美国algebra1 Chapter04

美国 algebra 1 数学经典教材

CHAPTER

4CHAPTER TABLE OF CONTENTS4-1 Solving Equations Using More Than One Operation 4-2 Simplifying Each Side of an Equation 4-3 Solving Equations That Have the Variable in Both Sides 4-4 Using Formulas to Solve Problems 4-5 Solving for a Variable in Terms of Another Variable 4-6 Transforming Formulas 4-7 Properties of Inequalities 4-8 Finding and Graphing the Solution Set of an Inequality 4-9 Using Inequalities to Solve Problems Chapter Summary Vocabulary Review Exercises Cumulative Review

FIRST DEGREE EQUATIONS AND INEQUALITIES IN ONE VARIABLEAn equation is an important problem-solving tool. A successful business person must make many decisions about business practices. Some of these decisions involve known facts, but others require the use of information obtained from equations based on expected trends. For example, an equation can be used to represent the following situation. Helga sews hand-made quilts for sale at a local craft shop. She knows that the materials for the last quilt that she made cost$76 and that it required 44 hours of work to complete the quilt. If Helga received$450 for the quilt, how much did she earn for each hour of work, taking into account the cost of the materials? Most of the problem-solving equations for business are complex. Before you can cope with complex equations, you must learn the basic principles involved in solving any equation.

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Solving Equations Using More Than One Operation117

4-1 SOLVING EQUATIONS USING MORE THAN ONE OPERATION

Some Terms and Definitions

An equationis a sentence that states that two algebraic expressions are equal.

For example,x 3 9 is an equation in which x 3 is called the left side,or

left member,and 9 is the right side,or right member.

An equation may be a true sentence such as 5 2 7,a false sentence such

as 6 3 4,or an open sentence such as x 3 9.The number that can replace

the variable in an open sentence to make the sentence true is called a root,or a

solution,of the equation.For example,6 is a root of x + 3 9.

As discussed in Chapter 3,the replacement set or domain is the set of pos-

sible values that can be used in place of the variable in an open sentence.If no

replacement set is given,the replacement set is the set of real numbers.The set

consisting of all elements of the replacement set that are solutions of the open

sentence is called the solution setof the open sentence.For example,if the

replacement set is the set of real numbers,the solution set of x 3 9 is {6}.If

no element of the replacement set makes the open sentence true,the solution

set is the empty or null set,лor {}.If every element of the domain satisfies an

equation,the equation is called an identity.Thus,5 x x ( 5) is an iden-

tity when the domain is the set of real numbers because every element of the

domain makes the sentence true.

Two equations that have the same solution set are equivalent equations.To

solve an equationis to find its solution set.This is usually done by writing sim-

pler equivalent equations.

If not every element of the domain makes the sentence true,the equation is

called a conditional equation,or simply an equation.Therefore,x 3 9 is aconditional equation.

Properties of Equality

When two numerical or algebraic expressions are equal,it is reasonable to

assume that if we change each in the same way,the resulting expressions will be

equal.For example:

5 7 12

(5 7) 3 12 3

(5 7) 8 12 8

2(5 7) 2(12)

5These examples suggest the following properties of equality:

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118First Degree Equations and Inequalities in One Variable

Properties of Equality

1.The addition property of equality.If equals are added to equals,the sums

are equal.

2.The subtraction property of equality.If equals are subtracted from equals,

the differences are equal.

3.The multiplication property of equality.If equals are multiplied by equals,

the products are equal.

4.The division property of equality.If equals are divided by nonzero equals,

the quotients are equal.

5.The substitution principle.In a statement of equality,a quantity may be

substituted for its equal.

To solve an equation,you need to work backward or “undo”what has been

done by using inverse operations.To undo the addition of a number,add its

opposite.For example,to solve the equation x 7 19,use the addition prop-

erty of equality.Add the opposite of 7 to both sides.

x 1 75

27

5

19 27

The variable xis now alone on one side and it is easy to read the solution,

x 12.

