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Ch.2. Solving Equations and Inequalities/2.1.Solving Equations The Addition Principle .ppt

Chapter 2

Solving Equations and Inequalities

SOLVING EQUATIONS: THE ADDITION PRINCIPLE

a. Determine whether a given number is a solution of a given equation.

b. Solve equations using the addition principle.

2.1

Equation

An equation is a number sentence that says that the expressions on either side of the equals sign, =, represent the same number.

Example

Determine whether the equation is true, false, or neither.

a. 4 + 6 = 10

b. 8 – 3 = 4

c. x + 9 = 21

True

False

Neither, we do not know what number x represents

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.

Example

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

Solution x + 12 = 21 Writing the equation

8 + 12 | 21 Substituting 8 for x

20  21 False

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

Equivalent Equations

Equations with the same solutions are called equivalent equations.

The Addition Principle For Equations

For any real numbers a, b, and c,

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

Example

Solve: x + 6 = –9.

Solution

x + 6 = –9

x + 6 – 6 = –9 – 6

x + 0 = –15

x = –15

Using the addition principle: adding –6 to both sides or subtracting 6 on both sides

Simplifying

Identity property of 0

Check

x + 6 = –9

–15 + 6 = | –9

–9 True

Example

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.

Solve: 8.3 = y  17.9

Example

Solve:

510

-+=

510

-+=

510

=×+

1010

Ch.2. Solving Equations and Inequalities/2.2.Solving Equations The Multiplication Principle.ppt

Chapter 2

Solving Equations and Inequalities

SOLVING EQUATIONS: THE MULTIPLICATION PRINCIPLE

a. Solve equations using the multiplication
principle.

2.2

The Multiplication Principle For Equations

For any real numbers a, b, and c with c  0,

a = b is equivalent to a • c = b • c.

Example

Solve 6x = 96.

Solution 6x = 96

6 6

x = 16

Check: 6x = 96

6(16) = 96

96 TRUE

The solution is 16.

Dividing by 6 on both sides

Example

Solution 7x = 84

Check: 7x = 84

7(12) | 84

84 | True

The solution is 12.

Solve: 7x = 84

Dividing both sides by 7.

Example

Solve x = 6.

Solution x = 6

(1)(x) = (1)6

1(1)x = 6

x = 6

The solution is 6.

Multiplying by 1 on both sides

Example

Solve:

Solution

Multiplying by the reciprocal of ¾ on both sides.

Example

Solve: 3.2y = 9600

3.2y = 9600

112

×=-

××

120

3.23.2

.29600

3000

Ch.2. Solving Equations and Inequalities/2.3.Using the Principles Together.ppt

Chapter 2

Solving Equations and Inequalities

USING THE PRINCIPLES TOGETHER

a. Solve equations using both the addition and multiplication principles.

b. Solve equations in which like terms need to be collected.

c. Solve equations by first removing parentheses and collecting like terms; solve equations with no solutions and equations with an infinite number of solutions.

2.3

Example

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 TRUE The solution is 3.

Solve: 9 + 8x = 33

Subtracting 9 from both sides

Dividing both sides by 8

Example

Solve: 35 – x = 27.

Solution

35 – x = 27

35 – 35 – x = 27 – 35 Adding –35 on both sides

–x = 8

–1(–x) = –1(–8) Multiplying by –1 on both sides

x = 8

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.

Example

Solution 5x + 4x = 36

9x = 36

x = 4

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

20 + 16 | 36

The solution is 4.

Solve. 5x + 4x = 36

Collecting like terms

Dividing by 9 on both sides

Example

Solve: 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

Collecting like terms

Subtracting 7 from both sides

Simplifying

Subtracting 3x from both sides

Dividing both sides by 5

Clearing Fractions and Decimals

In general, equations are easier to solve if they do not contain fractions or decimals.

The easiest way to clear an equation of fractions is to multiply every term on both sides by the least common multiple of all the denominators.

To clear an equation of decimals, we count the greatest number of decimals places in any one number and multiply on both sides by that multiple of 10.

