Slide 1- 1 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley.

Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Equations, Inequalities, and Problem Solving

2.1 Solving Equations

2.2 Using the Principles Together

2.3 Formulas

2.4 Applications with Percent

2.5 Problem Solving

2.6 Solving Inequalities

2.7 Solving Applications with Inequalities

2


Solving Equations

Equations and Solutions

The Addition Principle

The Multiplication Principle

Selecting the Correct Approach

2.1


Solution of an Equation

Any replacement for the variable that makes an equation true is called a solution of the equation. To solve an equation means to find all of its solutions.


Determine whether 8 is a solution of x + 12 = 21.

Since the left-hand and right-hand sides differ, 8 is not a solution.

20 21 False

8 + 12 | 21 Substituting 8 for x

Solution x + 12 = 21 Writing the equation

Example


Equivalent EquationsEquations with the same solutions are called equivalent equations.

The Addition Principle For any real numbers a, b, and c,

a = b is equivalent to a + c = b + c.


Solve: 8.3 = y 17.9

Solution 8.3 = y 17.9

8.3 + 17.9 = y 17.9 + 17.9

9.6 = y

Check: 8.3 = y 17.9

8.3 | 9.6 17.9

8.3 = 8.3

The solution is 9.6.

Example


The Multiplication Principle For any real numbers a, b, and c with c 0,

a = b is equivalent to a • c = b • c.

Solve: 3

154

x

315

4x

315

4

4 4

3 3x

1 20x

20x

Multiplying both sides by 4/3.

Solution

Example


Solve: 7x = 84

Solution 7x = 84

7 84

7 7

x

1 12x

12x

Dividing both sides by 7.

The solution is –12.

Check: 7x = 84 7(12) | 84

84 = 84

Example


Using the Principles Together

Applying Both Principles

Combining Like Terms

Clearing Fractions and Decimals

Contradictions and Identities

2.2


Solve: 9 + 8x = 33

Solution 9 + 8x = 33

9 + 8x 9 = 33 9

9 + ( 9) + 8x = 24

8x = 24

x = 3

Check: 9 + 8x = 33

9 + 8(3) | 33

9 + 24 | 33

33 = 33 The solution is 3.

8 48 8

2x

Subtracting 9 from both sides

Dividing both sides by 8

Example


Solution

Solve: 2 5 113

x

52

5 11 53

x

25 11

3x

216

3x

62 2

2

3

3 31x

8 21 x

3

2

24x

Adding 5 to both sides

Multiplying both sides by 3/2.

Simplifying

Example


Solve: 36 t = 17

Solution 36 t = 17

36 t 36 = 17 36

t = 19

(1) t = (19)(1)

t = 19

Check: 36 t = 17

36 19 | 17

17 = 17

The solution is 19.


Multiplying both sides by 1

Example


Combining Like Terms

If like terms appear on the same side of an equation, we combine them and then solve.

Should like terms appear on both sides of an equation, we can use the addition principle to rewrite all like terms on one side.


Solve. 5x + 4x = 36

Solution 5x + 4x = 36

9x = 36

x = 4

Check: 5x + 4x = 36 5(4) + 4(4) | 36

20 + 16 | 36

36 = 36

The solution is 4.

9 6

9 9

3x

Combining like terms


Simplifying

Example


Solve. 4x + 7 6x = 10 + 3x + 12

Solution: 4x + 7 6x = 10 + 3x + 12

2x + 7 = 22 + 3x

2x + 7 7 = 22 + 3x 7

2x = 15 + 3x

2x 3x = 15 + 3x 3x

5x = 15

x = 3

5 15

5 5

x

Combining like terms


Simplifying

Subtracting 3x from both sides


Example


Check: 4x + 7 6x = 10 + 3x + 12

4(3) + 7 6(3) | 10 + 3(3) + 12

12 + 7 + 18 | 10 9 + 12

13 = 13

The solution is 3.


Solve.

Solution:

4(6 1) 8

5x

4(6 1) 8

5x

4(6

5

448

5

51) x

6 1 10x

11 10 16x 6 9x 6 9

6 6

x

3

2x

Example


An Equation-Solving Procedure1. Use the multiplication principle to clear any

fractions or decimals. (This is optional, but can ease computations.