To solve an equation in which the variable has been multiplied by a num-

ber,either divide by that number or multiply by its reciprocal.(Remember

multiplying by the reciprocal is the same as dividing by the number.) To solve

6x 24,divide both sides by 6 or multiply both sides by .

6x 24

6x6x 24or(6x)5245(24)x 4

(3)x 4To solve 55,multiply each side by the reciprocal of which is 3.555(3)5x 15

In the equation 2x 3 15,there are two operations in the left side:mul-

tiplication and addition.In forming the left side of the equation,xwas first mul-

tiplied by 2,and then 3 was added to the product.To solve this equation,we

must undo these operations by using the inverse elements in the reverse order.

Since the last operation was to add 3,the first step in solving the equation is to

add its opposite, 3,to both sides of the equation or subtract 3 from both sides

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Solving Equations Using More Than One Operation119

of the equation.Here we are using either the addition or the subtraction prop-

erty of equality.

2x 1 3515

2x 1 3 1 (23)515 1 (23)

2x512or2x 1 3515 23 235

Now we have a simpler equation that has the same solution set as the original

and includes only multiplication by 2.To solve this simpler equation,we multi-

ply both sides of the equation by ,the reciprocal of 2,or divide both sides of

the equation by 2.Here we can use either the multiplication or the division

property of equality.

2x512(2x)5(12)

x56or2x52x5x512126

After an equation has been solved,we checkthe equation,that is,we verify

that the solution does in fact make the given equation true by replacing the vari-

able with the solution and performing any computations.

Check:2x 3 15

2(6) 3 15

12 3 15

15 15

To find the solution of the equation 2x 3 15,we used several properties

of the four basic operations and of equality.The solution below shows the math-

ematical principle that we used in each step.

2x 3 15

(2x 3) ( 3) 15 ( 3)

2x [3 ( 3)] 15 ( 3)

2x 0 12

2x 12

(2x)

C(2)DxGivenAddition property of equalityAssociative property of additionAdditive inverse propertyAdditive identity propertyMultiplication property of equalityAssociative property of multiplication

Multiplicative inverse property

Multiplicative identity property (12) (12)1x 6x 6

These steps and properties are necessary to justify the solution of an equation

of this form.However,when solving an equation,we do not need to write each

of the steps,as shown in the examples that follow.

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120First Degree Equations and Inequalities in One Variable

EXAMPLE 1

Solve and check:7x 15 71

SolutionHow to Proceed

(1)Write the equation:

(2)Add 15,the opposite of 15 to

each side:

(3)Since multiplication and division are

inverse operations,divide each

side by 7:

(4)Check the solution.Write the solution

in place of xand perform the

computations:7x 15 71 15 157x 567x56 x 87x 15 71?7(8) 1 15 5 71?56 1 15 571

71 71

Answerx 8

Note:The check is based on the substitution principle.

EXAMPLE 2

Find the solution set and check:x 2 6 18

Solutionx 2 65 218

16 16Addition property of equality 12x

535AxB (212)Multiplication property of equality

x 20Check 6 18 ??(220) 2 6 5218212 2 6 5218 18 18

AnswerThe solution set is { 20}.

EXAMPLE 3

Solve and check:7 x 9

SolutionMETHOD1.Think of 7 xas 7 ( 1x).

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Solving Equations Using More Than One Operation121

71(2x) 5 9

27 27

5 Addition property of equality

Division property of equalityCheck7 x 9?7 2 ( 2) 5 97 1 2 5 9

9 9 ?5x 2

positive coefficient.

How to Proceed

(1)Write the equation:

(2)Add xto each side of the equation:

(3) Add 9 to each side of the equation:

The check is the same as for Method 1.METHOD2.Add xto both sides of the equation so that the variable has a7 x 97 x x 9 x7 9 x 9 7 9 9 x 2 x

Answer{ 2} or x 2

EXERCISES

Writing About Mathematics

1.Is it possible for the equation 2x 5 0 to have a solution in the set of positive real num-bers? Explain your answer.

