a. Determine whether a given number is a solution to an equation

b. Solve equations using the addition and multiplication principles

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

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

a. 4 + 6 = 10 - true

b. 8 – 3 = 4 - false

c. x + 9 = 21 - Neither, we do not know what number x represents.

The replacements for the variable that make an equation true are called the solutions of the equation. The set of all solutions is called the solution set of the equation. When we find all the solutions, we say we have solved the equation.

Example:

Is 8 a solution to the equation

x + 5 = 13

?

Apply 8 to the unknown variable x, we have 8 + 5 = 13, which is a true statement. Therefore 8 is a solution to this equation.

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 ¹ 21 False

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

Equations with the same solutions are called equivalent equations.

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For any real numbers a, b, and c,

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

For example:

Solve: x + 6 = –9.

x + 6 = –9

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

x + 6 – 6 = –9 – 6

x + 0 = –15

x = –15

Example: Solve: -8.3 = y - 17.9

Solution -8.3 = y - 17.9

-8.3 + 17.9 = y - 17.9 + 17.9

9.6 = y

For any real numbers a, b, and c, c ¹ 0,

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

Example:

Solve:

¾ x = 15

¾ x = 15

1x = 20

x = 20

Example:

Solve 6x = 96.

Solution

6x = 96

6 6

x = 16

Example:

Solve -7x = 84.

Solution

-7x = 84

-7 -7

x = -12

Example:

Solve: 9 + 8x = 33

Solution 9 + 8x = 33

9 + 8x - 9 = 33 - 9

9 + (- 9) + 8x = 24

8x = 24

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

-5x = 15

-5 -5

x = -3

Example:

Solve 18.4 – 6.2y = 7.24

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: 8x – 18 = 6 + 8(x – 3)

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.

Example:

Solve: 3 – 4(x + 6) = –4(x – 1) – 3.

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.

Example:

Solve: 3 – 8(x + 6) = 4(x – 1) – 5.

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

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1. Clear the equation of fractions or decimals if that is needed.

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

3. Collect like terms on each side of the equation, if necessary.

4. Use the addition principle to get all like terms with letters on one side and all other terms on the other side.

5. Collect like terms on each side again, if necessary.

6. Use the multiplication principle to solve for the variable.

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Exercise Set 1.1 (Page 84) (13 - 73) odds only

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1.3 Application and Problem Solving (Text Book Page 97)

a. Solve applied problems by translating to equations.

b. Solve basic motion problems.

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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.

¨If a problem is given in words, read it carefully.

¨Reread the problem, perhaps aloud. Try to verbalize the problem to yourself. Make notes as you read.

¨List the information given and the question to be answered. Choose a variable (or variables) to represent the unknown(s) and clearly state what the variable represents. Be descriptive!

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.

¨ Find further information if necessary. Look up a formula at the back of this book or in a reference book. Talk to a reference librarian or look up the topic using an Internet search engine.

¨ Make a table that lists all the information you have collected. Look for patterns that may help in the translation to an equation.

¨ Think of a possible answer and check the guess. Observe 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.

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

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:

The faculty discount at a bookstore is 15%. If a sweatshirt after the discount was $32.30. What was the original price of the sweatshirt?

Solution

1. Familiarize. Note that the discount is calculated from and then subtracted from the original price.

Guess $40.00. The discount is (0.15)(40) = $6. Subtracting: $40 – (0.15)($40) = $36.

Let S = the sweatshirt’s original price.

2. Translate.

The sweatshirt price minus the discount is the sale price.

S – (0.15)S = 32.30

3. Carry out. Solve the equation.

S – (0.15)S = 32.30

0.85S = 32.30

S = 38

4. Check. Note that 15% of $38 would be

(0.15)($38) = $5.70. When this is subtracted from $38 we have

$38 - $5.70 = $32.30

Thus $38 checks with the original problem.

5. State. The original price of the sweatshirt was $38.

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.

Measure of first angle + measure of second angle + measure of third angle = 180 degrees

x + 3x + (x + 10) = 180

3. Carry out.

x + 3x + (x + 10) = 180

5x + 10 = 180

5x = 170

x = 34

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 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?

1. Familarize. The apartment numbers are consecutive integers.

Let x = Wanda’s apartment

Let x + 1 = neighbor’s apartment

2. Translate.

Rewording: first integer plus second integer is 723

Translating: x + (x + 1) = 723

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:

The price of a sweater is $21.85 after a 5% discount. What is its original price?

$21.85 is 95% of the original price, therefore

x C 0.95 = 21.85

x = 21.85/0.95

x = 23

The original price of the sweater before discount is $23

Example:

One side of a 1,800 sq ft, rectangular yard is twice as long as the other. What is its dimension?

2x C x = 1800

2x² = 1800

x² = 900

x = 30

2x = 60

Therefore the yard is 30 ft by 60 ft.

Example:

An airplane has been instructed to climb from its present altitude of 12,000 feet to a cruising altitude of 31,200 feet. The plane ascends at a rate of 3200 feet per minute. How long will it take the plane to reach the cruising altitude?

