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6 Algebra: Equations and Inequalities > 6.2 Linear Equations in One Variable and Proportions

6.2 Linear Equations in One Variable and Proportions

What am I Supposed to Learn?After you have read this section, you should be able to:

1 Solve linear equations.

2 Solve linear equations containing fractions.

3 Solve proportions.

4 Solve problems using proportions.

5 Identify equations with no solution or infinitely many solutions.

THE BELIEF THAT HUMOR AND LAUGHTER can have positive benefits on our lives is not new. The graphs in Figure 6.2 indicate that persons with a low sense ofhumor have higher levels of depression in response to negative life events than those with a high sense of humor. These graphs can be modeled by the followingformulas:

dFIGURE 6.2 Source: Steven Davis and Joseph Palladino, Psychology, 5th Edition. Prentice Hall, 2007.

In each formula, x represents the intensity of a negative life event (from 1, low, to 10, high) and D is the level of depression in response to that event.

Suppose that the low-humor group averages a level of depression of 10 in response to a negative life event. We can determine the intensity of that event by substituting10 for D in the low-humor model,

The two sides of an equation can be reversed. So, we can also express this equation as

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Low-Humor Group

▼

D = x +10

9

53

9

High-Humor Group

▼

D = x + .19

26

9

D = x + :10

9

53

9

10 = x + .10

9

53

9

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Notice that the highest exponent on the variable is 1. Such an equation is called a linear equation in one variable. In this section, we will study how to solve suchequations. We return to the models for sense of humor and depression later in the section.

Solving Linear Equations in One Variable

1 Solve linear equations.

We begin with the general definition of a linear equation in one variable.

Definition of a Linear EquationA linear equation in one variable x is an equation that can be written in the form

where a and b are real numbers, and

An example of a linear equation in one variable is

Solving an equation in x involves determining all values of x that result in a true statement when substituted into the equation. Such values are solutions, or roots, ofthe equation. For example, substitute for x in We obtain

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x + = 10.10

9

53

9

ax + b = 0,

a ≠ 0.

4x + 12 = 0.

−3 4x + 12 = 0.

4 (−3) + 12 = 0, or − 12 + 12 = 0.



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51


This simplifies to the true statement Thus, is a solution of the equation We also say that satisfies the equation because when we substitute for x, a true statement results. The set of all such solutions is called the equation's solution set. For example, the solution set of theequation is

Two or more equations that have the same solution set are called equivalent equations. For example, the equations

are equivalent equations because the solution set for each is To solve a linear equation in x, we transform the equation into an equivalent equation one or moretimes. Our final equivalent equation should be of the form

The solution set of this equation is the set consisting of the number.

To generate equivalent equations, we will use the following properties:

The Addition and Multiplication Properties of EqualityThe Addition Property of Equality

The same real number or algebraic expression may be added to both sides of an equation without changing the equation's solution set.

The Multiplication Property of Equality

The same nonzero real number may multiply both sides of an equation without changing the equation's solution set.

Because subtraction is defined in terms of addition, the addition property also lets us subtract the same number from both sides of an equation without changing theequation's solution set. Similarly, because division is defined in terms of multiplication, the multiplication property of equality can be used to divide both sides of anequation by the same nonzero number to obtain an equivalent equation.

Table 6.1 illustrates how these properties are used to isolate x to obtain an equation of the form

TABLE 6.1 Using Properties of Equality to Solve Equations Equation How to Isolate x Solving the Equation The Equation's Solution Set

Add 3 to both sides. {11}

Subtract 7 from both sides.

Divide both sides by 6 (or multiply both sides by ). {5}

Multiply both sides by 5. {45}




0 = 0. −3 4x + 12 = 0. −3 4x + 12 = 0,−3

4x + 12 = 0 {−3} .

4x + 12 = 0 and 4x = −12 and x = −3

{−3} .

x = a number.

a = b and a + c = b + c are equivalent equations.

a = b and ac = bc are equivalent equations as long as c ≠ 0.

x = a number.

►

These equations are

solved using the

Addition Property

of Equality.

x − 3 = 8x − 3 + 3

x

= 8 + 3

= 11

x + 7 = −15x + 7 − 7

x

= −15 − 7

= −22{−22}

►

These equations are

solved using the

Multiplication

Property of Equality.

6x = 30 16

6x

6

x

= 30

6

= 5

= 9x

5

5 ⋅ x

5

x

= 5 ⋅ 9

= 45












Page 352


Example 1 Using Properties of Equality to Solve an EquationSolve and check:

SOLUTION

Our goal is to obtain an equivalent equation with x isolated on one side and a number on the other side.