2.Max wants to solve the equation 7x 15 71.He begins by multiplying both sides of theequation by the reciprocal of the coefficient of x.

a.Is it possible for Max to solve the equation if he begins in this way? If so,what

would be the result of multiplying by and what would be his next step?

b.In this section you learned to solve the equation 7x 15 71 by first adding

the opposite of 15, 15,to both sides of the equation.Which method do you

think is better? Explain your answer.

Developing Skills

In 3 and 4,write a complete solution for each equation,listing the property used in each step.

3.3x 5 354.x21 15

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122First Degree Equations and Inequalities in One Variable

In 5–32,solve and check each equation.

5.55 6a 7

9.15 a 3

13.x 8

17.7.2 21.4a 0.2 5

25.t 1 7 47

29.514 2 x6.17 8c 710.11 6d 114.12 y18. 1 9 522.4 3t 0.226.0.04c 1.6 030.0.8r 19 207.9 1x 711.8 y = 115.5y8.11 15t 1612. 1216.2m 3020.9d 2 51724.13 5 2 y19. 2 1 323.x 1 11 527.15x 14 1931.w 1 652228.8 18c 132.842 162m 616Applying Skills

33.The formula F C132gives the relationship between the Fahrenheit temperature Fand

the Celsius temperature C.Solve the equation 59 C132to find the temperature in

degrees Celsius when the Fahrenheit temperature is 59°.

34.When Kurt orders from a catalog,he pays $3.50 for shipping and handling in addition to the

cost of the goods that he purchases.Kurt paid $33.20 when he ordered six pairs of socks.Solve the equation 6x 3.50 33.20 to find x,the price of one pair of socks.

35.When Mattie rents a car for one day,the cost is $29.00 plus $0.20 a mile.On her last trip,

Mattie paid $66.40 for the car for one day.Find the number of miles,m,that Mattie droveby solving the equation 29 0.20x 66.40.

36.On his last trip to the post office,Hal paid $4.30 to mail a package and bought some 39-cent

stamps.He paid a total of $13.66.Find s,the number of stamps that he bought,by solvingthe equation 0.39s 4.30 13.66.

4-2 SIMPLIFYING EACH SIDE OF AN EQUATIONAn equation is often written in such a way that one or both sides are not in sim-

plest form.Before starting to solve the equation by using additive and multi-

plicative inverses,you should simplify each side by removing parentheses if

necessary and adding like terms.

Recall that an algebraic expression that is a number,a variable,or a prod-

uct or quotient of numbers and variables is called a term.First-degree equations

in one variablecontain two kinds of terms,terms that are constants and termsthat contain the variable to the first power only.

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Simplifying Each Side of an Equation123

Like and Unlike Terms

Two or more terms that contain the same variable or variables,with corre-

sponding variables having the same exponents,are called like termsor similar

terms.For example,the following pairs are like terms.

6kand k5x2and 7x29aband 0.4ab923xy23and 2xy

Two terms are unlike termswhen they contain different variables,or the

same variable or variables with different exponents.For example,the following

pairs are unlike terms.

3xand 4y5x2and 5x39ab and 0.4a32xy23and xy

To add like terms,we use the distributive property of multiplication over

addition.

9x 2x (9 2)x 11x

16d 3d (–16 3)d 13d

Note that in the above examples,when like terms are added:

1.The sum has the same variable factor as the original terms.

2.The numerical coefficient of the sum is the sum of the numerical coeffi-

cients of the terms that were added.

The sum of like terms can be expressed as a single term.The sum of unlike

terms cannot be expressed as a single term.For example,the sum of 2xand 3

cannot be written as a single term but is written 2x 3.

EXAMPLE 1

Solve and check:2x 3x 4 6

SolutionHow to Proceed

(1)Write the equation:

(2)Simplify the left side by

combining like terms:

(3)Add 4,the additive

inverse of 4,to

each side:

(4)Multiply by ,the

multiplicative inverse

of 5:

(5)Simplify each side.2x 3x 4 65x 4 65x(5x)Check2x 3x 4 6?2(22) 1 3(22) 1 4 52624 2 6 1 4 526 6 6 ? 4 4 105(210)x 2

Answer 2

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124First Degree Equations and Inequalities in One Variable

Note:When solving equations,remember to check the answer in the original

equation and not in the simplified one.