1. Familiarize. Read the problem carefully and try and organize the information in a table.

2. Translate. To translate, we use the motion formula d = rt and substitute 19,200 for d and 3200 ft/min for r:

d = rt

19200 = 3200t

3. Solve. We solve the equation.

19200 = 3200t

19200 = 3200t

3200 3200

6 = t

4. Check. Ascending at a rate of 3200 feet per minute in a time of 6 minutes the plane would travel 19,200 feet. The answer checks.

5. State. It will take the plane 6 minutes to reach the cruising altitude.

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Exercise Set 1.3 (Page 105) (1 - 31) odds only

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

b. Write interval notation for the solution set of an inequality

c. Solving inequalities and applied problems

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An inequality is a sentence containing

<, >, #, $, …

Examples of Inequalities

3x + 2 > 7, c # 7, and 4x – 6 … 3

Any replacement or value for the 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 have been found, we say that we have solved the inequality.

A solution to an inequality is one where when it replaces the unknown variable(s), makes the inequality a true statement.

Example:

Is 6 a solution to the inequality

x – 5 # 3

?

Replacing unknown variable x with 6, we have

6 – 5 = 1, and

1 # 3, which is a true statement

Therefore 6 is a solution to x – 5 # 3

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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.

Example:

Graph x < 2 on a number line.

Note that in set-builder notation the solution is

Another way to write solutions of an inequality in one variable is to use 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,

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

Thus,

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

We use the symbols 4 and -4 to represent positive and negative infinity, respectively. Thus the notation (a, 4 ) represents the set of all real numbers greater than a, and ( -4 , a) represents the set of all numbers less than a.

The notations (–4 , a] and [a, 4 ) are used when we want to include the endpoint a.

In summary:

[a,b] denotes all values between a and b, including points a and b.

[a,b) denotes all values between a and b, including a but excluding b.

(a,b] denotes all values between a and b, excluding a but including b.

(a,b) denotes all values between a and b, but excluding a and b.

Example:

Write interval notation for the given set.

a. {x|3 < x < 8}

b. {x|x ³ 4}

a. {x|3 < x < 8} = (3, 8)

b. {x|x ³ 4} = [4, ¥)

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;

Similar statements hold for £ and ³.

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}.

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 ³.

Example:

Solve. 3x - 3 > x + 7

Solution

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}.

Example:

Solve: 5(x - 3) - 7x ³ 4(x - 3) + 9

Solution

5(x - 3) - 7x ³ 4(x - 3) + 9

5x - 15 - 7x ³ 4x - 12 + 9

-2x - 15 ³ 4x - 3

-2x - 15 + 3 ³ 4x - 3 + 3

-2x - 12 ³ 4x

-2x + 2x - 12 ³ 4x + 2x

-12 ³ 6x

-2 ³ x

The solution set is {x|x £ - 2}.

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The quantity x is at least some amount q: x ³ 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.)

Save a lot offers two different checking-account plans. Plan A charges $0.25 per check where as Plan B costs $3 per month plus $0.05 per check. For what numbers of checks per month will Plan A cost less?

1. Familiarize. Listing the given information in a table will be helpful.

Suppose you wrote 20 checks per month, which is the better plan?

Plan A: $0.25(20) = $5.00
Plan B: $3 + $0.05(20) = $4

2. Translate. We want to find the number of checks
for which Plan A will cost less.

Plan A £ Plan B

$0.25x £ $3 + 0.05x

3. Solve.

0.25x £ 3 + 0.05x

0.20x £ 3

x £ 15

4. Check.

Plan A: $0.25(15) = $3.75

Plan B: $3 + 0.05(15) = $3.75

5. State. Writing less than 15 checks per month will
make Plan A the better deal.

Example:

Alice averaged 87% in her first three quizzes in her Math class. How many points out of 100 must she score in her last quiz to get an A for the class, which is 90% - 100% of the average quiz scores.

(3 C 87 + x)/4 $ 90

(261 + x)/4 $ 90

261 + x $ 360

x $ 99

Alice must score a 99 or above to get an A for the class.

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Exercise Set 1.4 (Page 123) (41 - 85) odds only

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a. Find the intersection and union of two sets

b. Solve applied problems

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Two inequalities joined by the word "and" or the word "or" are called compound inequalities.

Example:

3x – 9 # 0 and x $ -5

7x + 1 # 8 or x > 8

The intersection of two sets A and B is the set of all members that are common to A and B. We denote the intersection of sets A and B as A 1 B.

The intersection of {2,3,4,6,8} and {1,4,6,7,9}

{2,3,4,6,8} 1 {1,4,6,7,9} = {4,6}

Conjunction of inequalities:

When two or more sentences are joined by the word and to make a compound sentence, the new sentence is called a conjunction of the sentences.

-1 < x and x < 3

A number is a solution of a conjunction if it is a solution of both inequalities. For example, 0 is a solution because it is a solution of –1 < x as well as x < 3.

Note that the solution set of a conjunction is the intersection of the solution sets of the individual sentences.

Note that for a < b, a < x and x < b can be abbreviated a < x < b;

and, equivalently,

b > x and x > a can be abbreviated b > x > a.