Now we check the proposed solution, 7, by replacing x with 7 in the original equation.

Because the check results in a true statement, we conclude that the solution set of the given equation is

Check Point 1Solve and check:

Here is a step-by-step procedure for solving a linear equation in one variable. Not all of these steps are necessary to solve every equation.

Solving a Linear Equation1. Simplify the algebraic expression on each side by removing grouping symbols and combining like terms.

2. Collect all the variable terms on one side and all the constants, or numerical terms, on the other side.

3. Isolate the variable and solve.

4. Check the proposed solution in the original equation.

Example 2 Solving a Linear EquationSolve and check:

SOLUTION

Step 1 Simplify the algebraic expression on each side.

d




2x + 3 = 17.

2x + 3

2x + 3− 3

2x

2x

2

x

= 17

= 17 − 3

= 14

= 142

= 7

This is the given equation.

Subtract 3 from both sides.

Simplify.

Divide both sides by 2.

Simplify : = 1x = x and = 7.2x

2

14

2

2x + 3

2 ⋅ 7 + 3

14 + 3

►17This statement is true.

= 17

17=?

17=?

= 17

This is the original equation.

Substitute 7 for x.The question mark indicates that we

do not yet know if the two sides are equal.

Multiply: 2 ⋅ 7 = 14.

Add: 14 + 3 = 17.

{7} .

4x + 5 = 29.

2 (x − 4) − 5x = −5.












Page 353


Step 2 Collect variable terms on one side and constants on the other side. The only variable term in is and is already on theleft side. We will collect constants on the right side by adding 8 to both sides.

Step 3 Isolate the variable and solve. We isolate the variable, x, by dividing both sides of by

Step 4 Check the proposed solution in the original equation. Substitute for x in the original equation.



Great Question!What are the differences between what I'm supposed to do with algebraic expressions and algebraic equations?

We simplify algebraic expressions. We solve algebraic equations. Although basic rules of algebra are used in both procedures, notice the differences between theprocedures:

Simplifying an Algebraic Expression Solving an Algebraic Equation

Simplify: Solve:

Solution Solution

The solution set is




−3x − 8 = −5 −3x, −3x

−3x − 8+ 8

−3x

= −5+ 8

= 3

Add 8 to both sides.

Simplify.

−3x = 3 −3.

−3x

−3

x

= 3

−3

= −1

Divide both sides by − 3.

Simplify : = 1x = x and = −1.−3x

−33

−3

−1

2(x − 4) − 5x

2 (−1 − 4) − 5(−1)

2 (−5) − 5(−1)

−10 − (−5)

►− 5This statement is true.

= −5

−5=?

−5=?

−5=?

= −5


Substitute − 1 for x.

Simplify inside parentheses:

−1 − 4 = −1 + (−4) = −5.

Multiply : 2 (−5) = −10 and

5(−1) = −5.

−10 − (−5) = −10 + 5 = −5

{−1} .

6 (x − 3) − 10x = −10.

3(x − 7) − (5x − 11).

▲

This is not an equation.

There is no equal sign.

3(x − 7) − (5x − 11) = 14.

▲

This is an equation.

There is an equal sign.

3 (x − 7) − (5x − 11)

= 3x − 21 − 5x + 11

= (3x − 5x) + (−21 + 11)

= −2x + (−10)

= −2x − 10

▲

Stop! Further simplication is not

possible. Avoid the common error of

setting − 2x − 10 equal to 0.

3 (x − 7) − (5x − 11)

3x − 21 − 5x + 11

−2x − 10

►− 2x − 10 + 10Add 10 to both sides.

−2x

►Divide both sides by − 2.−2x

−2

x

=

=

=

=

=

=

=

14

14

14

14 + 10

2424−2

−12{−12}.














SOLUTION

Step 1 Simplify the algebraic expression on each side. Neither side contains grouping symbols or like terms that can be combined. Therefore, we can skip thisstep.

Step 2 Collect variable terms on one side and constants on the other side. One way to do this is to collect variable terms on the left and constants on the right.This is accomplished by subtracting 8x from both sides and adding 12 to both sides.

Step 3 Isolate the variable and solve. We isolate the variable, x, by dividing both sides of by

Step 4 Check the proposed solution in the original equation. Substitute for x in the original equation.


Check Point 3Solve the equation:

Great Question!Do I have to solve by collecting variable terms on the left and numbers on the right?

No. If you prefer, you can solve the equation by collecting variable terms on the right and numbers on the left. To collect variable terms on the right, subtract 5x fromboth sides:

To collect numbers on the left, subtract 24 from both sides:

Now isolate x by dividing both sides by 3:

This is the same solution that we obtained in Example 3.