The algebraic expression that is on one side of an equation may contain

e the distributive property to remove the parentheses solving

the equation.The following examples illustrate how the distributive and asso-

ciative properties are used to do this.

EXAMPLE 2

Solve and check:27x 3(x 6) 6

SolutionSince 3(x 6) means that (x 6) is to be multiplied by 3,we will use the

distributive property to remove parentheses and then combine like terms.Note

that for this solution,in the first three steps the left side is being simplified.

These steps apply only to the left side and only change the form but not the

numerical value.The next two steps undo the operations of addition and multi-

plication that make up the expression 24x 18.Since adding 18 and dividing

by 24 will change the value of the left side,the right side must be changed in the

same way to retain the equality.

How to Proceed

(1)

(2)

(3)

(4)Write the equation:Use the distributive property:Combine like terms:Use the addition property of

equality.Add 18,the additive

inverse of 18,to each side:

(5)Use the division property of

equality.Divide each side by 24:

(6)Simplify each side:

Check

(1)Write the equation:

(2)Replace xby 2(3)Perform the indicated computation:27A2B 2 3A2 2 6B 5 6??

?27x 3(x 6) 627x 3x 18 624x 18 6 18 1824x 1224x 1252x 227x 3(x 6) 627A2B 2 3A26 5 62 1 18 5 6

2 1 5 6?

?

5 6

6 6

Answerx 2

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Simplifying Each Side of an Equation125

Representing Two Numbers with

the Same Variable

Problems often involve finding two or more different numbers.It is useful to

express these numbers in terms of the same variable.For example,if you know

the sum of two numbers,you can express the second in terms of the sum and the

first number.

If the sum of two numbers is 12 and one of the numbers is 5,then the

other number is 12 5 or 7.

If the sum of two numbers is 12 and one of the numbers is 9,then the

other number is 12 9 or 3.

If the sum of two numbers is 12 and one of the numbers is x,then the

other number is 12 x.

A problem can often be solved algebraically in more than one way by writ-

ing and solving different equations,as shown in the example that follows.The

methods used to obtain the solution are different,but both use the facts stated

in the problem and arrive at the same solution.

EXAMPLE 3

The sum of two numbers is 43.The larger number minus the smaller number is

5.Find the numbers.

SolutionThis problem states two facts:

FACT 1

FACT 2The sum of the numbers is 43.The larger number minus the smaller number is 5.In other words,the

larger number is 5 more than the smaller.

(1)Represent each number in terms of the same variable using Fact 1:

the sum of the numbers is 43.

Let x the larger number.

Then,43 x the smaller number.

(2)Write an equation using Fact 2:

|The larger number minus the smaller number is 5.___________________________________↓↓↓↓↓x (43 x) 5|||

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126First Degree Equations and Inequalities in One Variable

(3)Solve the equation.

(a) Write the equation:

(b) To subtract (43 x),add its opposite:

(c)Combine like terms:

(d) Add the opposite of 43 to each side:

(e) Divide each side by 2:

(4)Find the numbers.

The larger number x 24.

The smaller number 43 x 43 24 19.x (43 x) 5x ( 43 x) 52x 43 5 43 432x 482x548x 24

CheckA word problem is checked by comparing the proposed solution with the facts

stated in the original wording of the problem.Substituting numbers in the

equation is not sufficient since the equation formed may not be correct.

The sum of the numbers is 43:24 19 43.

The larger number minus the smaller number is 5:24 19 5.

AlternateReverse the way in which the facts are used.

Solution(1)Represent each number in terms of the same variable using Fact 2:

the larger number is 5 more than the smaller.

Let x the smaller number.

Then,x 5 the larger number.

(2)Write an equation using the first fact.

The sum of the numbers is 43.______________________↓↓↓x (x 5) 43||

(3)Solve the equation.

(a) Write the equation:

(b) Combine like terms:

(c) Add the opposite of 5 to each side:

(d) Divide each side by 2:

(4)Find the numbers.