2. If necessary, use the distributive law to remove parentheses. Then combine like terms on each side.

3. Use the addition principle, as needed, to isolate all variable terms on one side. Then combine like terms.

4. Multiply or divide to solve for the variable, using the multiplication principle.

5. Check all possible solutions in the original equation.


Contradictions and Identities

An identity is an equation that is true for all replacements that can be used on both sides of the equation.

A contradiction is an equation that is never true.

A conditional equation is sometimes true and sometimes false, depending on what the replacement is.


a) Solve: 8 3 2 3(2 1)x x x

The equation is true regardless of the choice for x, so all real numbers are solutions. The equation is an identity.

8 3 2 3(2 1)x x x

8 3 2 6 3x x x

8 3 8 3x x

Example

Solution:


b) Solve: 3 2 3( 5)x x x

The equation is false for any choice of x, so there is no solution for this equation. The equation is a contradiction.

3 2 3 15x x x

3 2 3( 5)x x x

3 15x x

3 15

Example

Solution:


c) Solve: 11 3( 21)x x x

There is one solution, 7. For all other choices of x, the equation is false. The equation is a conditional equation.

11 3( 21)x x x 11 3 63x x x 8 63x x

9 63x 7x

Example

Solution:


Formulas

Evaluating Formulas

Solving for a Variable

2.3


Many applications of mathematics involve relationships among two or more quantities. An equation that represents such a relationship will use two or more letters and is known as a formula.


The formula can be used to determine how far M, in miles, you are from lightening when its thunder takes t seconds to reach your ears. If it takes 5 seconds for the sound of thunder to reach you after you have seen the lightening, how far away is the storm?

Solution We substitute 5 for t and calculate M.

The storm is 1 mile away.

1

5M t

1

5M t

55

1M

1M

Example


Circumference of a circle.

The formula C = d gives the circumference C of a circle with diameter d. Solve for d.

Solution C = d

C d

Cd

d

Example


To Solve a Formula for a Given Variable1. If the variable for which you are solving

appears in a fraction, use the multiplication principle to clear fractions.

2. Isolate the term(s), with the variable for which you are solving on one side of the equation.

3. If two or more terms contain the variable for which you are solving, factor the variable out.

4. Multiply or divide to solve for the variable in question.


Solve for y: x = wy + zy 6

Solution x = wy + zy 6

x = wy + zy 6

x + 6 = wy + zy

x + 6 = y(w + z)

6x

w zy

We want this letter alone.


Factoring

Dividing both sides by w + z

Example


Applications with Percent

Converting Between Percent Notation and Decimal Notation

Solving Percent Problems

2.4


Percent Notation

n% means 1

, or , or 0.01.100 100

nn n


Convert to decimal notation: a) 34% b) 7.6%

Solution

a) 34% = 34 0.01

= 0.34

b) 7.6% = 7.6 0.01

= 0.076

Example


To convert from percent notation to decimal notation, move the decimal point two places to the left and drop the percent symbol.

Convert the percent notation in the following sentence to decimal notation: Marta received a 75% on her first algebra test.

Solution

75% 75.0% 0.75.0 75% = 0.75

Move the decimal point two places to the left.

Example


To convert from decimal notation to percent notation, move the decimal point two places to the right and write a percent symbol.

Convert to percent notation:

a) 3.24 b) 0.2 c)

Solution

a) We first move the decimal point two places to the right: 3.24. and then write a % symbol: 324%

3

8

Example


Solution continued

b) We first move the decimal point two places to the right (recall that 0.2 = 0.20): 0.20.

and then write a % symbol: 20%

c) Note that 3/8 = 0.375. We move the decimal point two places to the right: 0.37.5

and then write a % symbol: 37.5%


Key Word in Percent Translations“Of” translates to “•” or “”.

“What” translates to a variable.

“Is” or “Was” translates to “=”.

% translates to 1

" ", or " 0.01".100


What is 13% of 72?

