1 final exam review math 002 college algebra. 2 write sets using set notation a set is a collection...

Final Exam Review

Math 002 College Algebra

Write sets using set notation

A set is a collection of objects called the elements or members of the set. Set braces { } are usually used to enclose the elements. In Algebra, the elements of a set are usually numbers.

• Example 1: 3 is an element of the set {1,2,3} Note: This is referred to as a Finite Set since we can count the elements of the set.

• Example 2: N= {1,2,3,4,…} is referred to as a Natural Numbers or Counting Numbers Set.

• Example 3: W= {0,1,2,3,4,…} is referred to as a Whole Number Set.

Two sets are equal if they contain exactly the same elements. (Order doesn’t matter)

• Example 1: {1,12} = {12,1}

• Example 2: {0,1,3} {0,2,3}

In Algebra, letters called variables are often used to represent numbers or to define sets of numbers. (x or y). The notation {x|x has property P}is an example of “Set Builder Notation” and is read as:

{x x has property P}

the set of all elements x such that x has a property P

• Example 1: {x|x is a whole number less than 6} Solution: {0,1,2,3,4,5}

In Algebra, letters called variables are often used to represent numbers or to define sets of numbers. (x or y). The notation {x|x has property P}is an example of “Set Builder Notation” and is read as:

{x x has property P}

the set of all elements x such that x has a property P

• Example 2: {x|x is a natural number greater than 12} Solution: {13,14,15,…}

Using a number line

One way to visualize a set a

numbers is to use a “Number Line”.• Example 1: The set of numbers shown above includes

positive numbers, negative numbers and 0. This set is part

of the set of “Integers” and is written:

I = {…, -2, -1, 0, 1, 2, …}

-2 -1 0 1 2 3 4 5

Using a number line

Each number on a number line is called the

coordinate of the point that it labels, while

the point is the graph of the number.• Example 1: The fractions shown above are examples of rational

numbers. A rational number is one than can be expressed as the

quotient of two integers, with the denominator not 0.

-2 -1 0 1 2 3 4 5

coordinate

Graph of -1

Finding Additive inverses

For any real number x, the number –x

is the additive inverse of x.

Example 1:Number

Inverse Additive

6 - 6- 4 4

- 8.7 8.70 0

Using the Absolute Value

Examples: Find :

To find the absolute value of a signed number:

Caution: The absolute value of a number is always positive

X > 0, then = X

X = 0, then = 0

X < 0, then = -X

Using Inequality Symbols

Equality/Inequality Symbols:

Caution: With the symbol , if either the or the = part is true, then the inequality is true. This is also the case for the symbol.

Symbol Meaning Example

= is equal to 4 = 4

is not equal to 4 5

is less than 4 5

is less than or equal -4 -3

is greater than -4 -5

is greater than or equal -8 - 10

Graphing Sets of Real Numbers

A parenthesis ( or ) is used to indicate a number is not an element of a set. A bracket [ or ] is used to indicate a number is a member of a set.

Example 1: Write in interval notation and graph: {x|x 3}Solution: Interval Notation (-,3)

Example 2: Write in interval notation and graph: {x|x 0}Solution: Interval Notation [0, )

-2 -1 0 1 2 3 4 5

Graphing Sets of Real Numbers

A parenthesis ( or ) is used to indicate a number is not an element of a set. A bracket [ or ] is used to indicate a number is a member of a set.

Example 3: Write in interval notation and graph: {x| -2 x 3}

Solution: Interval Notation [-2,3)

-2 -1 0 1 2 3 4 5

Adding Real Numbers

Example: Add (–5.6) + (-2.1) =

Example:

To add signed numbers: 1) If the numbers are alike, add their absolute values and use the common

sign.2) If the numbers are not alike, subtract the smaller absolute value from the

larger absolute value. Use the sign of the larger absolute value.

(-5.6) + (-2.1) = -( 5.6 + 2.1 ) = -7.7

2 5ADD: + (- )

2 5 4 5 5 4 1 + (- ) = + (- ) = -( - )

3 6 6 6 6 6 6

Subtracting Real Numbers

Example: Subtract (–56) - (-70) =

To subtract signed numbers:

1) Rewrite as an addition problem by adding the opposite of the number to be subtracted. Find the sum.

