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A. Solving Linear Equations Using Properties of Equality

# A. Solving Linear Equations Using Properties of Equality

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Guide to Solving Linear Equations in One Variable

• Eliminate parentheses using the distributive property, then combine any like terms.

• Use the additive property of equality to write the equation with all variable terms

on one side, and all constants on the other. Simplify each side.

• Use the multiplicative property of equality to obtain an equation of the form

x ϭ constant.

• For applications, answer in a complete sentence and include any units of measure

indicated.

For our first example, we’ll use the equation 31x Ϫ 12 ϩ x ϭ Ϫx ϩ 7 from our

initial discussion.

EXAMPLE 1

Solving a Linear Equation Using Properties of Equality

Solve for x: 31x Ϫ 12 ϩ x ϭ Ϫx ϩ 7.

Solution

31x Ϫ 12 ϩ x ϭ Ϫx ϩ 7

3x Ϫ 3 ϩ x ϭ Ϫx ϩ 7

4x Ϫ 3 ϭ Ϫx ϩ 7

5x Ϫ 3 ϭ 7

5x ϭ 10

xϭ2

original equation

distributive property

combine like terms

multiply both sides by 15 or divide both sides by 5

(multiplicative property of equality)

As we noted in Table R.1, the solution is x ϭ 2.

Now try Exercises 7 through 12

To check a solution by substitution means we substitute the solution back into the

original equation (this is sometimes called back-substitution), and verify the lefthand side is equal to the right. For Example 1 we have:

31x Ϫ 12 ϩ x ϭ Ϫx ϩ 7

original equation

312 Ϫ 12 ϩ 2 ϭ Ϫ2 ϩ 7

substitute 2 for x

3112 ϩ 2 ϭ 5

5 ϭ 5✓

simplify

solution checks

If any coefficients in an equation are fractional, multiply both sides by the least

common denominator (LCD) to clear the fractions. Since any decimal number can be

written in fraction form, the same idea can be applied to decimal coefficients.

EXAMPLE 2

Solution

A. You’ve just seen how

we can solve linear equations

using properties of equality

Solving a Linear Equation with Fractional Coefficients

Solve for n: 14 1n ϩ 82 Ϫ 2 ϭ 12 1n Ϫ 62.

1

4 1n ϩ 82

1

4n ϩ 2

Ϫ 2 ϭ 12 1n Ϫ 62

Ϫ 2 ϭ 12n Ϫ 3

1

1

4n ϭ 2n Ϫ 3

1

41 4n2 ϭ 41 12n Ϫ 32

n ϭ 2n Ϫ 12

Ϫn ϭ Ϫ12

n ϭ 12

original equation

distributive property

combine like terms

multiply both sides by LCD ϭ 4

distributive property

subtract 2n

multiply by Ϫ1

Verify the solution is n ϭ 12 using back-substitution.

Now try Exercises 13 through 30

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Section R.3 Solving Linear Equations and Inequalities

Example 1 illustrates what is called a conditional equation, since the equation is true for

x ϭ 2, but false for all other values of x. The equation in Example 2 is also conditional. An

identity is an equation that is always true, no matter what value is substituted for the variable. For instance, 21x ϩ 32 ϭ 2x ϩ 6 is an identity with a solution set of all real numbers, written as 5x|x ʦ ‫ޒ‬6, or x ʦ 1Ϫq, q 2 in interval notation. Contradictions are

equations that are never true, no matter what real number is substituted for the variable.

The equations x Ϫ 3 ϭ x ϩ 1 and Ϫ3 ϭ 1 are contradictions. To state the solution set

for a contradiction, we use the symbol “л” (the null set) or “{ }” (the empty set). Recognizing these special equations will prevent some surprise and indecision in later chapters.

EXAMPLE 3

Solving Equations (Special Cases)

Solve each equation and state the solution set.

a. 21x Ϫ 42 ϩ 10x ϭ 8 ϩ 413x ϩ 12

b. 8x Ϫ 16 Ϫ 10x2 ϭ 24 ϩ 613x Ϫ 52

Solution

a. 21x Ϫ 42 ϩ 10x ϭ 8 ϩ 413x ϩ 12

2x Ϫ 8 ϩ 10x ϭ 8 ϩ 12x ϩ 4

12x Ϫ 8 ϭ 12x ϩ 12

Ϫ8 ϭ 12

original equation

distributive property

combine like terms

Since Ϫ8 is never equal to 12, the original equation is a contradiction.