Solution

Translate:

Thus, 9.36 is 13% of 72. The answer is 9.36.

What is 13% of 72?

0.13

72

a

9.36a

Example


9 is 12 percent of what?

Solution

Translate:

9 i

s 12% of what?

9 = 0.12 w

9

0.12w

75 w

Example


What percent of 60 is 27?

Solution

Translate: What percent of 60 is 27?

60 27n

27

60n

0.45 45%n

Example


To complete her water safety course instruction, Catherine must complete 45 hours of instruction. If she has completed 75% of her requirement, how many hours has Catherine completed?

Solution

Rewording: What is 75% of 45?

Translating: a = 0.75 45

a = 33.75

Catherine has completed 33.75 hours of instruction.

Example


Problem Solving

Five Steps for Problem Solving

Applying the Five Steps

2.5


Five Steps for Problem Solving in Algebra

1. Familiarize yourself with the problem.

2. Translate to mathematical language. (This often means writing an equation.)

3. Carry out some mathematical manipulation. (This often means solving an equation.)

4. Check your possible answer in the original problem.

5. State the answer clearly, using a complete English sentence.


To Become Familiar with a Problem1. Read the problem carefully. Try to visualize the

problem.2. Reread the problem, perhaps aloud. Make sure

you understand all important words.3. List the information given and the question(s) to

be answered. Choose a variable (or variables) to represent the unknown and specify what the variable represents. For example, let L = length in centimeters, d = distance in miles, and so on.

4. Look for similarities between the problem and other problems you have already solved.


To Become Familiar with a Problem (continued)

5. Find more information. Look up a formula in a book, at the library, or online. Consult a reference librarian or an expert in the field.

6. Make a table that uses all the information you have available. Look for patterns that may help in the translation.

7. Make a drawing and label it with known and unknown information, using specific units if given.

8. Think of a possible answer and check the guess. Note the manner in which the guess is checked.


The apartments in Wanda’s apartment house are consecutively numbered on each floor. The sum of her number and her next door neighbor’s number is 723. What are the two numbers?

Solution1. Familiarize. The apartment numbers are consecutive integers.

Let x = Wanda’s apartment

Let x + 1 = neighbor’s apartment

Example


2. Translate.

Rewording:

Translating:

3. Carry out. x + (x + 1) = 723

2x + 1 = 723

2x = 722

x = 361

If x is 361, then x + 1 is 362.

First integer plus second integer is 723

1 723x x


4. Check. Our possible answers are 361 and 362. These are consecutive integers and the sum is 723.

5. State. The apartment numbers are 361 and 362.


Digicon prints digital photos for $0.12 each plus $3.29 shipping and handling. Your weekly budget for the school yearbook is $22.00. How many prints can you have made if you have $22.00?

Solution 1. Familiarize. Suppose the yearbook staff takes 220 digital photos. Then the cost to print them would be the shipping charge plus $0.12 times 220.

$3.29 + $0.12(220) which is $29.69. Our guess of 220 is too large, but we have familiarized ourselves with the way in which the calculation is made.

Example


2. Translate.

Rewording:

Translating:

3. Carry out.

4. Check. Check in the original problem. $3.29 + 155(0.12) = $21.89, which is less than $22.00.

5. State. The yearbook staff can have 155 photos printed per week.

Shipping plus photo cost is $22

$3.29 0.12( ) 22x

3.29 0.12 22x

0.12 18.71x

155.9 155x


You are constructing a triangular kite. The second angle of the kite is three times as large as the first. The third angle is 10 degrees more than the first. Find the measure of each angle.

Solution1. Familiarize. Make a drawing and write in the given information.

2. Translate. To translate, we need to recall that the sum of the measures of the angles in a triangle is 180 degrees.

3x

x

x + 10

Example


2. Translate (continued).

3. Carry out.

The measures for the angles appear to be:

first angle: x = 34second angle: 3x = 3(34) = 102; third angle: x + 10 = 34 + 10 = 44

Measure of measure of measure of

first angle + second angle + third angle is 180

3 10 180x x x

3 10 180x x x

5 10 180x 5 170x

34x


4. Check. Consider 34, 102 and 44 degrees. The sum of these numbers is 180 degrees and the second angle is three times the first angle. The third angle is 10 degrees more than the first. These numbers check.