Example

Solve 18.4 – 6.2y = 7.24

Solution

18.4 – 6.2y = 7.24

100(18.4 – 6.2y) = 100(7.24) Multiplying by 100

(100)(18.4) – 100(6.2y) = 100(7.24) Using the distributive law

1840 – 620y = 724 Simplifying

1840 – 1840 – 620y = 724 – 1840 Subtracting 1840

–620y = – 1116 Collecting like terms

–620 – 620 Dividing by – 620

y = 1.8

Example

Solve: 9x = 3(15 – 2x)

Solution

9x = 3(15 – 2x)

9x = 45 – 6x Using the distributive law

9x + 6x = 45 – 6x + 6x Subtracting 6x to get all x-terms on one side

15x = 45 Collecting like terms

15 15 Dividing by 15

x = 3

An Equation-Solving Procedure

1. Multiply on both sides to clear the equation of fractions or decimals. (This is optional, but can ease computations.)

2. If parentheses occur, multiply to remove them using the distributive laws.

3. Collect like terms on each side, if necessary.

4. Get all terms with variables on one side and all numbers (constant terms) on the other side, using the addition principle.

5. Collect like terms again, if necessary.

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

7. Check all possible solutions in the original equation.

Example

Solve: 3 – 8(x + 6) = 4(x – 1) – 5.

Solution

3 – 8(x + 6) = 4(x – 1) – 5

3 – 8x – 48 = 4x – 4 – 5

–45 – 8x = 4x – 9

–8x – 45 + 45 = 4x – 9 + 45

–8x = 4x + 36

–8x – 4x = 4x + 36 – 4x

–12x = 36

–12 –12

x = – 3

Using the distributive law to multiply and remove parentheses

Collecting like terms

Adding 45

Subtracting 4x

Collecting like terms

Dividing by 12

Example

Solve: 8x – 18 = 6 + 8(x – 3)

Solution

8x – 18 = 6 + 8(x – 3)

8x – 18 = 6 + 8x – 24

8x – 18 = 8x – 18

8x – 8x – 18 = 8x – 8x – 18

–18 = –18

Every real number is a solution. There are infinitely many solutions.

Using the distributive law to multiply and remove parentheses

Collecting like terms

Adding 8x

Example

Solve: 3 – 4(x + 6) = –4(x – 1) – 3.

Solution

3 – 4(x + 6) = –4(x – 1) – 3

3 – 4x – 24 = –4x + 4 – 3

–4x – 21 = –4x + 1

–4x + 4x – 21 = –4x + 4x + 1

–21 = 1 FALSE

There are no solutions.

Ch.2. Solving Equations and Inequalities/2.4 Formulas.ppt

Chapter 2

Solving Equations and Inequalities

FORMULAS

a. Evaluate a formula.

b. Solve a formula for a specified variable.

2.4

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.

Example

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.

Example

Solve for r: d = rt.

Solution

d = rt We want this letter alone.

Dividing by t

Simplifying

Example

Solve for x: y = x + b.

Solution

y = x + b We want this letter alone.

y – b = x + b – b Subtracting b

y – b = x Simplifying

Example

Solve for y: wy + x = z

Solution

wy + x = z We want this letter alone.

wy + x – x = z – x Subtracting x

wy = z – x Simplifying

wy = z – x Dividing by w

w w

Simplifying

A Formula-Solving Procedure

1. Multiply on both sides to clear fractions or decimals, if that is needed.

2. Collect like terms on each side, if necessary.

3. Get all terms with the letter to be solved for on one side of the equation and all other terms on the other side.

4. Collect like terms again, if necessary.

5. Solve for the letter in question.

Example

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

Solution

C = d

=×

Ch.2. Solving Equations and Inequalities/2.5.Applications of Percent.ppt

Chapter 2

Solving Equations and Inequalities

APPLICATIONS OF PERCENT

a. Solve applied problems involving percent.

2.5

To solve a problem involving percents, it is helpful to translate first to an equation.

For example, “23% of 5 is what?”

23% of 5 is what?

    

0.23  5 = a

Note how the key words are translated.

Key Word in Percent Translations

“Of” translates to “•” or “”.

“What number” or “what percent” translates to any letter.

“Is” translates to “=”.

% translates to

Example

Translate: What is 19% of 82?

Solution

What is 19% of 82?

    

a = 0.19  82

Example

Translate: 7 is 10% of what?

Solution

7 is 10% of what?

    

7 = 0.10  b

Example

Translate: 18 is what percent of 38?

Solution

18 is what percent of 38

    

18 = p  38

3 Types of Percent Problems

1. Finding the amount (the result of taking the percent)

Example: What is 25% of 60?

Translation: a = 0.25  60

2. Finding the base (the number you are taking the
percent of)

Example: 15 is 25% of what number?

Translation: 15 = 0.25  b

3. Finding the percent number (the percent itself)

Example: 15 is what percent of 60?