The smaller number x 19.

The larger number x 5 19 5 24.

(5)Check.(See the first solution.)x (x 5) 432x 5 43 5 52x 382x538x 19

AnswerThe numbers are 24 and 19.

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Simplifying Each Side of an Equation127

EXERCISES

Writing About Mathematics

1.Two students are each solving a problem that states that the difference between two num-bers is 12.Irene represents one number by xand the other number by x 12.Henry repre-sents one number by xand the other number by x 12.Explain why both students are

correct.

2.A problem states that the sum of two numbers is 27.The numbers can be represented by xand 27 x.Is it possible to determine which is the larger number and which is the smallernumber? Explain your answer.

Developing Skills

In 3–28,solve and check each equation.

3.x (x 6) 20

5.(15x 7) 12 4

7.x (4x 32) 12

9.5(x 2) 20

11.8(2c 1) 56

13.30 2(10 y)

15.25 2(t 5) 19

17.55 4 3(m 2)

19.3(2b 1) 7 50

21.7r (6r 5) 7

23.5m 2(m 5) 17

25.3(a 5) 2(2a 1) 0

27.0.3a (0.2a 0.5) 0.2(a 2) 1.34.x (12 x) 386.(14 3c) 7c 948.7x (4x 39) 010.3(y 9) 3012.6(3c 1) 4214.4(c 1) 3216.18 6x 4(2x 3)18.5(x 3) 30 1020.5(3c 2) 8 4322.8b 4(b 2) 2424.28y 6(3y 5) 4026.0.04(2r 1) 0.03(2r 5) 0.2928.(8 1 4x) 2 (6x 1 3) 5 9Applying Skills

In 29–33,write and solve an equation for each problem.Follow these steps:

a.List two facts in the problem.

b.Choose a variable to represent one of the numbers to be determined.

e one of the facts to write any other unknown numbers in terms of the chosen e the second fact to write an equation.

e.Solve the equation.

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128First Degree Equations and Inequalities in One Variable

f.Answer the question.

g.Check your answer using the words of the problem.

29.Sandi bought 6 yards of material.She wants to cut it into two pieces so that the difference

between the lengths of the two pieces will be 1.5 yards.What should be the length of eachpiece?

30.The Tigers won eight games more than they lost,and there were no ties.If the Tigers played

78 games,how many games did they lose?

31.This month Erica saved $20 more than last month.For the two months,she saved a total of

$70.How much did she save each month?

32.On a bus tour,there are 100 passengers on three buses.Two of the buses each carry four

fewer passengers than the third bus.How many passengers are on each bus?

33.For a football game,of the seats in the stadium were filled.There were 31,000 empty seats

at the game.What is the stadium’s seating capacity?

4-3 SOLVING EQUATIONS THAT HAVE THE

VARIABLE IN BOTH SIDESA variable represents a number.As you know,any number may be added to

both sides of an equation without changing the solution set.Therefore,the same

variable (or the same multiple of the same variable) may be added to or sub-

tracted from both sides of an equation without changing the solution set.

For instance,to solve 8x 30 5x,write an equivalent equation that has

only a constant in the right side.To do this,eliminate 5xfrom the right side by

adding its opposite, 5x,to each side of the equation.

METHOD1METHOD2Check

8x 30 5x?8(10) 5 30 1 5(10)?80 5 30 1 50

80 80 8x 30 5x3x 30x 108x 30 5x8x ( 5x) 30 5x ( 5x)3x 30x 10

Answer:x 10

To solve an equation that has the variable in both sides,transform it into an

equivalent equation in which the variable appears in only one side.Then,solve

the equation.

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Solving Equations That Have the Variable in Both Sides129

EXAMPLE 1

Solve and check:7x 63 2x

SolutionHow to Proceed

(1)Write the equation:

(2)Add 2xto each side of

the equation:

(3)Divide each side of the

equation by 9:

(4)Simplify each side:7x 63 2x 2x 2x9x 639xCheck7x 63 2x?7(7) 5 63 2 2(7)?49 563 2 1449 49 563x 7

Answerx 7

To solve an equation that has both a variable and a constant in both sides,

first write an equivalent equation with only a variable term on one side.Then

solve the simplified equation.The following example shows how this can be

done.