(-56) - (-70) = (-56) + (70 ) = +(70-56) = +14

Finding the distance between two points on a number line.

To find the distance between two points on a number line, find the absolute value of the difference between the two points.

Example 1: Find the distance between the points: –2 and 3 Solution: |(-2) – (3)| = |-2 -3| = |-5| = 5

or |(3) – (-2)| = |3 +2| = |5| =5

-2 -1 0 1 2 3 4 5

Multiplying Real Numbers To find the product of a positive and negative signed number:

Find the product of the absolute values. Make the sign negative.

Example: Multiply (–12)(5) =

Example:

To find the product of two negative signed numbers:Find the product of the absolute values. Make the sign positive.

(-12) (5) = -(12 5) = -60

7 8Multiply: (- ) (- ) =

12 257 8 7 8 14

(- ) (- ) = +( ) = +12 25 12 25 75

Dividing Real Numbers To find the quotient of a positive and negative signed number:

Find the quotient of the absolute values. Make the sign negative.

Example: Divide (–12)(5) =

Example:

To find the quotient of two negative signed numbers:Find the quotient of the absolute values. Make the sign positive.

Caution: Division by zero is “undefined”:

(-12) (5) = - 12 5 = - 2.4 2.4

7 8Divide: (- ) (- ) =

7 8 7 8 7 25 175(- ) (- ) = +( ) = +( ) = +

12 25 12 25 12 8 96

6 0 , but = 0

if , x 2x-2

is undefined

Using Exponents

If “a” is a real number and “n” is a natural number, then an = a•a•a•••a•a (n factors of a).

where n is the exponent, a is the base, and an is an exponential expression. Exponents are also called powers.

To find the value of a whole number exponent: 100 = 1, 20 = 1, 80 = 1, #0 = 1101 = 10, 21 = 2, 81 = 8, #1 = #102 = 10 x 10 = 100, 22 = 2 x 2 = 4, 82 = 8 x 8 = 64103 = 10 x 10 x 10 = 1000, 23 = 2 x 2 x 2 = 8104 = 10 x 10 x 10 x 10 = 10,000 24 = 2 x 2 x 2 x 2 = 16(-10)3 = (-10)(-10)(-10) (12).5 = 12

Order of Operations

To evaluate an expression:

12 - 9 ÷ 3 = 12 – 3 = 9

(12 – 9) ÷ 3 = (3) ÷ 3 = 1

(43 – 120 ÷ 2)2 + 82 = (64 – 60)2 + 64 = 42 + 64 =

16 + 64 = 80

Order of OperationsPerform left to right

HighestInside Parenthesis

ExponentiationMultiplication/DivisionAddition/Subtraction

Lowest

Evaluating Algebraic Expressions

if w = 4, x = -12, y = 64, z = -3

Find: 2 32w z

2 3 2 34 2( 3) 16 2( 27) 16 54 32 8w z

Using the Distributive Property

For any real numbers a, b , and c

a(b + c) = ab + ac or (b + c)a = ab +acNote: This is often referred to as “removing parenthesis”

• Example 1: -4(p – 5) = -4p – 20

• Example 2: -6m +2m = m(-6 + 2) = m(-4) = -4m

Note: This is often referred to as “factoring out m”

Using the Inverse Properties

For any real number a, there is a single real number –a, such that

a +(-a) = 0 Note: This is referred to as the “additive inverse”

For any nonzero real number a, there is a single real number , such

that: a • = 1 Example 2: 4 • = 1

Note: This is referred to as the “reciprocal or inverse”

•Example 1: 4 + (– 4) = 0

Using the Identity Properties

Zero is the only number that can be added to any number to get that number.

0 is called the “identity element for addition” a + 0 = a Example 1: 4 + 0 = 4

Note: This is referred to as the “additive identity”

One is the only number that can be multiplied by any number to get that number.