Translation: 15 = p  60

Example

What is 8% of 34?

Solution

Translate: a = 0.08  34

Solve: The variable is by itself. To solve the equation, we just multiply.

a = 0.08(34)

a = 2.72

Thus, 2.72 is 8% of 34. The answer is 2.72.

Example

15 is 16% of what?

Solution

Translate: 15 is 16% of what?

15 = 0.16  b

Solve: To solve we divide both sides

of the equation by 0.16:

Example

27 is what percent of 36?

Solution

Translate: 27 is what percent of 36?

27 = p  36

Solve: To solve we divide both sides by 36 and convert the answer to percent notation:

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

A family pays a monthly electric bill of $172.00. With careful monitoring they can reduce their bill to $163.40. What is the percent of decrease?

Solution We find the amount of decrease.

We rephrase and translate.

8.60 is what percent of 172.00?

8.60 = p  172.00

continued

Solve. We divide both sides by 172:

The percent of decrease of the electric bill is 5%.

"", or "0.01".

100

´´

0.08

2.7

150.16

0.160.16

93.75

0.1615.00.00

144

120

112

93.75

2736

=×

272

366

0.75

75%

.75

3627.00

252

180

172.00

163.40

Original

8.60

bill

New bill

Dec

eas

172172

.60172

0.05

Ch.2. Solving Equations and Inequalities/2.6.Applications and Problem Solving.ppt

Chapter 2

Solving Equations and Inequalities

APPLICATIONS AND PROBLEM SOLVING

a. Solve applied problems by translating to equations.

2.6

Five Steps for Problem Solving in Algebra

1. Familiarize yourself with the problem situation.

2. Translate the problem to an equation.

3. Solve the equation.

4. Check the answer in the original problem.

5. State the answer to the problem clearly.

To Familiarize Yourself with a Problem

 If a problem is given in words, read it carefully. Reread the problem, perhaps aloud. Try to verbalize the problem as if you were explaining it to someone else.

 Choose a variable (or variables) to represent the unknown and clearly state what the variable represents. Be descriptive! Make sure you understand all important words.

For example, let L = length in centimeters, d = distance in miles, and so on.

 Make a drawing and label it with known information, using specific units if given. Also, indicate the unknown information.

To Become Familiar with a Problem (continued)

 Find more information. Look up formulas or definitions with which you are not familiar. Consult the Internet or a reference librarian.

 Create a table that lists all the information you have available. Look for patterns that may help in the translation to an equation.

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

Example

A 480-in. piece of pipe is cut into two pieces. One piece is three times the length of the other. Find the length of each piece of pipe.

Solution

1. Familiarize. Make a drawing. Noting the lengths.

480 in

2. Translate. From the statement of the problem.

One piece is three times the length of the other the total is 480 inches.

x + 3x = 480

3. Solve.

x + 3x = 480

4x = 480

4 4

x = 120 inches

4. Check. Do we have an answer to the problem?

No, we need the lengths of both pieces of pipe.

If x = 120 the length of one piece

3x = the length of the other piece. 3(120) = 360 inches

Since 120 + 360 = 480 our answer checks.

5. State. One section of pipe is 120 inches and the other section is 360 inches.

Example

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. Familarize. 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.

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.

Consecutive integers:

16, 17, 18, 19, 20; and so on

x, x + 1, x + 2; and so on.

Consecutive even integers:

16, 18, 20, 22, and so on

x, x + 2, x + 4, and so on.

Consecutive odd integers:

21, 23, 25, 27, 29, and so on

x, x + 2, x + 4, and so on.

Example

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?

Solution

1. Familarize. The apartment numbers are
consecutive integers.

Let x = Wanda’s apartment

Let x + 1 = neighbor’s apartment

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.

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.

Example

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.

Solution

1. 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.

x + 10

2. Translate (continued).

3. Carry out.

The measures for the angles appear to be:

first angle: x = 34

second angle: 3x = 3(34) = 102;

third angle: x + 10 = 34 + 10 = 44

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.

Example

The Hicksons are planning to sell their home. If they want to be left with $142,00 after paying 6% of the selling price to a realtor as a commission, for how much must they sell the house?

Familiarize. Suppose the Hicksons sell the house for $150,000. A 6% commission can be determined by finding 6% of 150,000.

6% of 150,000 = 0.06(150,000) = 9000.

Subtracting this commission from $150,000 would leave the Hicksons with 150,000 – 9,000 = 141,000.