The solution set is empty: { }

b. 8x Ϫ 16 Ϫ 10x2 ϭ 24 ϩ 613x Ϫ 52

8x Ϫ 6 ϩ 10x ϭ 24 ϩ 18x Ϫ 30

18x Ϫ 6 ϭ 18x Ϫ 6

Ϫ6 ϭ Ϫ6

original equation

distributive property

combine like terms

subtract 18x; identity

The result shows that the original equation is an identity, with an infinite

number of solutions: 5x | x ʦ ‫ޒ‬6 . You may recall this notation is read, “the set

of all numbers x, such that x is a real number.”

Now try Exercises 31 through 36

B. You’ve just seen how

we can recognize equations

that are identities or

In Example 3(a), our attempt to solve for x ended with all variables being eliminated,

leaving an equation that is always false —a contradiction (Ϫ8 is never equal to 12).

There is nothing wrong with the solution process, the result is simply telling us the

original equation has no solution. In Example 3(b), all variables were again eliminated

but the end result was always true —an identity (Ϫ6 is always equal to Ϫ6). Once

again we’ve done nothing wrong mathematically, the result is just telling us that the

original equation will be true no matter what value of x we use for an input.

C. Solving Linear Inequalities

A linear inequality resembles a linear equality in many respects:

Linear Inequality

Related Linear Equation

(1)

x 6 3

xϭ3

(2)

3

p Ϫ 2 Ն Ϫ12

8

3

p Ϫ 2 ϭ Ϫ12

8

A linear inequality in one variable is one that can be written in the form ax ϩ b 6 c,

where a, b, and c ʦ ‫ ޒ‬and a 0. This definition and the following properties also apply

when other inequality symbols are used. Solutions to simple inequalities are easy to spot.

For instance, x ϭ Ϫ2 is a solution to x 6 3 since Ϫ2 6 3. For more involved inequalities we use the additive property of inequality and the multiplicative property of

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inequality. Similar to solving equations, we solve inequalities by isolating the variable

on one side to obtain a solution form such as variable 6 number.

If A, B, and C represent algebraic expressions and A 6 B,

then A ϩ C 6 B ϩ C

Like quantities (numbers or terms) can be added to both sides of an inequality.

While there is little difference between the additive property of equality and the

additive property of inequality, there is an important difference between the multiplicative property of equality and the multiplicative property of inequality. To illustrate, we

begin with Ϫ2 6 5. Multiplying both sides by positive three yields Ϫ6 6 15, a true

inequality. But notice what happens when we multiply both sides by negative three:

Ϫ2 6 5

original inequality

Ϫ21Ϫ32 6 51Ϫ32

6 6 Ϫ15

multiply by negative three

false

The result is a false inequality, because 6 is to the right of Ϫ15 on the number line.

Multiplying (or dividing) an inequality by a negative quantity reverses the order relationship between two quantities (we say it changes the sense of the inequality). We

must compensate for this by reversing the inequality symbol.

6 7 Ϫ15

change direction of symbol to maintain a true statement

For this reason, the multiplicative property of inequality is stated in two parts.

The Multiplicative Property of Inequality

EXAMPLE 4

If A, B, and C represent algebraic

expressions and A 6 B,

then AC 6 BC

If A, B, and C represent algebraic

expressions and A 6 B,

then AC 7 BC

if C is a positive quantity

(inequality symbol remains the same).

if C is a negative quantity

(inequality symbol must be reversed).

Solving an Inequality

Solve the inequality, then graph the solution set and write it in interval notation:

Ϫ2

1

5

3 x ϩ 2 Յ 6.

Solution

WORTHY OF NOTE

As an alternative to multiplying or

dividing by a negative value, the

be used to ensure the variable term

will be positive. From Example 4,

the inequality Ϫ4x Յ 2 can be

written as Ϫ2 Յ 4x by adding 4x to

both sides and subtracting 2 from

both sides. This gives the solution

Ϫ12 Յ x, which is equivalent to

Ϫ12 .