SOLUTION



5x − 12 = 8x + 24.

5x − 12

5x − 12 − 8x

−3x − 12

−3x − 12 + 12

−3x

= 8x + 24

= 8x + 24 − 8x

= 24

= 24 + 12

= 36


Subtract 8x from both sides.

Simplify : 5x − 8x = −3x.

Add 12 to both sides and collect

constants on the right side.

Simplify.

−3x = 36 −3.

−3x

−3

x

= 36

−3

= −12


Simplify.

−12

5x − 12

5(−12) − 12

−60 − 12

►− 72This statement is true.

= 8x + 24

8 (−12) + 24=?

−96 + 24=?

= −72


Substitute − 12 for x.

Multiply: 5(−12) = −60 and

8(−12) = −96.

Add: − 60 + (−12) = −72 and

−96 + 24 = −72.

{−12} .

2x + 9 = 8x − 3.

5x − 12 = 8x + 24

5x − 12 − 5x

−12

= 8x + 24 − 5x

= 3x + 24.

−12 − 24

−36

= 3x + 24 − 24

= 3x.

−36

3

−12

= 3x

3

= x.

2 (x − 3) − 17 = 13 − 3(x + 2) .







Page 354

Step 1 Simplify the algebraic expression on each side.

d









Page 355

6 Algebra: Equations and Inequalities > 6.2 Linear Equations in One Variable and Proportions > Linear Equations with Fractions

Step 2 Collect variable terms on one side and constants on the other side. We will collect variable terms of on the left by adding 3x toboth sides. We will collect the numbers on the right by adding 23 to both sides.

Step 3 Isolate the variable and solve. We isolate the variable, x, by dividing both sides of by 5.

Step 4 Check the proposed solution in the original equation. Substitute 6 for x in the original equation.

The true statement verifies that the solution set is


Linear Equations with Fractions

2 Solve linear equations containing fractions.

Equations are easier to solve when they do not contain fractions. How do we remove fractions from an equation? We begin by multiplying both sides of the equation bythe least common denominator of any fractions in the equation. The least common denominator is the smallest number that all denominators will divide into. Multiplyingevery term on both sides of the equation by the least common denominator will eliminate the fractions in the equation. Example 5 shows how we “clear an equation offractions.”

Example 5 Solving a Linear Equation Involving FractionsSolve and check:

SOLUTION

The denominators are 2 and 5. The smallest number that is divisible by 2 and 5 is 10. We begin by multiplying both sides of the equation by 10, the least commondenominator.

d



2x − 23 = −3x + 7

2x − 23 + 3x

5x − 23

5x − 23 + 23

5x

= −3x + 7+ 3x

= 7

= 7 + 23

= 30

Add 3x to both sides.

Simplify: 2x + 3x = 5x.

Add 23 to both sides.

Simplify.

5x = 30

5x

5

x

= 30

5

= 6


Simplify.

2 (x − 3) − 17

2 (6 − 3) − 17

2 (3) − 17

6 − 17

−11

= 13 − 3(x + 2)

13 − 3(6 + 2)=?

13 − 3(8)=?

13 − 24=?

= −11


Substitute 6 for x.

Simplify inside parentheses.

Multiply.

Subtract.

−11 = −11 {6} .

4 (2x + 1) = 29 + 3(2x − 5) .

= − 4.3x

2

8x

5










6 Algebra: Equations and Inequalities > 6.2 Linear Equations in One Variable and Proportions > Linear Equations with Fractions

At this point, we have an equation similar to those we have previously solved. Collect the variable terms on one side and the constants on the other side.

Isolate x by multiplying or dividing both sides of this equation by

Check the proposed solution. Substitute 40 for x in the original equation. You should obtain This true statement verifies that the solution set is


Example 6 An Application: Responding to Negative Life EventsIn the section opener, we introduced line graphs, repeated in Figure 6.2, indicating that persons with a low sense of humor have higher levels of depression inresponse to negative life events than those with a high sense of humor. These graphs can be modeled by the following formulas:

dFIGURE 6.2 (repeated)

In each formula, x represents the intensity of a negative life event (from 1, low, to 10, high) and D is the average level of depression in response to that event. If thehigh-humor group averages a level of depression of 3.5, or in response to a negative life event, what is the intensity of that event? How is the solution shown onthe red line graph in Figure 6.2?

SOLUTION

We are interested in the intensity of a negative life event with an average level of depression of for the high-humor group. We substitute for D in the high-humormodel and solve for x, the intensity of the negative life event.



15x − 16x

−x

▲

We're not finished. A negative

sign should not precede x.