Translate. Reword the problem and translate.

Selling price less Commission is Amount remaining

x  0.06x = 142,000

Solve.

x  0.06x = 142,000

1x – 0.06x = 142,000

0.94x = 142,000

0.94 0.94

x = 151,063.83

Check.

Find 6% of 151,063.83:

6% of 151,063.83 = (0.06)(151,063.83) = 9063.83

Subtract the commission to find the remaining amount.

151,063.83 – 9063.83 = 142,000.

State.

To be left with $142,000 the Hicksons must sell the house for $151,063.83.

Shipping plus photo cost is

$22

¯¯¯¯¯

$3.29 0.12()

3.290.1222

0.1218.71

155.9155

First integer plus second integ

er is 723

¯¯¯¯¯

(

)

723

++=

Measure of measure of

measure of

first angle + second angle

+ third angle is 180

¯¯¯¯

¯¯¯

(

)

10 180

xxx

+++=

(

)

310180

xxx

+++=

510180

5170

Ch.2. Solving Equations and Inequalities/2.7.Solving Inequalities.ppt

Chapter 2

Solving Equations and Inequalities

SOLVING INEQUALITIES

a. Determine whether a given number is a solution of an inequality.

b. Graph an inequality on a number line.

c. Solve inequalities using the addition principle.

d. Solve inequalities using the multiplication principle.

e. Solve inequalities using the addition and multiplication principles together.

2.7

Solutions of Inequalities

An 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).

Solution of an Inequality

A replacement that makes an inequality true is called a solution. The set of all solutions is called the solution set. When we have found the set of all solutions of an inequality, we say that we have solved the inequality.

Example

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.

Graphs of Inequalities

Because solutions of inequalities like x < 4 are too numerous to list, it is helpful to make a drawing that represents all the solutions.

The graph of an inequality is a drawing that represents its solutions. Graphs of inequalities in one variable can be drawn on a number line by shading all the points that are solutions.

Example

Graph each inequality:

a) x < 3, b) y  4; c) 3 < x  5

Solution

a) The solutions of x < 3 are those numbers less than 3.

Shade all points to the left of 3.

The parenthesis at 3 and the shading to the left indicate that 3 is not part of the graph, but numbers like 1 and 2 are.

)

Solution

b) The solutions of y  4 are shown on the number line by shading the point for 4 and all points to the right of 4. The bracket at 4 indicates that 4 is part of the graph.

c) The inequality 3 < x  5 is read “3 is less than x and x is less than or equal to 5.”

[

(

]

The Addition Principle for Inequalities

For 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;

Example

Solve x + 6 > 2 and then graph the solution.

Solution

x + 6 > 2

x + 6  6 > 2  6

x > 4

Any number greater than 4 makes the statement true.

(

Example

Solve 4x  1  3x  4 and then graph the solution.

Solution

4x  1  3x  4

4x  1 + 1  3x  4 + 1

4x  3x  3

4x  3x  3x  3x  3

x  3

The solution set is {x|x  3}.

Adding 1 to both sides

Subtracting 3x from both sides

Simplifying

]

The Multiplication Principle for Inequalities

For 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 .

Example

Solve and graph each inequality:

a) b) 4y < 20

Solution

Dividing by 7

Simplifying

]

The symbol stays the same.

Solution

b) 4y < 20

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

The symbol must be reversed!

Dividing both sides by 4

(

Example

3x  3 > x + 7

3x  3 + 3 > x + 7 + 3

3x > x + 10

3x  x > x  x + 10

2x > 10

x > 5

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

Solve. 3x  3 > x + 7

Adding 3 to both sides

Simplifying

Subtracting x from both sides

Simplifying

Dividing both sides by 2

Simplifying

(

Example

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

Dividing both sides by 6

Adding 3 to both sides

]

Example

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

Remember to reverse the symbol!

735

420

2216

320

Ch.2. Solving Equations and Inequalities/2.8.Applications and Problem Solving with Inequalities.ppt

Chapter 2

Solving Equations and Inequalities

APPLICATIONS AND PROBLEM SOLVING WITH INEQUALITIES

a. Translate number sentences to inequalities.

b. Solve applied problems using inequalities.

2.8

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.)

Example

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. Familarize. Suppose the copier was worked on for
4 hours. The cost would be $65 + 4($45), or $245. A bill of
$150 shows that the copier was worked on for less
than 4 hours. Let h = the number of hours.