1

5

Ϫ2

xϩ Յ

3

2

6

Ϫ2

1

5

6 a x ϩ b Յ 162

3

2

6

Ϫ4x ϩ 3 Յ 5

Ϫ4x Յ 2

Ϫ4x

2

Ն

Ϫ4

Ϫ4

1

xՆϪ

2

original inequality

clear fractions (multiply by LCD)

simplify

subtract 3

divide by Ϫ4, reverse inequality sign

result

Ϫ1

2

• Graph:

Ϫ3 Ϫ2 Ϫ1

[

0

1

2

3

• Interval notation: x ʦ 3Ϫ12, q 2

4

Now try Exercises 37 through 46

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29

To check a linear inequality, you often have an infinite number of choices—any

number from the solution set/interval. If a test value from the solution interval results

in a true inequality, all numbers in the interval are solutions. For Example 4, using

x ϭ 0 results in the true statement 12 Յ 56 ✓.

Some inequalities have all real numbers as the solution set: 5x| x ʦ ‫ޒ‬6, while other

inequalities have no solutions, with the answer given as the empty set: { }.

EXAMPLE 5

Solving Inequalities

Solve the inequality and write the solution in set notation:

a. 7 Ϫ 13x ϩ 52 Ն 21x Ϫ 42 Ϫ 5x

b. 31x ϩ 42 Ϫ 5 6 21x Ϫ 32 ϩ x

Solution

a. 7 Ϫ 13x ϩ 52 Ն 21x Ϫ 42 Ϫ 5x

7 Ϫ 3x Ϫ 5 Ն 2x Ϫ 8 Ϫ 5x

Ϫ3x ϩ 2 Ն Ϫ3x Ϫ 8

2 Ն Ϫ8

original inequality

distributive property

combine like terms

Since the resulting statement is always true, the original inequality is true for

all real numbers. The solution is all real numbers ‫ޒ‬.

b. 31x ϩ 42 Ϫ 5

3x ϩ 12 Ϫ 5

3x ϩ 7

7

6

6

6

6

21x Ϫ 32 ϩ x

2x Ϫ 6 ϩ x

3x Ϫ 6

Ϫ6

original inequality

distribute

combine like terms

subtract 3x

Since the resulting statement is always false, the original inequality is false for

all real numbers. The solution is { }.

C. You’ve just seen how

we can solve linear inequalities

Now try Exercises 47 through 52

D. Solving Compound Inequalities

In some applications of inequalities, we must consider more than one solution interval.

These are called compound inequalities, and these require us to take a close look at

the operations of union and intersection ă. The intersection of two sets A and

B, written A ă B, is the set of all elements common to both sets. The union of two sets

A and B, written A ´ B, is the set of all elements that are in either set. When stating the

union of two sets, repetitions are unnecessary.

EXAMPLE 6

Finding the Union and Intersection of Two Sets

Solution

A ă B is the set of all elements in both A and B:

A ʝ B ϭ 51, 2, 36.

A ´ B is the set of all elements in either A or B:

A ´ B ϭ 5Ϫ2, Ϫ1, 0, 1, 2, 3, 4, 56.

WORTHY OF NOTE

For the long term, it may help to

rephrase the distinction as follows.

The intersection is a selection of

elements that are common to two

sets, while the union is a collection

of the elements from two sets (with

no repetitions).

For set A ϭ 5Ϫ2, Ϫ1, 0, 1, 2, 36 and set B ϭ 51, 2, 3, 4, 56,

determine A ă B and A B.

Now try Exercises 53 through 58

Notice the intersection of two sets is described using the word “and,” while the

union of two sets is described using the word “or.” When compound inequalities are

formed using these words, the solution is modeled after the ideas from Example 6. If

“and” is used, the solutions must satisfy both inequalities. If “or” is used, the solutions

can satisfy either inequality.

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EXAMPLE 7

Solving a Compound Inequality

Solve the compound inequality, then write the solution in interval notation:

Ϫ3x Ϫ 1 6 Ϫ4 or 4x ϩ 3 6 Ϫ6.