EXAMPLE 2

Solve and check:3y 7 5y 3

SolutionMETHOD1METHOD2Check

3y 7 5y 3?3(5) 7 5 5(5) 2 3?15 7 5 25 2 3

22 22 3y 7 5y 3 22y

3y 7 5y 37 2y 310 2

y52y

y 5y 5

Answery 5

A graphing calculator can be used to check an equation.The calculator can

determine whether a given statement of equality or inequality is true or false.If

the statement is true,the calculator will display 1;if the statement is false,thecalculator will display 0.The symbols for equality and inequality are found inTESTthe menu.

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130First Degree Equations and Inequalities in One Variable

To check that y 5 is the solution to the equation 3y 7 5y 3,first

store 5 as the value of y.then enter the equation to be checked.

ENTER:

5 STO ALPHAENTER

TEST2ndALPHAENTERALPHAENTERDISPLAY:The calculator displays 1 which indi-cates that the statement of equality istrue for the value that has beenstored for y.

EXAMPLE 3

The larger of two numbers is 4 times the smaller.If the larger number exceeds

the smaller number by 15,find the numbers.

Note:When srepresents the smaller number and 4s represents the larger

number,“the larger number exceeds the smaller by 15”has the following

e any one of them.

1.The larger equals 15 more than the smaller,written as 4s = 15 s.

2.The larger decreased by 15 equals the smaller,written as 4s 15 s.

3.The larger decreased by the smaller is 15,written as 4s s 15.

SolutionLet s = the smaller number.

Then 4s = the larger number.

|The larger is 15 more than the smaller._____________________________↓↓↓↓↓4s 15 s

4s 15 s|||||

4s

s

4s 15 s5

4(5) 20

CheckThe larger number,20,is 4 times the smaller number,5.The larger number,20,

exceeds the smaller number,5,by 15.

AnswerThe larger number is 20;the smaller number is 5.

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Solving Equations That Have the Variable in Both Sides131

EXAMPLE 4

In his will,Uncle Clarence left $5,000 to his two nieces.Emma’s share is to be

$500 more than Clara’s.How much should each niece receive?

Solution(1)Use the fact that the sum of the two shares is $5,000 to express each share

in terms of a variable.

Let x Clara’s share.

Then 5,000 x Emma’s share.

(2)Use the fact that Emma’s share is $500 more than Clara’s share to write

an equation.

Emma’s share is $500 more than Clara’s share.___________________________________↓↓↓↓↓5,000 x 500 x||||||

(3)Solve the equation to find Clara’s share.

5,000 x 500 x5,000 500 2x4,500 2x

2,250 x

Clara’s share is x $2,250.

(4)Find Emma’s share:5,000 x 5,000 2,250 $2,750.

Alternate(1) Use the fact that Emma’s share is $500 more than Clara’s share to express

each share in terms of a variable.SolutionLet x Clara’s share.

Then x 500 Emma’s share.

(2)Use the fact that the sum of the two shares is $5,000 to write an equation.

Clara’s share plus Emma’s share is $5,000.__________________________↓↓↓↓↓x (x 500) 5,000||||

(3)Solve the equation to find Clara’s share

x (x 500) 5,000

2x 500 5,0002x 4,500

x 2,250

Clara’s share is x $2,250.

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132First Degree Equations and Inequalities in One Variable

(4)Find Emma’s share:x 500 2250 500 $2,750.

Check$2,750 is $500 more than $2,250,and $2,750 $2,250 $5,000.

AnswerClara’s share is $2,250,and Emma’s share is $2,750.

EXERCISES

Writing About Mathematics

1.Milus said that he finds it easier to work with integers than with fractions.Therefore,inorder to solve the equation a 2 751 3,he began by multiplying both sides of theequation by 4.

4Aa 2 7B54Aa 1 3B3a 28 2a 12

Do you agree with Milus that this is a correct way of obtaining the solution? If so,whatmathematical principle is Milus using?

