x + 2 = 7(x − 1) + 2x, {1, 3, 5, 7, 9}

4. 2x − 1 = 3(x + 2) − 4x − 7, {0, 1, 2, 3, 4}

5. m 2 + 4 = 5, {−2, −1, 0, 1, 2}

6. b − (−3b − 3) = 4b + 3, {1, 2, 3, 4}

7. t − 1 = 7 − t, {0, 4, 8, 12}

8. 1.5n + (−2 + 0.5) = 3 − 2(4.5 + 0.5), {−5, −4, −3, −2, −1}

9. −10v + 10 = −10, {0, 1, 2, 3, 4, 5}

10. | z + 1 | = | 2z + 1 − z | , {−5, 0, 5, 10}

11. y − 1 = 0, {1, 2, 3, 4, 5, 6}

12. 2f + 23 = 3( f − 7), {14, 15, 16, 17, 18}

13. 12n − 40 = −20 − (−160), {0, 5, 10, 15}

14. | a __ 2 + 10 | = 20 + (−10 − 5), {−30, −10, 10, 30}

15. 4n + 2 = 2(1 + 2n), {11, 12, 13, 14}

16. x 2 + 2 = x(4 − x) + x, {0, 1, 2, 3, 4)

17. t + 17 = 13, {−12, −8, −4, 0, 4}

18. 2 ( | 100 ____ c | ) = (20 − |c|) + | c __ 2 | , {−20, −10, 10, 20}

19. 4 − t = 1, {1, 2, 3, 4}

20. 3v − (7 − 1) = 5 + (2 + 1), {−1, 0, 1, 2}

REPLACEMENT SETS 27

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28 UNIT 1 ALGEBRA BASICS

Solve.

21. Merrie wants to paint an accent wall in her living room. She knows that the wall is 18 feet long, but she cannot remember the exact height. She knows that it is somewhere around 10 feet high, and she previously calculated that she would need paint to cover 162 square feet. Find the solution set for 18h = 162 given the replacement set {8, 9, 10, 11} to determine the height of her ceiling.

22. Charo is buying food for her friends and is ordering from the $1 and $2 value menus. She knows that she needs three $1 sandwiches, two $2 parfaits and two $2 salads. Drinks are $1, and she has $13 in her wallet. Find the solution set for 3 · 1 + 2 · 2 + 2 · 2 + d · 1 = 13 given the replacement set {0, 1, 2, 3} to determine how many drinks she can purchase.

23. Jacob is making two replicas of his favorite team’s triangular pennant. He only has 17 feet of the special trim that he wants to use to border both pennants. He wants the base of each pennant to be 1.5 feet. He also wants the sides of each pennant to be longer than the base. Find the solution set for 2(1.5 + 2s) = 17 given the replacement set {2, 2.5, 3, 3.5, 4} to determine what length each side of the two pennants should be.

24. A movie theater customer bought two $3.50 popcorns, four $2.50 drinks, and four cups of mixed fruit for a total cost of $29. Find the solution set for 2 · 3.5 + 4 · 2.5 + 4c = 29 given the replacement set {1.5, 2, 2.5, 3} to determine the cost of each cup of mixed fruit.

25. Joaquim and three of his friends are planting fl owers in a circular garden. Each of them guessed the radius of the garden. They know the circumference of the garden is 157 feet. Find the solution set for 3.14 · 2r = 157 given the replacement set {20, 25, 30, 35} to determine the radius of the garden.

26. Travis decides to participate in a local charity event and fi ll a shoebox with presents for local children who are less fortunate. He goes to the store and buys two $5 toys, one $3 toy, three $4 toys, and three games that cost the same amount. The cashier rings up the total, then realizes that the games are supposed to be on sale, so she subtracts the original cost of one game to cover the price difference. The total cost is $35. Find the solution set for 2 · 5 + 1 · 3 + 3(4 + g) − g = 35 given the replacement set {2, 3, 4, 5} to determine the original cost of each game.

27. A park maintenance crew is laying recycled rubber mulch in a rectangular ground area to provide padding for a play area. They want the length of the area that is covered to be twice as long as the width that is covered. They have enough mulch to cover 750 square feet. Find the solution set for 2w · w = 800 given the replacement set {10, 15, 20, 25, 30} to determine how wide they can make the area covered by the rubber mulch.

28. Three thirsty friends drink an entire 64-ounce bottle of juice. Benito drank twice as much as Tala, and Amy drank 11 ounces less than Benito. Find the solution set for t + 2t + (2t − 11) = 64 given the replacement set {10, 15, 20, 25} to fi nd out how many ounces of juice Tala drank.