5. State. The measures of the angles are 34, 44 and 102 degrees.


Solving Inequalities

Solutions of Inequalities Graphs of Inequalities Set Builder and Interval Notation Solving Inequalities Using the Addition

Principle Solving Inequalities Using the

Multiplication Principle Using the Principles Together

2.6


Solutions of InequalitiesAn inequality is a number sentence containing > (is greater than), < (is less than), (is greater than or equal to), or (is less than or equal to).

Determine whether the given number is a solution of x < 5: a) 4 b) 6

Solution

a) Since 4 < 5 is true, 4 is a solution.

b) Since 6 < 5 is false, 6 is not a solution.

6-3 -1 1 3 5-4 0 4-2-4 2

Example


Solutions of InequalitiesAn inequality is any sentence containing

, , , , or .

Any value for a variable that makes an inequality true is called a solution. The set of all solutions is called the solution set. When all solutions of an inequality are found, we say that we have solved the inequality.

Examples:3 2 7, 7, and 4 6 3.x c x


Solution

Determine whether 5 is a solution to 3 2 7.x

We substitute to get 3(5) + 2 > 7, or 17 >7, a true statement. Thus, 5 is a solution.

Example


The graph of an inequality is a visual representation of the inequality’s solution set. An inequality in one variable can be graphed on a number line.

Graph x < 2 on a number line.

Solution

Note that in set-builder notation the solution is

| 2 .x x

Example

4 5 3 1 -1 0 -5 -6 -4 -3 -2 2 6)


Interval Notation We can write solutions of an inequality in one variable using interval notation. Interval notation uses parentheses, ( ), and brackets, [ ].

If a and b are real numbers such that a < b, we define the open interval (a, b) as the set of all numbers x for which a < x < b. Thus,

( , ) | .a b x a x b (a, b)

a b

)(


Interval Notation The closed interval [a, b] is defined as the set of all numbers x for which

Thus,

[ , ] | .a b x a x b

.a x b

a b

[a, b]

[ ]


Interval Notation There are two types of half-open intervals, defined as follows:

1. ( , ] | .a b x a x b

2. [ , ) | .a b x a x b

(a, b]

a b

( ]

a b

[a, b)

[ )


Interval Notation We use the symbols to represent positive and negative infinity, respectively. Thus the notation (a, ) represents the set of all real numbers greater than a, and ( , a) represents the set of all numbers less than a.

and

The notations (– , a] and [a, ) are used when we want to include the endpoint a.

( , )a

( , )a

a

a

(

)


The Addition Principle for InequalitiesFor any real numbers a, b, and c:

a < b is equivalent to a + c < b + c;

a b is equivalent to a + c b + c;

a > b is equivalent to a + c > b + c;

a b is equivalent to a + c b + c.


Solution

Solve and graph x – 2 > 7.

x – 2 > 7

x – 2 + 2 > 7 + 2

x > 9

The solution set is | 9 , or (9, ).x x

Example

5 6 7 8 9 10 11 12 13 14 15 16 17

(


Solve 4x 1 x 10 and then graph the solution.

Solution

4x 1 x 10

4x 1 + 1 x 10 + 1

4x x 9 4x x x x 9 3x 9 x 3The solution set is {x|x 3}.



Subtracting x from both sides

Simplifying

Simplifying

Example


The Multiplication Principle for InequalitiesFor any real numbers a and b, and for any positive number c:

a < b is equivalent to ac < bc, and

a > b is equivalent to ac > bc.

For any real numbers a and b, and for any negative number c:

a < b is equivalent to ac > bc, and

a > b is equivalent to ac < bc.

Similar statements hold for and .


Solve and graph each inequality:

a) b) 4y < 20

Solution

a)

The solution set is {x|x 28}. The graph is shown below.

1 47

x

14

7x

7 41

34 x

28x

3015 255 2010 30

Multiplying both sides by 4

Simplifying

Example

]


b) 4y < 20

The solution set is {y|y > 5}. The graph is shown below.