2.Katie said that Example 3 could be solved by letting equal the smaller number and xequal the larger number.Is Katie correct? If so,what equation would she write to solve theproblem?

Developing Skills

In 3–36,solve and check each equation.

3.7x 10 2x

6.y 4y 30

9.0.8m 0.2m 24

12.2x 1 2453x

15.x 9x 72

18.7r 10 3r 50

21.x 4 9x 4

24.c 20 55 4c

27.3m (m 1) 6m 1

30.t 2 1154(16 2 t) 2 t4.9x 44 2x7.2d 36 5d10.8y 90 2y13.5a 40 3a16.0.5m 30 1.1m19.4y 20 5y 922.9x 3 2x 4625.2d 36 3d 5428.x 3(1 x) 47 x5.5c 28 c8.2y51y 2 811.2.3x 36 0.3x14.5c 2c 8117.4c59c 1 4420.7x 8 6x 123.y 30 12y 1426.7y 5 9y 2929.3b 8 10 (4 8b)31.18 4n 8 2(1 8n)

33.8a 3(5 2a) 85 3a

35.3m 5m 12 7m 88 532.8c 1 7c 2(7 c)34.4(3x 5) 5x 2( x 15)

36.5 3(a 6) a 1 8a

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Solving Equations That Have the Variable in Both Sides133

In 37–42,a.write an equation to represent each problem,and b.solve the equation to find each number.

37.Eight times a number equals 35 more than the number.Find the number.

38.Six times a number equals 3 times the number,increased by 24.Find the number.

39.If 3 times a number is increased by 22,the result is 14 less than 7 times the number.Find the

number.

40.The greater of two numbers is 1 more than twice the smaller.Three times the greater

exceeds 5 times the smaller by 10.Find the numbers.

41.The second of three numbers is 6 more than the first.The third number is twice the first.

The sum of the three numbers is 26.Find the three numbers.

42.The second of three numbers is 1 less than the first.The third number is 5 less than the sec-

ond.If the first number is twice as large as the third,find the three numbers.

Applying Skills

In 43–50,use an algebraic solution to solve each problem.

43.It took the Gibbons family 2 days to travel 925 miles to their vacation home.They traveled

75 miles more on the first day than on the second.How many miles did they travel eachday?

44.During the first 6 month of last year,the interest on an investment was $130 less than dur-

ing the second 6 months.The total interest for the year was $1,450.What was the interest foreach 6-month period?

45.Gemma has 7 more five-dollar bills than ten-dollar bills.The value of the five-dollar bills

equals the value of the ten-dollar bills.How many five-dollar bills and ten-dollar bills doesshe have?

46.Leonard wants to save $100 in the next 2 months.He knows that in the second month he will

be able to save $20 more than during the first month.How much should he save each month?

47.The ABC Company charges $75 a day plus $0.05 a mile to rent a car.How many miles did

Mrs.Kiley drive if she paid $92.40 to rent a car for one day?

48.Kesha drove from Buffalo to Syracuse at an average rate of 48 miles per hour.On the

return trip along the same road she was able to travel at an average rate of 60 miles perhour.The trip from Buffalo to Syracuse took one-half hour longer than the return trip.Howlong did the return trip take?

49.Carrie and Crystal live at equal distances from school.Carries walks to school at an average

rate of 3 miles per hour and Crystal rides her bicycle at an average rate of 9 miles per hour.It takes Carrie 20 minutes longer than Crystal to get to school.How far from school do

Crystal and Carrie live?

50.Emmanuel and Anthony contributed equal amounts to the purchase of a gift for a friend.

Emmanuel contributed his share in five-dollar bills and Anthony gave his share in one-

dollar bills.Anthony needed 12 more bills than Emmanuel.How much did each contributetoward the gift?