1 is called the “identity element for multiplication”

a • 1 = a Example 2: 4 • 1 = 4

Note: This is referred to as the “multiplicative identity”

Using the Commutative and Associate Properties

For any real numbers a, b, and c,

a + b = b + a

and ab = ba Note: These are referred to as the “commutative properties”

For any real numbers a, b, and c,

a + (b + c) = (a + b) + c

and a(bc) = (ab)c Note: These are referred to as the “associative properties”

Using the Properties of Real Numbers

Example 1: Simplify 12b – 9 + 4b – 7 b +1 =Solution: 12b + 4b – 7b + (-9 + 1) =

b(12 +4 -7) + (-8) = 9b - 8

Example 2: Simplify 6 – (2x + 7) –3 =Solution: -2x -7 + 6 – 3 = -2x -4

Using the Multiplication Property of 0

For any real number a: a • 0 = 0

Example 1: Simplify (6 – (2x + 7) –3)(0) =

Solution: 0

Fractions and Mixed Numbers

Proper and Improper Fractions; Mixed Numbers

1) Writing a fraction to describe the parts of a unit

2) Writing a fraction from a ruler or unit line.

Numerator Number of Shaded Parts

=Denominator Total Number of Parts in One Unit

Writing a fraction to describe the parts of a unit

Selecting Proper and Improper Fractions

Definitions:

1) A proper fraction is one in which the numerator is less than the denominator. (ie : 5/9)

2) An improper fraction is one in which the numerator is not less than the denominator. (ie: 9/5 or 9/9)

1) To determine if a fraction is proper or improper: Compare the numerator and

denominator, if the numerator is less than the denominator, the fraction is

proper.

Fraction List:

Proper Fractions Improper FractionsAnswer 13/16, 5/10, 1/21, 8/8, 11/8, 8/4, 25/1, 11/11

Select Proper or Improper fractions from the following list

8/8, 13/16, 5/10, 11/8, 8/4, 1/21, 25/1, 11/11

Changing Improper Fractions to mixed numbers

1) To change an improper fraction to a mixed number: Divide the numerator by the denominator. Write the whole number and then the fraction created by dividing the remainder by the denominator.

Fraction Strategy Answer

Change 20/7 to a mixed number

27 20 R 6

Changing mixed numbers to Improper Fractions

1) To change a mixed number to an improper fraction: Multiply the denominator by the whole number. Take the product and add to the numerator of the fraction. Write this sum over the denominator.

Mixed Number Strategy Answer

Change to an improper fraction

= 142 7 6 + = 20x

Simplifying Fractions

1) To simplify a fraction completely: Eliminate all common factors (other than 1) in the numerator and denominator.

Simplify

2 x 3 x 3

3 x 3 x 5

Multiplying and Dividing Fractions

1) To multiply fractions: Simplify, eliminate common factors, and multiply the numerators and denominators.

Fractions Strategy Answer

Multiply

4 8 3 3

5 5 3 0

2 x 2 x 2 x 2 x 3 3 x 11

5 x 11 2 x 3 x 5 24

4 8 3 3

5 5 3 0

Multiplying and Dividing Fractions: Reciprocal

1) To find the reciprocal of a fraction: Interchange the numerator and denominator.

Note: The product of the original fraction and the reciprocal should be one. (1/1)

Find the reciprocal of 1 4

Multiplying and Dividing Fractions: Dividing

1) To divide fractions: Multiply the first fraction by the reciprocal of the second fraction (divisor).

Divide

4 8 3 3

5 5 3 0

2 x 2 x 2 x 2 x 3 2 x 3 x 5

5 x 11 3 x 11 96

1214 8 3 3

5 5 3 0

=55 33

Multiplying and Dividing Mixed Numbers:

1) To multiple mixed numbers: Change to improper fractions, simplify, and multiply.

Note: Answer is written as a mixed number.

Multiply and write as a mixed number

21 10 3 x 7 2 x 5 35 = =

4 3 2 x 2 3 2

1 15 3

5 34 3

Multiplying and Dividing Mixed Numbers:

1) To divide mixed numbers: Change to improper fractions, and divide.