2. Translate.

Rewording:

Translating:

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.

Average or mean

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

Example

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?

Solution

1. Familarize. 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.

2. Translate.

Rewording:

Translating:

3. Carry out.

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.

Initial fee plus hours is less tha

n 150

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65 45

150

68878

84.25

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868878

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868878

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Ch.2. Solving Equations and Inequalities/kin12_ppt_02.zip

kin12_ppt_02/kin12_02_01.ppt

Chapter 2

Solving Equations and Inequalities

SOLVING EQUATIONS: THE ADDITION PRINCIPLE

a. Determine whether a given number is a solution of a given equation.

b. Solve equations using the addition principle.

2.1

Equation

An equation is a number sentence that says that the expressions on either side of the equals sign, =, represent the same number.

Example

Determine whether the equation is true, false, or neither.

a. 4 + 6 = 10

b. 8 – 3 = 4

c. x + 9 = 21

True

False

Neither, we do not know what number x represents

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.

Example

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

Solution x + 12 = 21 Writing the equation

8 + 12 | 21 Substituting 8 for x

20  21 False

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

Equivalent Equations

Equations with the same solutions are called equivalent equations.

The Addition Principle For Equations

For any real numbers a, b, and c,

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

Example

Solve: x + 6 = –9.

Solution

x + 6 = –9

x + 6 – 6 = –9 – 6

x + 0 = –15

x = –15

Using the addition principle: adding –6 to both sides or subtracting 6 on both sides

Simplifying

Identity property of 0

Check

x + 6 = –9

–15 + 6 = | –9

–9 True

Example

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.

Solve: 8.3 = y  17.9

Example

Solve:

510

-+=

510

-+=

510

=×+

1010

kin12_ppt_02/kin12_02_02.ppt

Chapter 2

Solving Equations and Inequalities

SOLVING EQUATIONS: THE MULTIPLICATION PRINCIPLE

a. Solve equations using the multiplication
principle.

2.2

The Multiplication Principle For Equations

For any real numbers a, b, and c with c  0,

a = b is equivalent to a • c = b • c.

Example

Solve 6x = 96.

Solution 6x = 96

6 6

x = 16

Check: 6x = 96

6(16) = 96

96 TRUE

The solution is 16.

Dividing by 6 on both sides

Example

Solution 7x = 84

Check: 7x = 84

7(12) | 84

84 | True

The solution is 12.

Solve: 7x = 84

Dividing both sides by 7.

Example

Solve x = 6.

Solution x = 6

(1)(x) = (1)6

1(1)x = 6

x = 6

The solution is 6.

Multiplying by 1 on both sides

Example

Solve:

Solution

Multiplying by the reciprocal of ¾ on both sides.

Example

Solve: 3.2y = 9600

3.2y = 9600

112

×=-

××

120

3.23.2

.29600

3000

kin12_ppt_02/kin12_02_03.ppt

Chapter 2

Solving Equations and Inequalities

USING THE PRINCIPLES TOGETHER

a. Solve equations using both the addition and multiplication principles.

b. Solve equations in which like terms need to be collected.

c. Solve equations by first removing parentheses and collecting like terms; solve equations with no solutions and equations with an infinite number of solutions.

2.3

Example

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 TRUE The solution is 3.

Solve: 9 + 8x = 33

Subtracting 9 from both sides

Dividing both sides by 8

Example

Solve: 35 – x = 27.

Solution

35 – x = 27

35 – 35 – x = 27 – 35 Adding –35 on both sides

–x = 8

–1(–x) = –1(–8) Multiplying by –1 on both sides

x = 8

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.

Example

Solution 5x + 4x = 36

9x = 36

x = 4

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

20 + 16 | 36

The solution is 4.

Solve. 5x + 4x = 36

Collecting like terms

Dividing by 9 on both sides

Example

Solve: 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

Collecting like terms

Subtracting 7 from both sides

Simplifying

Subtracting 3x from both sides

Dividing both sides by 5

Clearing Fractions and Decimals

In general, equations are easier to solve if they do not contain fractions or decimals.

The easiest way to clear an equation of fractions is to multiply every term on both sides by the least common multiple of all the denominators.

To clear an equation of decimals, we count the greatest number of decimals places in any one number and multiply on both sides by that multiple of 10.