=

=

16x − 40 − 16x

−40

Subtract 16x from both sides to get the variable

terms on the left.

Simplify.

−1.

−x

−1

x

= −40

−1

= 40


Simplify.

60 = 60. {40} .

= 7 − .2x

3x

2

Low-Humor Group

▼

D = x +10

9

53

9

High-Humor Group

▼

D = x + .19

26

9

,7

2

7

2

7

2








Page 356

d









6 Algebra: Equations and Inequalities > 6.2 Linear Equations in One Variable and Proportions > Proportions

The formula indicates that if the high-humor group averages a level of depression of 3.5 in response to a negative life event, the intensity of that event is or 5.5.This is illustrated on the line graph for the high-humor group in Figure 6.3.

dFIGURE 6.3

Check Point 6Use the model for the low-humor group given in Example 6 on the previous page to solve this problem. If the low-humor group averages a level of depression of 10 inresponse to a negative life event, what is the intensity of that event? How is the solution shown on the blue line graph in Figure 6.2?

Proportions

3 Solve proportions.

A ratio compares quantities by division. For example, a group contains 60 women and 30 men. The ratio of women to men is We can express this ratio as afraction reduced to lowest terms:

This ratio can be expressed as 2:1, or 2 to 1.

A proportion is a statement that says that two ratios are equal. If the ratios are and then the proportion is

We can clear this equation of fractions by multiplying both sides by bd, the least common denominator:

We see that the following principle is true for any proportion:

The Cross-Products Principle for Proportions

The cross products ad and bc are equal.



,112

.60

30

= = .60

30

2 ⋅ 30

1 ⋅ 30

2

1

a

b,c

d

= .a

b

c

d

a

b

bd ⋅ a

b

ad

= c

d

= bd ⋅ c

d

= bc.

This is the given proportion.

Multiply both sides by bd (b ≠ 0 and d ≠ 0). Then simplify.

On the left, ⋅ = da = ad. On the right, ⋅ = bc.db

1a

b

b d

1c

d

If = , then ad = bc. (b ≠ 0 and d ≠ 0)a

b

c

d








Page 357









6 Algebra: Equations and Inequalities > 6.2 Linear Equations in One Variable and Proportions > Proportions

For example, since we see that or We can also use and conclude that When using the cross-productsprinciple, it does not matter on which side of the equation each product is placed.

If three of the numbers in a proportion are known, the value of the missing quantity can be found by using the cross-products principle. This idea is illustrated in Example7(a).

Example 7 Solving ProportionsSolve each proportion and check:

a.

b.

SOLUTION

a.

The solution set is

Check

b. d

The solution set is

Check



= ,23

6

92 ⋅ 9 = 3 ⋅ 6, 18 = 18. =2

3

6

93 ⋅ 6 = 2 ⋅ 9.

=63x

7

5

= .20

x−10

30x

63x

63 ⋅ 5

315315

7

45

= 7

5

= 7x

= 7x

= 7x

7

= x

This is the given proportion.

Apply the cross-products principle.

Simplify.


Simplify.

{45} .

63

45

7⋅ 9

5⋅ 9

7

5

=? 7

5

=? 7

5

= 7

5

Substitute 45 for x in = .63x

75

Reduce to lowest terms.6345

This true statement verifies that the solution set is {45} .

{30} .

20

30−10

20

20

1

=? 30

30

=? 30

30

= 1

Substitute 30 for x in = .20x−10

30x

Subtract : 30 − 10 = 20.

This true statement verifies that the solution is 30.








Page 358









6 Algebra: Equations and Inequalities > 6.2 Linear Equations in One Variable and Proportions > Applications of Proportions

Check Point 7Solve each proportion and check:

a.

b.

Applications of Proportions

4 Solve problems using proportions.

We now turn to practical application problems that can be solved using proportions. Here is a procedure for solving these problems:

Solving Applied Problems Using Proportions1. Read the problem and represent the unknown quantity by x (or any letter).

2. Set up a proportion by listing the given ratio on one side and the ratio with the unknown quantity on the other side. Each respective quantity should occupy thesame corresponding position on each side of the proportion.

3. Drop units and apply the cross-products principle.

4. Solve for x and answer the question.

Example 8 Applying Proportions: Calculating TaxesThe property tax on a house with an assessed value of $480,000 is $5760. Determine the property tax on a house with an assessed value of $600,000, assuming thesame tax rate.

SOLUTION

Step 1 Represent the unknown by x. Let tax on the $600,000 house.

Step 2 Set up a proportion. We will set up a proportion comparing taxes to assessed value.