29. Chelsea is starting to train for a triathlon and must run 1 mile every day. The rectangular path that she runs on is 1 mile long and its width is 0.125 mile. Find the solution set for 2 · 0.125 + 2l = 1 given the replacement set {0.125, 0.25, 0.375, 0.5} to determine the length of the path.

30. A swimmer swims 30 laps along the length of a rectangular pool. The distance that he swims is the same as if he swam 10 times around the perimeter of a pool with the same length and a width of 25 m. Find the solution set for 30l = 10(2l + 2 · 25) given the replacement set {40, 50, 60, 70} to determine the length of each lap.

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Problem Solving

When using math to solve real-world problems, it can help to have a plan. The following suggested problem-solving strategy has five steps.

PROBLEM SOLVING 29

Step 1 Identify Read the problem and identify the unknowns. What are you

being asked to fi nd? Write it down in words. If you can, estimate the answer.

Step 2 Strategize Select and defi ne variables and variable expressions for all

the unknowns.Step 3 Set Up Write an equation, inequality, system of equations, or whatever

other tools you need that model the problem that is being solved.Step 4 Solve Solve the model (equation, inequality, system, etc.).Step 5 Check Check your answer for reasonableness and accuracy with the

original problem statement.

PROBLEM-SOLVING PLAN

Application: Groups of PeopleExample 1 There are 5 times as many students as teachers. There are 12 teachers. How many students are there?Solution Work through the five steps of the problem-solving plan.Step 1 Identify The unknown is the number of students. You are asked to

find the number of students. Estimate: There are more students than teachers, so the answer will be greater than 12. Twelve is close to 10, so a possible estimate is 50 students because 10 · 5 = 50.

Step 2 Strategize Define variables. Number of teachers: t Number of students: sStep 3 Set Up Write an equation to represent the fact that the number of

students is five times the number of teachers: s = 5t.

Algebraic equations can capture key relationships among quantities in the world.

RECONNECT TO THE BIG IDEA

(continued)

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30 UNIT 1 ALGEBRA BASICS

Step 4 Solve Substitute the number of teachers into the equation. s = 5t s = 5 · 12 s = 60 There are 60 students.Step 5 Check The answer of 60 is close to the estimate of 50 and is

reasonable. Because 60 ÷ 5 = 12, the answer is correct. ■

Application: Perimeter and AreaYou can also use the problem-solving plan to solve geometry problems.Example 2 The area of a rectangle is 70 square units and the width of the rectangle is 5 units. Find the perimeter of the rectangle. Solution You can use the problem-solving plan to solve problems involving distance, rate, and time.Step 1 Identify You are asked to find the perimeter. To find the

perimeter, you must know both the length and width of the rectangle. The width is known, but the length is not known.

Step 2 Strategize Write and solve an equation for the area of a rectangle to find the value of the length. Then find the perimeter of the rectangle. Define variables. Use w for width, l for length, A for area, and P for perimeter.

Step 3 Set Up Write an equation to represent the area. A = lw 70 = l · 5Step 4 Solve Think: “What number times 5 equals 70?” You can guess

and check, or divide: 70 ÷ 5 = 14. The length is 14 units. To find the perimeter, substitute 5 and 14 into P = 2l + 2w.

P = 2 · 14 + 2 · 5 = 29 + 10 = 39Step 5 Check 5 · 14 = 70. The given area is 70, so the length is correct.

The perimeter is the sum of the sides: 5 + 14 + 5 + 14 = 38. The answer is correct. ■

One way to check for accuracy is to use a diff erent method than the one used to solve the problem. In this check, we add all four side lengths instead of fi nding the sum of twice the length and twice the width.

TIP

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Application: Distance, Rate, and TimeExample 3 LeAnn drove 125 miles at a rate of 50 miles per hour. How long did it take her to drive this distance?Solution

Step 1 Identify You are given the distance and rate and are asked to find the time. Because 50 · 2 = 100, the answer should be greater than 2 hours.

Step 2 Strategize Use d for distance, r for rate, and t for time. Step 3 Set Up Distance is the product of the rate and the time, or d = rt.

Substitute the known information into the general equation: 125 = 50t.

Step 4 Solve To determine how many times 50 divides into 125, divide 125 by 50: 125 ÷ 50 = 2.5.