Example

Solve 18.4 – 6.2y = 7.24

Solution

18.4 – 6.2y = 7.24

100(18.4 – 6.2y) = 100(7.24) Multiplying by 100

(100)(18.4) – 100(6.2y) = 100(7.24) Using the distributive law

1840 – 620y = 724 Simplifying

1840 – 1840 – 620y = 724 – 1840 Subtracting 1840

–620y = – 1116 Collecting like terms

–620 – 620 Dividing by – 620

y = 1.8

Example

Solve: 9x = 3(15 – 2x)

Solution

9x = 3(15 – 2x)

9x = 45 – 6x Using the distributive law

9x + 6x = 45 – 6x + 6x Subtracting 6x to get all x-terms on one side

15x = 45 Collecting like terms

15 15 Dividing by 15

x = 3

An Equation-Solving Procedure

1. Multiply on both sides to clear the equation of fractions or decimals. (This is optional, but can ease computations.)

2. If parentheses occur, multiply to remove them using the distributive laws.

3. Collect like terms on each side, if necessary.

4. Get all terms with variables on one side and all numbers (constant terms) on the other side, using the addition principle.

5. Collect like terms again, if necessary.

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

7. Check all possible solutions in the original equation.

Example

Solve: 3 – 8(x + 6) = 4(x – 1) – 5.

Solution

3 – 8(x + 6) = 4(x – 1) – 5

3 – 8x – 48 = 4x – 4 – 5

–45 – 8x = 4x – 9

–8x – 45 + 45 = 4x – 9 + 45

–8x = 4x + 36

–8x – 4x = 4x + 36 – 4x

–12x = 36

–12 –12

x = – 3

Using the distributive law to multiply and remove parentheses

Collecting like terms

Adding 45

Subtracting 4x

Collecting like terms

Dividing by 12

Example

Solve: 8x – 18 = 6 + 8(x – 3)

Solution

8x – 18 = 6 + 8(x – 3)

8x – 18 = 6 + 8x – 24

8x – 18 = 8x – 18

8x – 8x – 18 = 8x – 8x – 18

–18 = –18

Every real number is a solution. There are infinitely many solutions.

Using the distributive law to multiply and remove parentheses

Collecting like terms

Adding 8x

Example

Solve: 3 – 4(x + 6) = –4(x – 1) – 3.

Solution

3 – 4(x + 6) = –4(x – 1) – 3

3 – 4x – 24 = –4x + 4 – 3

–4x – 21 = –4x + 1

–4x + 4x – 21 = –4x + 4x + 1

–21 = 1 FALSE

There are no solutions.

kin12_ppt_02/kin12_02_04.ppt

Chapter 2

Solving Equations and Inequalities

FORMULAS

a. Evaluate a formula.

b. Solve a formula for a specified variable.

2.4

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.

Example

Solution We substitute 5 for t and calculate M.

The storm is 1 mile away.

Example

Solve for r: d = rt.

Solution

d = rt We want this letter alone.

Dividing by t

Simplifying

Example

Solve for x: y = x + b.

Solution

y = x + b We want this letter alone.

y – b = x + b – b Subtracting b

y – b = x Simplifying

Example

Solve for y: wy + x = z

Solution

wy + x = z We want this letter alone.

wy + x – x = z – x Subtracting x

wy = z – x Simplifying

wy = z – x Dividing by w

w w

Simplifying

A Formula-Solving Procedure

1. Multiply on both sides to clear fractions or decimals, if that is needed.

2. Collect like terms on each side, if necessary.

3. Get all terms with the letter to be solved for on one side of the equation and all other terms on the other side.

4. Collect like terms again, if necessary.

5. Solve for the letter in question.

Example

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

Solution

C = d

=×

kin12_ppt_02/kin12_02_05.ppt

Chapter 2

Solving Equations and Inequalities

APPLICATIONS OF PERCENT

a. Solve applied problems involving percent.

2.5

To solve a problem involving percents, it is helpful to translate first to an equation.

For example, “23% of 5 is what?”

23% of 5 is what?

    

0.23  5 = a

Note how the key words are translated.

Key Word in Percent Translations

“Of” translates to “•” or “”.

“What number” or “what percent” translates to any letter.

“Is” translates to “=”.

% translates to

Example

Translate: What is 19% of 82?

Solution

What is 19% of 82?

    

a = 0.19  82

Example

Translate: 7 is 10% of what?

Solution

7 is 10% of what?

    

7 = 0.10  b

Example

Translate: 18 is what percent of 38?

Solution

18 is what percent of 38

    

18 = p  38

3 Types of Percent Problems

1. Finding the amount (the result of taking the percent)

Example: What is 25% of 60?