Step 3 Drop the units and apply the cross-products principle. We drop the dollar signs and begin to solve for x.

Step 4 Solve for x and answer the question.

The property tax on the $600,000 house is $7200.

Check Point 8The property tax on a house with an assessed value of $250,000 is $3500. Determine the property tax on a house with an assessed value of $420,000, assuming thesame tax rate.

Great Question!Are there other proportions that I can use in step 2 to model the problem's conditions?

Yes. Here are three other correct proportions you can use:

•



=10x

23

= .2260−x

2x

x = the

Tax on $480,000 house

Assessed value ($480,000)

▼

Given ratio { $5760

$480,000

equals

▼

=

Tax on $600,000 house

Assessed value ($600,000)

▼

$x

$600,000

←Unknown←Given quantity

5760

480,000

480,000x

480,000x

= x

600,000

= (5760) (600,000)

= 3,456,000,000

This is the proportion that models the problem's conditions.


Multiply.

480,000x

480,000

x

=3,456,000,000

480,000

= 7200

Divide both sides by 480,000.

Simplify.

=$480,000 value

$5760 tax

$600,000 value

$x tax








Page 359

•

•

Each proportion gives the same cross product obtained in step 3.


=$480,000 value

$600,000 value

$5760 tax

$x tax

=$600,000 value

$480,000 value

$x tax

$5760 tax








6 Algebra: Equations and Inequalities > 6.2 Linear Equations in One Variable and Proportions > Equations with No Solution orInfinitely Many Solutions

Example 9 Applying Proportions: Estimating Wildlife PopulationWildlife biologists catch, tag, and then release 135 deer back into a wildlife refuge. Two weeks later they observe a sample of 140 deer, 30 of which are tagged.Assuming the ratio of tagged deer in the sample holds for all deer in the refuge, approximately how many deer are in the refuge?

SOLUTION

Step 1 Represent the unknown by x. Let total number of deer in the refuge.

Step 2 Set up a proportion.

Steps 3 and 4 Apply the cross-products principle, solve, and answer the question.

There are approximately 630 deer in the refuge.

Check Point 9Wildlife biologists catch, tag, and then release 120 deer back into a wildlife refuge. Two weeks later they observe a sample of 150 deer, 25 of which are tagged.Assuming the ratio of tagged deer in the sample holds for all deer in the refuge, approximately how many deer are in the refuge?

Equations with No Solution or Infinitely Many Solutions

5 Identify equations with no solution or infinitely many solutions.

Thus far, each equation or proportion that we have solved has had a single solution. However, some equations are not true for even one real number. By contrast, otherequations are true for all real numbers.

If you attempt to solve an equation with no solution, you will eliminate the variable and obtain a false statement, such as If you attempt to solve an equation thatis true for every real number, you will eliminate the variable and obtain a true statement, such as

Example 10 Attempting to Solve an Equation with No SolutionSolve:

SOLUTION



x = the

Unknown→

Original number

of tagged deer

Total number

of deer

▼

135x

equals

▼

=

Number of tagged deer

in the observed sample

Total number of deer

in the observed sample

⎫

⎭⎬⎪⎪⎪⎪

▼

30

140

Known ratio

135x

(135) (140)

18,90018,900

30

630

= 30

140

= 30x

= 30x

= 30x

30

= x

This is the proportion that models the problem's conditions.


Multiply.


Simplify.

2 = 5.4 = 4.

2x + 6 = 2 (x + 4) .

2x + 6

2x + 6

2x + 6 − 2x

►6Keep reading. 6 = 8

is not the solution.

= 2 (x + 4)

= 2x + 8

= 2x + 8 − 2x

= 8


Use the distributive property.

Subtract 2x from both sides.

Simplify.








Page 360









6 Algebra: Equations and Inequalities > 6.2 Linear Equations in One Variable and Proportions > Concept and Vocabulary Check

The original equation, is equivalent to the statement which is false for every value of x. The equation has no solution. The solution setis the empty set.

Check Point 10Solve:

Example 11 Solving an Equation for Which Every Real Number Is a SolutionSolve:

SOLUTION

d

Can you see that the equation is true for every value of x? Let's continue solving the equation by subtracting 4x from both sides.

The original equation is equivalent to the statement which is true for every value of x. Thus, the solution set consists of the set of all real numbers, expressedin set-builder notation as Try substituting any real number of your choice for x in the original equation. You will obtain a true statement.

Check Point 11Solve:

Great Question!Do I have to use sets to write the solution of an equation?