Solution

Begin with the statement as given:

Ϫ3x Ϫ 1 6 Ϫ4

Ϫ3x 6 Ϫ3

x 7 1

4x ϩ 3 6 Ϫ6

4x 6 Ϫ9

9

x 6 Ϫ

4

or

or

or

original statement

isolate variable term

solve for x, reverse first inequality symbol

The solution x 7 1 or x 6 Ϫ94 is better understood by graphing each

interval separately, then selecting both intervals (the union).

WORTHY OF NOTE

x Ͼ 1:

The graphs from Example 7 clearly

show the solution consists of two

disjoint (disconnected) intervals.

This is reflected in the “or”

statement: x 6 Ϫ94 or x 7 1, and in

the interval notation. Also, note the

solution x 6 Ϫ94 or x 7 1 is not

equivalent to Ϫ94 7 x 7 1, as there

is no single number that is both

greater than 1 and less than Ϫ94 at

the same time.

x Ͻ Ϫ9 :

4

)

Ϫ8 Ϫ7 Ϫ6 Ϫ5 Ϫ4 Ϫ3 Ϫ2 Ϫ1

0

1

2

3

4

5

6

Ϫ9

4

)

Ϫ8 Ϫ7 Ϫ6 Ϫ5 Ϫ4 Ϫ3 Ϫ2 Ϫ1

x Ͻ Ϫ 9 or x Ͼ 1:

4

0

1

2

3

4

5

6

Ϫ9

4

)

)

Ϫ8 Ϫ7 Ϫ6 Ϫ5 Ϫ4 Ϫ3 Ϫ2 Ϫ1

0

1

2

3

4

5

6

9

Interval notation: x ʦ aϪq, Ϫ b ´ 11, q 2.

4

Now try Exercises 59 and 60

EXAMPLE 8

Solving a Compound Inequality

Solve the compound inequality, then write the solution in interval notation:

3x ϩ 5 7 Ϫ13 and 3x ϩ 5 6 Ϫ1.

Solution

Begin with the statement as given:

3x ϩ 5 7 Ϫ13

3x 7 Ϫ18

x 7 Ϫ6

3x ϩ 5 6 Ϫ1

3x 6 Ϫ6

x 6 Ϫ2

and

and

and

original statement

subtract five

divide by 3

The solution x 7 Ϫ6 and x 6 Ϫ2 can best be understood by graphing each interval

separately, then noting where they intersect.

WORTHY OF NOTE

x Ͼ Ϫ6:

x Ͻ Ϫ2:

x Ͼ Ϫ6 and x Ͻ Ϫ2:

)

Ϫ8 Ϫ7 Ϫ6 Ϫ5 Ϫ4 Ϫ3 Ϫ2 Ϫ1

)

Ϫ8 Ϫ7 Ϫ6 Ϫ5 Ϫ4 Ϫ3 Ϫ2 Ϫ1

)

The inequality a 6 b (a is less than

b) can equivalently be written as

b 7 a (b is greater than a). In

Example 8, the solution is read,

“x 7 Ϫ6 and x 6 Ϫ2,” but if we

rewrite the first inequality as

Ϫ6 6 x (with the “arrowhead” still

pointing at Ϫ62, we have Ϫ6 6 x

and x 6 Ϫ2 and can clearly see

that x must be in the single interval

between Ϫ6 and Ϫ2.

0

1

2

3

4

5

6

0

1

2

3

4

5

6

0

1

2

3

4

5

6

)

Ϫ8 Ϫ7 Ϫ6 Ϫ5 Ϫ4 Ϫ3 Ϫ2 Ϫ1

Interval notation: x ʦ 1Ϫ6, Ϫ22.

Now try Exercises 61 through 72

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Section R.3 Solving Linear Equations and Inequalities

The solution from Example 8 consists of the single interval 1Ϫ6, Ϫ22, indicating the original inequality could actually be joined and written as Ϫ6 6 x 6 Ϫ2,

called a joint inequality. We solve joint inequalities in much the same way as linear inequalities, but must remember they have three parts (left, middle, and right).