4 20

4 4

y

5y

At this step, we reverse the inequality, because 4 is negative.


4-5 -3 -1 1 3-6 -2 2-4-6 0(


Solution

3x 3 > x + 7

3x 3 + 3 > x + 7 + 3

3x > x + 10

3x x > x x + 10

2x > 10

x > 5The solution set is {x|x > 5}.

Solve. 3x 3 > x + 7


Simplifying

2 0

2 2

1x

Subtracting x from both sides

Simplifying


Simplifying

8-1 1 3 5 7-2 0 4 82-2 6

Example

(


Solution 15.4 3.2x < 6.76

100(15.4 3.2x) < 100(6.76)

100(15.4) 100(3.2x) < 100(6.76)

1540 320x < 676

320x < 676 1540

320x < 2216

x > 6.925

The solution set is {x|x > 6.925}.

Solve. 15.4 3.2x < 6.76

2216

320x

Remember to reverse the symbol!

Example


Solution

5(x 3) 7x 4(x 3) + 9

5x 15 7x 4x 12 + 9

2x 15 4x 3 2x 15 + 3 4x 3 + 3

2x 12 4x

2x + 2x 12 4x + 2x

12 6x

2 x

The solution set is {x|x 2}.

Solve: 5(x 3) 7x 4(x 3) + 9

Using the distributive law to remove parentheses

Simplifying

Adding 2x to both sides



2-7 -5 -3 -1 1-8 -6 -2 2-4-8 0

Example

]


Solving Applications with Inequalities

Translating to Inequalities

Solving Problems

2.7


Important Words Sample Sentence Translation

is at least Brian is at least 16 years old b 16

is at most At most 3 students failed the course s 3

cannot exceed To qualify, earnings cannot exceed $5000

e $5000

must exceed The speed must exceed 20 mph s > 20

is less than Nicholas is less than 60 lb. n < 60

is more than Chicago is more than 300 miles away.

c > 300

is between The movie is between 70 and 120 minutes.

70 < m < 120

no more than The calf weighs no more than 560 lb.

w 560

no less than Carmon scored no less than 9.4. c 9.4


Translating “at least” and “at most”The quantity x is at least some amount q: x q.

(If x is at least q, it cannot be less than q.)

The quantity x is at most some amount q: x q.

(If x is at most q, it cannot be more than q.)


Lazer Line charges $65 plus $45 per hour for copier repair. Jonas remembers being billed less than $150. How many hours was Jonas’ copier worked on?

Solution

1. Familiarize. Suppose the copier was worked on for 4 hours. The cost would be $65 + 4($45), or $245. This shows that the copier was worked on for less than 4 hours. Let h = the number of hours.

2. Translate.

Rewording:

Translating:

Initial fee plus hours less than 150

65 45 150h

Example


3. Carry out.

65 + 45h 150

45h 85

4. Check. Since the time represents hours, we round down to one hour. If the copier was worked on for one hour, the cost would be $110, and if worked on for two hours the cost would exceed $150.

5. State. Jonas’ copier was worked on for less than two hours.

85

45h

81

9h


Average, or MeanTo find the average or mean of a set of numbers, add the numbers and then divide by the number of addends.


Samantha has test grades of 86, 88, and 78 on her first three math tests. If she wants an average of at least 80 after the fourth test, what possible scores can she earn on the fourth test?

Solution1. Familiarize. Suppose she earned an 85 on her fourth test. Her test average would be

This shows she could score an 85. Let’s let x represent the fourth test score.

88 56 88 78 84.254

Example


2. Translate. Rewording:

Translating:

3. Carry out.

86 88 7880

4

x

86 88 7880

44 4

x

252 320x

68x

should be

Average test scores at least 80

86 88 78 80

4

x


4. Check. As a partial check, we show that Samantha can earn a 68 on | the fourth test and average 80 for the four tests.

5. State. Samantha’s test average will not drop below 80 if she earns at least a 68 on the fourth test.

6886 88 78 32080.

4 4

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