Note: answer is written as a proper fraction

Divide and write as a mixed number

25 125 25 8 5 x 5 2 x 2 x 2 4 = =

2 8 2 125 2 5 x 5 x 5 5

1 51 2 1 5

Building Fractions, Listing order;Inequalities

1) To rename a Fraction: Multiply both the numerator and denominator by a common factor of the form a/a; ie: 4/4 where a = 0

Write three fractions equivalent to 6/7

7 3 6 4

1) To find a missing numerator: Divide the target denominator by the original denominator. Multiply this quotient by the original numerator.

Problem Strategy Answer

Find the missing numerator :

8 x 3 = 24

56 = 8

1) To list fractions from smallest to largest: Build the fractions so they have a common denominator. List the fractions formed so the numerators go from smallest to largest. Simplify.

FractionsStrategy: Find the LCM of the

denominators, build the fractions. Answer

LCM is 2 x 3 x 7 = 42

List from smallest to largest

2 28 5 30 9 27 = , = , =

3 42 7 42 14 42

2 5 9, ,

3 7 1 4

2 5 9, ,

3 7 1 4

9 2 5 < <

14 3 7

Adding Fractions

1) To add like fractions: Add the numerators and write the sum over the denominator. Simplify.

Add and simplify:

1 1 3 5 1 + + = =

10 10 10 10 21

1 1 3 + +

10 10 10

1 1 3 + +

10 10 10

Adding Unlike Fractions

1) To add unlike fractions: Build the fractions so they have a common denominator, add, and Simplify.

Add and simplify:

1 2 95 2 15 8 23

+ = + = 12 9 36 36 36

Adding Mixed Numbers

1) To add mixed numbers: Add the whole numbers, add the fractions. If the sum of the fractions is >1, write as a mixed number and add again. Simplify

Add and simplify:

1 54 + 9

1 5 8 5 134 + 9 = 13 + ( + ) = 132 16 16 16 16

1 3 1 3

Subtracting Fractions

1) To subtract fractions: Build each fraction to a common denominator, subtract the numerators, and write the difference over the denominator. Simplify

Subtract and simplify:

1 3 9 -

1 8 2 4

1 3 9 -

1 8 2 4

13 9 52 27 25 - = - =

18 24 72 72 72

Subtracting Mixed Numbers

1) To subtract mixed numbers: Build each fraction to a common denominator, subtract the whole numbers, subtract the fractions. ( If the fractions can not be separated, borrow 1 from the 1st numbers whole number and add to the fractional component) Simplify

Subtract and simplify:

3 22 2 - 1 5

3 2 27 10 1722 - 15 = (22-15) + ( - ) 7

5 9 45 45 45 1 7

Combinations of Operations on Fractions; Average of a Set of Fractions

Order of OperationsPerform left to right

HighestInside Parenthesis

ExponentiationMultiplication/DivisionAddition/Subtraction

Lowest

Simplify:

7 3 8 -

7 3 8 7 24 7 2 21 16 5 - = - = - = - =

8 4 9 8 36 8 3 24 24 24

7 3 8 -

Average of a Set of Fractions

1) Find the average of a set of fractions: Add the fractions and divide the sum by the number of fractions.

Find the mean:

1 5 3, ,

1 5 3 4 15 18 37 + + = + + =

6 8 4 24 24 24 24

3 7 3 3 7 1 3 7 = =

2 4 1 2 4 3 7 2

Linear Equationsand

Inequalities

Linear Equations in One Variable

A linear equation in one variable can be written in the form Ax + B = C, where A,B, and C are real numbers with A 0.

• Example 1: x + 4 = -2 Note: x + 4 by itself is called an algebraic expression.

• Example 2: 2k + 5 = 10 Note: A linear equation is also called a first degree equation since the highest power of x is 1.

• The following are called non-linear equations: y = 3x2 +10

Decide whether a number is a solution of a linear equation.

• Example 3: 8 is a solution of the equation x – 3 = 5

An equation is solved by finding its “solution set”.