This means operations must be applied to all three parts in each step of the solution

process, to obtain a solution form such as smaller number 6 x 6 larger number.

The same ideas apply when other inequality symbols are used.

EXAMPLE 9

Solving a Joint Inequality

Solve the joint inequality, then graph the solution set and write it in interval

2x ϩ 5

Ն Ϫ6.

notation: 1 7

Ϫ3

Solution

2x ϩ 5

Ն Ϫ6

Ϫ3

Ϫ3 6 2x ϩ 5 Յ 18

Ϫ8 6 2x Յ 13

13

Ϫ4 6 x Յ

2

1 7

original inequality

multiply all parts by Ϫ3; reverse the inequality symbols

subtract 5 from all parts

divide all parts by 2

13

2

• Graph:

)

Ϫ5 Ϫ4 Ϫ3 Ϫ2 Ϫ1

[

0

• Interval notation: x ʦ 1Ϫ4,

D. You’ve just seen how

we can solve compound

inequalities

1

13

2 4

2

3

4

5

6

7

8

Now try Exercises 73 through 78

E. Solving Basic Applications of Linear

Equations and Inequalities

Applications of linear equations and inequalities come in many forms. In most cases,

you are asked to translate written relationships or information given verbally into an

equation using words or phrases that indicate mathematical operations or relationships.

Here, we’ll practice this skill using ideas that were introduced in Section R.1, where

we translated English phrases into mathematical expressions. Very soon these skills

will be applied in much more significant ways.

EXAMPLE 10

Translating Written Information into an Equation

Translate the following relationships into equations, then solve:

In an effort to lower the outstanding balance on her credit card, Laura paid \$10 less

than triple her normal payment. If she sent the credit card company \$350.75, how

much was her normal payment? (See Section R.1, Examples 2 and 3.)

Solution

Let p represent her normal payment. Then “triple her normal payment” would be

3p, and “ten less than triple” would be 3p Ϫ 10. Since “she sent the company

\$350.75,” we have

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3p Ϫ 10 ϭ 350.75

3p ϭ 360.75

p ϭ 120.25

equation form

divide by 3

Laura’s normal payment is \$120.25 per month.

A calculator check is shown in the figure.

Now try Exercises 83 through 90

Inequalities are widely used to help gather information, and to make comparisons

that will lead to informed decisions.

EXAMPLE 11

Using an Inequality to Compute Desired Test Scores

Justin earned scores of 78, 72, and 86 on the first three out of four exams. What

score must he earn on the fourth exam to have an average of at least 80?

Solution

The current scores are 78, 72, and 86. An average of at least 80 means A Ն 80.

In organized form:

Test 1

Test 2

Test 3

Test 4

Computed Average

Minimum

78

72

86

x

78 ϩ 72 ϩ 86 ϩ x

4

80

Let x represent Justin’s score on the fourth exam, then

represents his average score.

78 ϩ 72 ϩ 86 ϩ x

Ն 80

4

78 ϩ 72 ϩ 86 ϩ x Ն 320

236 ϩ x Ն 320

x Ն 84

E. You’ve just seen how

we can solve applications

of linear equations and

inequalities

78 ϩ 72 ϩ 86 ϩ x

4

average must be greater than or equal to 80

multiply by 4

simplify

solve for x (subtract 236)

Justin must score at least an 84 on the last exam to earn an 80 average.

Now try Exercises 91 through 100

F. Solving Applications of Basic Geometry

As your translation skills grow, your ability to solve a wider range of more significant

applications will grow as well. In many cases, the applications will involve some basic

geometry and the most often used figures and formulas appear here. For a more complete review of geometry, see Appendix II Geometry Review with Unit Conversions,

which is posted online at www.mhhe.com/coburn.

Perimeter and Area

Perimeter is a measure of the distance around a two dimensional figure. As this is a linear measure, results are stated in linear units as in centimeters (cm), feet (ft), kilometers (km), miles (mi), and so on. If no unit is specified, simply write the result as x

units. Area is a measure of the surface of a two dimensional figure, with results stated

in square units as in x units2. Some of the most common formulas involving perimeter

and area are given in Table R.2A.

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