Because of the fundamental role that sets play in mathematics, it's a good idea to use set notation to express an equation's solution. If an equation has no solution, itssolution set is the empty set. If an equation with variable x is true for every real number, its solution set is

Concept and Vocabulary CheckFill in each blank so that the resulting statement is true.

1. An equation in the form such as is called a/an ___________________ equation in one variable.

2. Two or more equations that have the same solution set are called ___________________ equations.

3. The addition property of equality states that if then ___________________.

4. The multiplication property of equality states that if and then ___________________.

5. The first step in solving is to ________________________________.

6. The algebraic expression can be ___________________, whereas the algebraic equation can be ___________________.

7. The equation

can be cleared of fractions by multiplying both sides by the ___________________________________ of which is _________________.

8. A statement that two ratios are equal is called a/an ___________________.

9. The cross-products principle states that if ( and ), then ___________________.



2x + 6 = 2 (x + 4) , 6 = 8,∅,

3x + 7 = 3 (x + 1) .

4x + 6 = 6 (x + 1) − 2x.

4x + 6 = 4x + 6

4x + 6 − 4x

►6Keep reading. 6 = 6

is not the solution.

= 4x + 6 − 4x

= 6

6 = 6,{x|x is a real number} .

7x + 9 = 9 (x + 1) − 2x.

∅, {x|x is a real number} .

ax + b = 0,a ≠ 0, 3x + 17 = 0,

a = b, a + c =

a = b c ≠ 0, ac =

7 + 3 (x − 2) = 2x + 10

7 (x − 4) + 2x 7 (x − 4) + 2x = 35

= 2 +x

4

x

3

and ,x

4x

3

=a

b

c

db ≠ 0 d ≠ 0








Page 361









6 Algebra: Equations and Inequalities > 6.2 Linear Equations in One Variable and Proportions > Exercise Set 6.2

10. In solving an equation, if you eliminate the variable and obtain a statement such as the equation has ___________________ solution. The solution setcan be expressed using the symbol ___________________.

11. In solving an equation with variable x, if you eliminate the variable and obtain a statement such as the equation is ___________________ for every valueof x. The solution set can be expressed in set-builder notation as ____________________________.

In Exercises 12–15, determine whether each statement is true or false. If the statement is false, make the necessary change(s) to produce a true statement.

12. The equation is equivalent to _____

13. The equation is equivalent to _____

14. The equation has no solution. _____

15. The equation has precisely one solution. _____

Exercise Set 6.2Practice ExercisesIn Exercises 1–58, solve and check each equation.

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

11.

12.

13.

14.

15.

16.

17.

18.

19.

20.

21.

22.

23.

24.

25.

26.

27.

28.

29.

30.



2 = 3,

6 = 6,

2x + 5 = 0 2x = 5.

x + =13

12

x + 2 = 3.

3x = 2x

3 (x + 4) = 3 (4 + x)

x − 7 = 3

x − 3 = −17

x + 5 = −12

x + 12 = −14

= 4x

3

= 3x

5

5x = 45

6x = 18

8x = −24

5x = −25

−8x = 2

−6x = 3

5x + 3 = 18

3x + 8 = 50

6x − 3 = 63

5x − 8 = 72

4x − 14 = −82

9x − 14 = −77

14 − 5x = −41

25 − 6x = −83

9 (5x − 2) = 45

10 (3x + 2) = 70

5x − (2x − 10) = 35

11x − (6x − 5) = 40

3x + 5 = 2x + 13

2x − 7 = 6 + x

8x − 2 = 7x − 5

13x + 14 = −5 + 12x

7x + 4 = x + 16

8x + 1 = x + 43








31.

32.

33.

34.

35.

36.

37.

38.

39.

40.

41.

42.

43.

44.

45.

46.

47.

48.

49.

50.

51.

52.

53.

54.

55.

56.

57.

58.

In Exercises 59–72, solve each proportion and check.

59.

60.

61.

62.

63.

64.

65.

66.

67.

68.

69.

8y − 3 = 11y + 9

5y − 2 = 9y + 2

2 (4 − 3x) = 2 (2x + 5)

3 (5 − x) = 4 (2x + 1)

8 (y + 2) = 2 (3y + 4)

3 (3y − 1) = 4 (3 + 3y)

3 (x + 1) = 7 (x − 2) − 3

5x − 4 (x + 9) = 2x − 3

5 (2x − 8) − 2 = 5 (x − 3) + 3

7 (3x − 2) + 5 = 6 (2x − 1) + 24

6 = −4 (1 − x) + 3 (x + 1)

100 = − (x − 1) + 4 (x − 6)

10 (z + 4) − 4 (z − 2) = 3 (z − 1) + 2 (z − 3)