Step 5 Check Two and one-half hours is a reasonable time to drive 125 miles at a rate of 50 mph. Multiply to check for accuracy: 50 · 2.5 = 125. ■

Problem Set

Solve. For each problem:A. Define variables for the unknowns.B. Write an equation to model the problem.C. Solve the equation.D. Give your answer in a complete sentence.

1. There are 10 more boys than girls in class. There are 14 girls. How many boys are there?

2. There are 3 fewer than twice the number of tiles in the box than on the fl oor. There are 100 tiles on the fl oor. How many tiles are in the box?

3. The area of a rectangle is 80 square units and the width of the rectangle is 4 units. Find the perimeter of the rectangle.

4. Sean deposited $200 into an account and earned $36 in interest after 3 years. Use the formula I = prt, where I represents simple interest, p represents principal, r represents interest rate, and t represents time in years to compute the interest rate.

5. Roxy drove 200 miles at a rate of 50 miles per hour. How long did it take her to drive this distance?

6. Alban drove 1200 miles in 30 hours. What was his rate of travel in miles per hour?

PROBLEM SOLVING 31

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32 UNIT 1 ALGEBRA BASICS

7. Amida earned $60 in interest on her deposit after 5 years with a 0.04 interest rate. Use the formula I = prt, where I represents simple interest, p represents principal, r represents interest rate, and t represents time in years to compute the principal amount deposited.

8. The number of boys at a movie theatre is 6 times greater than the number of girls at a movie theatre. The number of girls at the movie theatre is 25. How many boys are at the movie theatre?

9. All sides of an equilateral triangle have equal length. If the perimeter of an equilateral triangle is 39 centimeters, what is the length of each side?

10. The area of a square fi eld is 529 square feet. Find the length of each side of the square.

11. Zahira drove at a rate of 50 miles per hour for 3.5 hours. How many miles did she travel?

12. The number of girls who work at Super Shopper is 2 more than 3 times the number of boys. If 30 boys work at Super Shopper, how many girls work there?

13. Ms. Zambia has three more students in her class than Mr. Powan. If Mr. Powan has 22 students in his class, how many students are in Ms. Zambia’s class?

14. Laura deposits $150 at a 5% interest rate. Use the formula I = prt, where I represents simple interest, p represents principal, r represents interest rate, and t represents time in years to compute her simple interest after 2 years.

15. Benoit plays baseball 3 times as much as he plays soccer. If Benoit plays soccer 6 hours per week, how many hours does he play baseball?

16. The number of backpacks sold at Out Door Adventure equals 10 fewer than 3 times the number of backpacks in the store. There are 30 backpacks in the store. How many backpacks were sold?

17. The sum of two equal angles is 110°. What is the measure of each of these angles?

18. The number of concert tickets sold today is 800 more than 6 times the number of tickets sold last year. One thousand tickets were sold last year. How many tickets were sold today?

19. Sarabi drove 20 miles at 40 miles per hour. How long did she drive?

20. The area of a triangle is 60 square units and the base of the triangle is 5 units. Find the height of the triangle.

21. Gita is 3 times as old as her son, Ajeet. If Gita is 36 years old, how old is Ajeet?

22. Katya works 8 more hours per week than Suzi. If Suzi works 16 hours per week, how many hours does Katya work?

23. Janel needs to convert the temperature 80ºF

to Celsius. If C = 5 __ 9 (F − 32), where C represents

degrees Celsius and F represents degrees Fahrenheit, what is the temperature in degrees Celsius?

24. Jeb can paint a room in 3 hours and Matthew can paint the same size room in 2.5 hours. If Jeb paints 5 rooms and Matthew paints 6 rooms, what is the total amount of time spent painting?

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25. Jamal worked 20 hours per week and earned $5.25 per hour. How much money did he earn?

26. Antonio has $1.65 in change. Three of those coins are dimes, 5 are quarters, and the remaining coins are nickels. How many nickels does Anthony have?

27. The width of a rectangular garden is 5 feet. The length of the garden is 3 times the width. How many feet of fencing is needed to enclose the garden?

28. Two computers are for sale. Computer A costs $700 less than 2 times the price of Computer B. If Computer B costs $900, how much does Computer A cost?

*29. Challenge Cerise has $5.67. She has 5 one-dollar bills, 2 quarters, and 3 nickels. What other coins does Cerise have?

*30. Challenge The length of a rectangle is 3 more than 2 times its width. The perimeter of the rectangle is 24 units. Find the width and length of the rectangle.

PROBLEM SOLVING 33

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