美国 algebra 1 数学经典教材

134First Degree Equations and Inequalities in One Variable

4-4 USING FORMULAS TO SOLVE PROBLEMS

To solve for the subject of a formula,substitute the known values in the formula

and perform the required computation.For example,to find the area of a tri-

angle when b 4.70 centimeters and h 3.20 centimeters,substitute the given

values in the formula for the area of a triangle:

A bh (4.70 cm)(3.20 cm)

7.52 cm2

Now that you can solve equations,you will be able to find the value of any

variable in a formula when the values of the other variables are known.To do

this:

1.Write the formula.

2.Substitute the given values in the formula.

3.Solve the resulting equation.

The values assigned to the variables in a formula often have a unit of

measure.It is convenient to solve the equation without writing the unit of

measure,but the answer should always be given in terms of the correct unit

of measure.

EXAMPLE 1

The perimeter of a rectangle is 48 centimeters.If the length of the rectangle is

16 centimeters,find the width to the nearest centimeter.A is the subject of the formula.

SolutionYou know that the perimeter of a geometric figure is the sum of the lengths of

all of its sides.When solving a perimeter problem,it is helpful to draw and

label a figure to model the e the formula P 2l 2w.

P 2l 2w

48 2(16) 2w

48 32 2w16 2w

16 2w

8 w?? CheckP 2l 2w48 52(16) 1 2(8)48 532 1 1648 48 w16 cmw16 cm

Answer8 centimeters

美国 algebra 1 数学经典教材

Using Formulas to Solve Problems135

EXAMPLE 2

A garden is in the shape of an isosceles triangle,a triangle that has two sides of

equal measure.The length of the third side of the triangle is 2 feet greater than

the length of each of the equal sides.If the perimeter of the garden is 86 feet,

find the length of each side of the garden.

SolutionLet x the length of each of the two equal sides.

Then,x 2 the length of the third side.

The perimeter is the sum of the lengths of the sides._______________________________________________↓↓↓86 x x (x 2)||||xx + 2x

86 3x 2

84 3x

28 x

The length of each of the equal sides x 28.

The length of the third side x 2 28 2 30.

CheckPerimeter 28 28 30 86

AnswerThe length of each of the equal sides is 28 feet.The length of the third side

(the base) is 30 feet.

EXAMPLE 3

The perimeter of a rectangle is 52 feet.The length is 2 feet more than 5 times

the width.Find the dimensions of the rectangle.

SolutionUse the formula for the perimeter of a rectangle,P 2l 2w,to solve this

problem.

Let w the width,in feet,of the rectangle.

Then 5w 2 the length,in feet,of the rectangle.

P 2l 2w

52 2(5w 2) 2w

52 10w 4 2w

52 12w 4

48 12w

4 wCheckThe length,22,is 2 more than 5 times the width,4. P 2l 2w52 52(22) 1 2(4)52 544 1 852 52 ? ?

AnswerThe width is 4 feet;the length is 5(w) 2 5(4) 2 22 feet.

美国 algebra 1 数学经典教材

136First Degree Equations and Inequalities in One Variable

EXAMPLE 4

Sabrina drove from her home to her mother’s home which is 150 miles away.

For the first half hour,she drove on local roads.For the next two hours she

drove on an interstate highway and increased her average speed by 15 miles

per hour.Find Sabrina’s average speed on the local roads and on the interstate

highway.

SolutionList the facts stated by this problem:

FACT 2

Sabrina drove on local roads for hour or 0.5 hour.Sabrina drove on the interstate highway for 2 hours.Sabrina’s rate or speed on the interstate highway was 15 mph more

than her rate on local roads.

This problem involves rate,time,and e the distance formula,

d rt,where ris the rate,or speed,in miles per hour,tis time in hours,and dis

distance in miles.

(1)Represent Sabrina’s speed for each part of the trip in terms of r.

Let r Sabrina’s speed on the local roads.

Then r 15 Sabrina’s speed on the interstate highway.

(2)Organize the facts in a table,using the distance formula.

The distance on the local roads plus the distance on the highway is 150 miles._________________________________________________________________↓↓↓↓↓0.5r 2(r 15) 150||||||

(4)Solve the equation.

(a) Write the equation:

(b) Use the distributive property:0.5r 2(r 15) 1500.5r 2r 30 150

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