−2 (z − 4) − (3z − 2) = −2 − (6z − 2)

− 5 = 72x

3

− 9 = −63x

4

+ =x

3x

2

5

6

− = 1x

4x

5

20 − =z

3z

2

− =z

512

z

6

+ = −y

325

y

525

+ = −y

1216

y

214

− 3 = + 23x

4x

2

− = +3x

525

x

325

− x = −3x

5x

10

5

2

2x − = +2x

7x

2

17

2

− 1 =x−3

5

x−5

4

− 4 =x−2

3

x+1

4

=24x

127

=56x

8

7

=x

6

18

4

=x

32

3

24

=−3

8x

40

=−3

8

6x

= −x

12

3

4

= −x

64

9

16

=x−2

12

8

3

=x−4

10

3

5

=x

7

x+14

5



Page 362

70.

71.

72.

In Exercises 73–92, solve each equation. Use set notation to express solution sets for equations with no solution or equations that are true for all real numbers.

73.

74.

75.

76.

77.

78.

79.

80.

81.

82.

83.

84.

85.

86.


=x

5

x−3

2

=y+10

10

y−2

4

=2y−5

3

y+6

3x − 7 = 3 (x + 1)

2 (x − 5) = 2x + 10

2 (x + 4) = 4x + 5 − 2x + 3

3 (x − 1) = 8x + 6 − 5x − 9

7 + 2 (3x − 5) = 8 − 3 (2x + 1)

2 + 3 (2x − 7) = 9 − 4 (3x + 1)

4x + 1 − 5x = 5 − (x + 4)

5x − 5 = 3x − 7 + 2 (x + 1)

4 (x + 2) + 1 = 7x − 3 (x − 2)

5x − 3 (x + 1) = 2 (x + 3) − 5

3 − x = 2x + 3

5 − x = 4x + 5

+ 2 =x

3x

3

+ 3 =x

4x

4









87.

88.

89.

90.

91.

92.

Practice Plus93. Evaluate for the value of x satisfying

94. Evaluate for the value of x satisfying

95. Evaluate for x satisfying and y satisfying

96. Evaluate for x satisfying and y satisfying

In Exercises 97–104, solve each equation.

97.

98.

99.

100.

101.

102.

103.

104.

Application ExercisesThe latest guidelines, which apply to both men and women, give healthy weight ranges, rather than specific weights, for your height. The further you are above the upperlimit of your range, the greater are the risks of developing weight-related health problems. The bar graph shows these ranges for various heights for people between theages of 19 and 34, inclusive.

dSource: U.S. Department of Health and Human Services

The mathematical model



=x

3x

2

=x

4x

3

=x−2

5

3

10

=x+4

8

3

16

− + 4 = x + 4x

2x

4

+ + 3 = x + 3x

22x

3

− xx2 4 (x − 2) +2 = 4x − 2 (2 − x) .

− xx2 2 (x − 6) = 3x + 2 (2x − 1) .

− (xy − y)x2 − 2 =x

5x

3−2y − 10 = 5y + 18.

− (xy − y)x2 + = − 43x

2

3x

4x

45 − y = 7 (y + 4) + 1.

[ ÷ 3] ⋅ 4 = −54x(3 + 6)2

− [4 ] = −8x23 (5 − 3)3

5 − 12x = 8 − 7x − [6 ÷ 3 (2 + ) + 5x]53

2 (5x + 58) = 10x + 4 (21 ÷ 3.5 − 11)

0.7x + 0.4 (20) = 0.5 (x + 20)

0.5 (x + 2) = 0.1 + 3 (0.1x + 0.3)

4x + 13 − {2x − [4 (x − 3) − 5]} = 2 (x − 6)

−2 {7 − [4 − 2 (1 − x) + 3]} = 10− [4x − 2 (x − 3)]








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describes a weight, W, in pounds, that lies within the healthy weight range for a person whose height is H inches over 5 feet. Use this information to solve Exercises105–106.

105. Use the formula to find a healthy weight for a person whose height is (Hint: because this person's height is 6 inches over 5 feet.) How manypounds is this healthy weight below the upper end of the range shown by the bar graph at the bottom of the previous column?

106. Use the formula to find a healthy weight for a person whose height is (Hint: because this person's height is 12 inches over 5 feet.) How manypounds is this healthy weight below the upper end of the range shown by the bar graph at the bottom of the previous column?

In the years after warning labels were put on cigarette packs, the number of smokers dropped from approximately two in five adults to one in five. The bar graph showsthe percentage of American adults who smoked cigarettes for selected years from 1970 through 2010.

dSource: Centers for Disease Control and Prevention

The mathematical model

describes the percentage of Americans who smoked cigarettes, p, x years after 1970. Use this model to solve Exercises 107–108.