Translation: a = 0.25  60

2. Finding the base (the number you are taking the
percent of)

Example: 15 is 25% of what number?

Translation: 15 = 0.25  b

3. Finding the percent number (the percent itself)

Example: 15 is what percent of 60?

Translation: 15 = p  60

Example

What is 8% of 34?

Solution

Translate: a = 0.08  34

Solve: The variable is by itself. To solve the equation, we just multiply.

a = 0.08(34)

a = 2.72

Thus, 2.72 is 8% of 34. The answer is 2.72.

Example

15 is 16% of what?

Solution

Translate: 15 is 16% of what?

15 = 0.16  b

Solve: To solve we divide both sides

of the equation by 0.16:

Example

27 is what percent of 36?

Solution

Translate: 27 is what percent of 36?

27 = p  36

Solve: To solve we divide both sides by 36 and convert the answer to percent notation:

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

A family pays a monthly electric bill of $172.00. With careful monitoring they can reduce their bill to $163.40. What is the percent of decrease?

Solution We find the amount of decrease.

We rephrase and translate.

8.60 is what percent of 172.00?

8.60 = p  172.00

continued

Solve. We divide both sides by 172:

The percent of decrease of the electric bill is 5%.

"", or "0.01".

100

´´

0.08

2.7

150.16

0.160.16

93.75

0.1615.00.00

144

120

112

93.75

2736

=×

272

366

0.75

75%

.75

3627.00

252

180

172.00

163.40

Original

8.60

bill

New bill

Dec

eas

172172

.60172

0.05

kin12_ppt_02/kin12_02_06.ppt

Chapter 2

Solving Equations and Inequalities

APPLICATIONS AND PROBLEM SOLVING

a. Solve applied problems by translating to equations.

2.6

Five Steps for Problem Solving in Algebra

1. Familiarize yourself with the problem situation.

2. Translate the problem to an equation.

3. Solve the equation.

4. Check the answer in the original problem.

5. State the answer to the problem clearly.

To Familiarize Yourself with a Problem

 If a problem is given in words, read it carefully. Reread the problem, perhaps aloud. Try to verbalize the problem as if you were explaining it to someone else.

 Choose a variable (or variables) to represent the unknown and clearly state what the variable represents. Be descriptive! Make sure you understand all important words.

For example, let L = length in centimeters, d = distance in miles, and so on.

 Make a drawing and label it with known information, using specific units if given. Also, indicate the unknown information.

To Become Familiar with a Problem (continued)

 Find more information. Look up formulas or definitions with which you are not familiar. Consult the Internet or a reference librarian.

 Create a table that lists all the information you have available. Look for patterns that may help in the translation to an equation.

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

Example

A 480-in. piece of pipe is cut into two pieces. One piece is three times the length of the other. Find the length of each piece of pipe.

Solution

1. Familiarize. Make a drawing. Noting the lengths.

480 in

2. Translate. From the statement of the problem.

One piece is three times the length of the other the total is 480 inches.

x + 3x = 480

3. Solve.

x + 3x = 480

4x = 480

4 4

x = 120 inches

4. Check. Do we have an answer to the problem?

No, we need the lengths of both pieces of pipe.

If x = 120 the length of one piece

3x = the length of the other piece. 3(120) = 360 inches

Since 120 + 360 = 480 our answer checks.

5. State. One section of pipe is 120 inches and the other section is 360 inches.

Example

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. Familarize. 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.

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.

Consecutive integers:

16, 17, 18, 19, 20; and so on

x, x + 1, x + 2; and so on.

Consecutive even integers:

16, 18, 20, 22, and so on

x, x + 2, x + 4, and so on.

Consecutive odd integers:

21, 23, 25, 27, 29, and so on

x, x + 2, x + 4, and so on.

Example

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?

Solution

1. Familarize. The apartment numbers are
consecutive integers.

Let x = Wanda’s apartment

Let x + 1 = neighbor’s apartment

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.

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.

Example

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.

Solution

1. 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.

x + 10

2. Translate (continued).

3. Carry out.

The measures for the angles appear to be:

first angle: x = 34

second angle: 3x = 3(34) = 102;

third angle: x + 10 = 34 + 10 = 44

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

Example

The Hicksons are planning to sell their home. If they want to be left with $142,00 after paying 6% of the selling price to a realtor as a commission, for how much must they sell the house?