So 3 < 2x +1 < 7 can be solved as 3 < 2x +1 and 2x + 1 < 7

The word "and" corresponds to "intersection" and to the symbol "Ç". In order for a number to be a solution of a conjunction, it must make each part of the conjunction true.

Example:

Solve and graph:

2x + 1 $ 3 and 3x < 12

2x + 1 $ -3 and 3x < 12

2x $ -4 and 3x < 12

x $ -2 and x < 4

Sometimes two sets have no elements in common. In such a case, we say that the intersection of the two sets is the empty set, denoted { } or Æ. Two sets with an empty intersection are said to be disjoint.

Example:

Solve and graph:

5 + x > 10 and x - 4 < -3

5 + x > 10 and x - 4 < -3

x > 5 and x < 1

Since no number is greater than 5 and less than 1, the solution set is the empty set q

The union of two sets A and B is the collection of elements belonging to A and/or B. We denote the union of A and B by A U B.

Example:

The union of {2,3,4,6,8} and {1,4,6,7,9}

{2,3,4,6,8} U {1,4,6,7,9} = {1,2,3,4,6,7,8,9}

When two or more sentences are joined by the word or to make a compound sentence, the new sentence is called a disjunction of the sentences.

Example: x < 2 or x > 8

Note that the solution set of a disjunction is the union of the solution sets of the individual sentences.

The word "or" corresponds to "union" and to the symbol "U" In order for a number to be in the solution set of a disjunction, it must be in at least one of the solution sets of the individual sentences.

Example:

Solve and graph:

2x + 1 $ 3 or 3x < -3

2x + 1 $ 3 or 3x < -3

2x $ 2 or 3x < -3

x $ 1 or x < -1

Example:

The equation I = 2(s + 10) can be used to convert dress sizes s in the United States to dress sizes I in Italy. Which dress sizes in the United States correspond to dress sizes between 34 and 52 in Italy?

1. Familiarize. We can substitute into the formula and find a value. For a dress size of 8 in the United States, we get the corresponding dress size in Italy:

I =2(s + 10) = 2(8 + 10) = 2(18) = 36.

This tells us the size we are looking for is larger than a size 8 in the United States.

2. Translate. We want sizes between 34 and 52, so we want to find values of s for which 34 < I < 52 or

34 < 2(s + 10) < 52.

3. Solve. 34 < 2(s + 10) < 52

2 2 2 Dividing by 2

17 < s + 10 < 26

7 < s < 16 Subtracting 10

4. Check. Substitute some values as we did in the Familiarize step.

5. State. Dress sizes between 7 and 16 in the United States correspond to dress sizes between 34 and 52 in Italy.

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Exercise Set 1.5 (Page 137) (49 - 65) odds only

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1.6 Absolute Value Equations and Inequalities (Text Book Page 141)

a. Simplifying expressions containing absolute values.

b. Solve equations and inequalities with absolute value expressions

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The absolute value of x denoted |x|, is defined as follows:

for any real numbers a and b.

(The absolute value of a product is the product of the absolute values.

b)

for any real numbers a and b ¹ 0.

(The absolute value of a quotient is the quotient of the absolute values.)

c) |-a| = |a|,

for any real number a.

(The absolute value of the opposite of a number is the same as the absolute value of the number.)

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For any real numbers a and b, the distance between them is |a – b|.

Example:

Find the distance between -12 and -56 on the number line.

| -12 – (-56)|

= 44

or

| -56 – (-12)|

= | -44|

= 44

For any positive number p and any algebraic expression X:

a. The solutions of |X| = p are those numbers that satisfy X = –p or X = p.

b. The equation |X| = 0 is equivalent to the equation X = 0.

c. The equation |X| = –p has no solution.

Example:

|x| = 2, then x = 2 or x = -2

Example:

|x + 3| = 4, then x + 3 = 4 or x + 3 = -4

x = 1 or x = -7

Sometimes an equation has two absolute-value expressions. Consider |a| = |b|. This means that a and b are the same distance from zero.

If a and b are the same distance from zero, then either they are the same number or they are opposites.

Example:

|x – 6| = |2x + 5|, we can break this down to

x – 6 = 2x + 5, or x – 6 = -(2x + 5)

x = -11 or x = 1/3

For any positive number p and any expression X:

a) The solutions of |X| = p are those numbers that satisfy X = –p or X = p.

b) The solutions of |X| < p are those numbers that satisfy –p < X < p.

c) The solutions of |X| > p are those numbers that satisfy X < –p or p < X.

Example:

Solve: |3x + 7| < 8. Then graph.

Solution:

The number 3x + 7 must be less than 8 units from 0.

The solution set is {x|–5 < x < 1/3}. The graph is as follows:

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Exercise Set 1.6 (Page 148) (41 – 69, 83 - 103) odds only

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Discussion:

The Universe (or Universal Set) is defined as the collection of all possible elements.

If A and B are subsets within this Universe, could you come up with real world examples where A 1 B = q and A U B = Universe?

Your example must satisfy both conditions. Be sure to clearly define what your A, B and Universe are.

Click on DISCUSS below and submit your responses.