107.

a. Does the mathematical model underestimate or overestimate the percentage of American adults who smoked cigarettes in 2010? By how much?

b. Use the mathematical model to project the year when only 7% of American adults will smoke cigarettes.

108.

a. Does the mathematical model underestimate or overestimate the percentage of American adults who smoked cigarettes in 2000? By how much?

b. Use the mathematical model to project the year when only 2% of American adults will smoke cigarettes.

109. The volume of blood in a person's body is proportional to body weight. A person who weighs 160 pounds has approximately 5 quarts of blood. Estimate thenumber of quarts of blood in a person who weighs 200 pounds.


− 3H = 53W

2

5'6'' . H = 6

6'0'' . H = 12

p + = 37x

2









110. The number of gallons of water used when taking a shower is proportional to the time in the shower. A shower lasting 5 minutes uses 30 gallons of water. Howmuch water is used in a shower lasting 11 minutes?

111. An alligator's tail length is proportional to its body length. An alligator with a body length of 4 feet has a tail length of 3.6 feet. What is the tail length of an alligatorwhose body length is 6 feet?

112. An object's weight on the moon is proportional to its weight on Earth. Neil Armstrong, the first person to step on the moon on July 20, 1969, weighed 360 poundson Earth (with all of his equipment on) and 60 pounds on the moon. What is the moon weight of a person who weighs 186 pounds on Earth?

113. St. Paul Island in Alaska has 12 fur seal rookeries (breeding places). In 1961, to estimate the fur seal pup population in the Gorbath rookery, 4963 fur seal pupswere tagged in early August. In late August, a sample of 900 pups was observed and 218 of these were found to have been previously tagged. Estimate the totalnumber of fur seal pups in this rookery.

114. To estimate the number of bass in a lake, wildlife biologists tagged 50 bass and released them in the lake. Later they netted 108 bass and found that 27 of themwere tagged. Approximately how many bass are in the lake?

Writing in Mathematics115. What is the solution set of an equation?

116. State the addition property of equality and give an example.

117. State the multiplication property of equality and give an example.

118. What is a proportion? Give an example with your description.

119. Explain how to solve a proportion. Illustrate your explanation with an example.

120. How do you know whether an equation has one solution, no solution, or infinitely many solutions?

121. What is the difference between solving an equation such as and simplifying an algebraic expression such as If there isa difference, which topic should be taught first? Why?

122. Suppose that you solve by multiplying both sides by 20, rather than the least common denominator of and (namely, 10). Describe whathappens. If you get the correct solution, why do you think we clear the equation of fractions by multiplying by the least common denominator?

123. Suppose you are an algebra teacher grading the following solution on an examination:

You should note that 8 checks, and the solution set is The student who worked the problem therefore wants full credit. Can you find any errors in the solution? Iffull credit is 10 points, how many points should you give the student? Justify your position.

124. Although the formulas in Example 6 on page 356 are correct, some people object to representing the variables with numbers, such as a 1-to-10 scale for theintensity of a negative life event. What might be their objection to quantifying the variables in this situation?

Critical Thinking ExercisesMake Sense? In Exercises 125–128, determine whether each statement makes sense or does not make sense, and explain your reasoning.

125. Although I can solve by first subtracting from both sides, I find it easier to begin by multiplying both sides by 20, the least commondenominator.

126. Because I know how to clear an equation of fractions, I decided to clear the equation of decimals by multiplying both sides by 10.

127. The number 3 satisfies the equation so {3} is the equation's solution set.

128. I can solve by using the cross-products principle or by multiplying both sides by 18, the least common denominator.

129. Write three equations whose solution set is

130. If x represents a number, write an English sentence about the number that results in an equation with no solution.

131. A woman's height, h, is related to the length of the femur, f (the bone from the knee to the hip socket), by the formula Both h and f aremeasured in inches. A partial skeleton is found of a woman in which the femur is 16 inches long. Police find the skeleton in an area where a woman slightly over 5 feettall has been missing for over a year. Can the partial skeleton be that of the missing woman? Explain.



2 (x − 4) + 5x = 34 2(x − 4) + 5x?

− = 1x

5x

2x

5x

2

Solve : −3(x − 6)

Solution : −3x − 18

−2x − 18

−2x

x

= 2− x.

= 2 − x

= 2

= −16

= 8.

{8} .

3x + =15

14

15

0.5x + 8.3 = 12.4

7x + 9 = 9(x + 1) − 2x,

=x

946

{5} .

f = 0.432h − 10.44.









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