Familiarize. Suppose the Hicksons sell the house for $150,000. A 6% commission can be determined by finding 6% of 150,000.

6% of 150,000 = 0.06(150,000) = 9000.

Subtracting this commission from $150,000 would leave the Hicksons with 150,000 – 9,000 = 141,000.

Translate. Reword the problem and translate.

Selling price less Commission is Amount remaining

x  0.06x = 142,000

Solve.

x  0.06x = 142,000

1x – 0.06x = 142,000

0.94x = 142,000

0.94 0.94

x = 151,063.83

Check.

Find 6% of 151,063.83:

6% of 151,063.83 = (0.06)(151,063.83) = 9063.83

Subtract the commission to find the remaining amount.

151,063.83 – 9063.83 = 142,000.

State.

To be left with $142,000 the Hicksons must sell the house for $151,063.83.

Shipping plus photo cost is

$22

¯¯¯¯¯

$3.29 0.12()

3.290.1222

0.1218.71

155.9155

First integer plus second integ

er is 723

¯¯¯¯¯

(

)

723

++=

Measure of measure of

measure of

first angle + second angle

+ third angle is 180

¯¯¯¯

¯¯¯

(

)

10 180

xxx

+++=

(

)

310180

xxx

+++=

510180

5170

kin12_ppt_02/kin12_02_07.ppt

Chapter 2

Solving Equations and Inequalities

SOLVING INEQUALITIES

a. Determine whether a given number is a solution of an inequality.

b. Graph an inequality on a number line.

c. Solve inequalities using the addition principle.

d. Solve inequalities using the multiplication principle.

e. Solve inequalities using the addition and multiplication principles together.

2.7

Solutions of Inequalities

An 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).

Solution of an Inequality

Example

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.

Graphs of Inequalities

Because solutions of inequalities like x < 4 are too numerous to list, it is helpful to make a drawing that represents all the solutions.

The graph of an inequality is a drawing that represents its solutions. Graphs of inequalities in one variable can be drawn on a number line by shading all the points that are solutions.

Example

Graph each inequality:

a) x < 3, b) y  4; c) 3 < x  5

Solution

a) The solutions of x < 3 are those numbers less than 3.

Shade all points to the left of 3.

The parenthesis at 3 and the shading to the left indicate that 3 is not part of the graph, but numbers like 1 and 2 are.

)

Solution

b) The solutions of y  4 are shown on the number line by shading the point for 4 and all points to the right of 4. The bracket at 4 indicates that 4 is part of the graph.

c) The inequality 3 < x  5 is read “3 is less than x and x is less than or equal to 5.”

[

(

]

The Addition Principle for Inequalities

For 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;

Example

Solve x + 6 > 2 and then graph the solution.

Solution

x + 6 > 2

x + 6  6 > 2  6

x > 4

Any number greater than 4 makes the statement true.

(

Example

Solve 4x  1  3x  4 and then graph the solution.

Solution

4x  1  3x  4

4x  1 + 1  3x  4 + 1

4x  3x  3

4x  3x  3x  3x  3

x  3

The solution set is {x|x  3}.

Adding 1 to both sides

Subtracting 3x from both sides

Simplifying

]

The Multiplication Principle for Inequalities

For 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 .

Example

Solve and graph each inequality:

a) b) 4y < 20

Solution

Dividing by 7

Simplifying

]

The symbol stays the same.

Solution

b) 4y < 20

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

The symbol must be reversed!

Dividing both sides by 4

(

Example

3x  3 > x + 7

3x  3 + 3 > x + 7 + 3

3x > x + 10

3x  x > x  x + 10

2x > 10

x > 5

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

Solve. 3x  3 > x + 7

Adding 3 to both sides

Simplifying

Subtracting x from both sides

Simplifying

Dividing both sides by 2

Simplifying

(

Example

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

Dividing both sides by 6

Adding 3 to both sides

]

Example

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

Remember to reverse the symbol!

735

420

2216

320

kin12_ppt_02/kin12_02_08.ppt

Chapter 2

Solving Equations and Inequalities

APPLICATIONS AND PROBLEM SOLVING WITH INEQUALITIES

a. Translate number sentences to inequalities.

b. Solve applied problems using inequalities.

2.8

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.)

Example

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

2. Translate.

Rewording:

Translating:

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.

Average or mean

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

Example

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?

Solution

1. Familarize. 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.

2. Translate.

Rewording:

Translating:

3. Carry out.

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.

Initial fee plus hours is less tha

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