The solution set of x – 3 = 5 is {8}.

Note: Equivalent equations are equations that have the same solution set:

• Example 4: 5x +2 = 17, 5x =15 and x =3 are equivalent.

The solution set is {3}.

Addition Properties of Equality • For all real numbers A,B, and C, the

equations A = B and A + C = B + C are equivalent.

Note: The same number can be added to each side of an expression without changing its solution set.

• Example 5: Solution of the equation x – 3 = 5

x – 3 +3 = 5 + 3 x = 8

The solution set of x – 3 = 5 is {8}. • Example 6: 5x +2 = 17, 5x + 2 + (-2) =17 + (-2) and 5x =15

are equivalent. The solution set is {3}.

Multiplication Properties of Equality

• For all real numbers A,B, and for

C 0, the equations A = B and AC = BC are equivalent.

Note: Each side of an equation may be multiplied by the same non-zero number without changing its solution set.

• Example 7: Solution of the equation

The solution set of 5x = 15 is {3}.

5x = 15

1 1( ) 5 ( ) 155 5

Example 8: Solve: -7 + 3x – 9x = 12x -5

-7 + 7 –6x = 12x –5 + 7

-6x –12x = 12x –12x +2

-18x = 2

The solution set is { }.

1 1( ) (-18) ( ) 2

18 181

x = -9

Solving a Linear Equation in One Variable:

• Step 1: Clear fractions. (Eliminate any fractions by multiplying each side by the least common denominator.)

• Step 2: Simplify each side separately. (Use the distributive property to clear parenthesis and combine like terms.)

• Step 3: Isolate the variable terms on the left side of the equal sign. (Use the addition property to get all the variable terms on the left side and all the numbers on the right side.)

• Step 4: Isolate the variable. (Use the multiplication property to get an equation where the coefficient of the variable is 1.)

• Step 5: Check. (Substitute the proposed solution into the original equation and verify both sides of the equal sign are equivalent.)

Example 8: Solve: 6 – (4 +x) = 8x – 2(3x + 5)

Step 1: Does not applyStep 2: 6 – 4 –x = 8x – 6x –10

-x +2 = 2x – 10Step 3: -x + (-2x) +2 +(-2) = 2x + (-2x) –10 + (-2)

-3x = -12 Step 4:

Step 5: 6 – (4 + 4) = 8(4) – 2(3(4) + 5) 6 – 8 = 32 - 34 -2 = -2

The solution set is {4}.

1 1( ) (-3) ( ) (-12)

3 3 x = 4

Linear Inequalities in One Variable

Interval Notation is used to write solution sets of inequalities. Note: A parenthesis is used to indicate an endpoint in not included. A square bracket indicates the endpoint is included.

Type of Interval Set Interval Notation Graph

{x a x } (a, )

{x a x b} (a, b)

{x x b} (-, b)

{x x is a real number} (-, )

Interval Notation

Open Interval

{x a x} [a, )

{x a x b} (a, b]

{x a x b} [a, b)

{x x b} (-, b]

Interval Notation

Half-open Interval

Closed Interval {x a x b} [a, b]

Interval Notation

A linear inequality in on variable can be written in the form Ax + B = C, where A, B, and C are real numbers with A 0. Note: The next examples include definitions and rules for < , >, , and .

• Examples of linear inequalities:

x + 5 < 2 x – 3 > 5 2k +4 10

Solving linear inequalities using the Addition Property:

For all real numbers A, B, and C, the inequalities A < B and A + C < B + Ca are equivalent.Note: As with equations, the addition property can be used to add negative values or to subtract the same number from each side .

• Example 1: Solve k – 5 > 1

k – 5 +5 > 1 + 5k > 6

Solution set: (6,)• Example 2: Solve 5x + 3 4x – 1 and graph the solution

set.5x – 4x -1 –3x -4

Solution set: [-4,)

Solving linear inequalities using the Multiplication Property:For all real numbers A, B, and C, with C 0 ,1) if C > 0, then the inequalities A < B and AC < BC are equivalent.2) if C < 0, then the inequalities A < B and AC > BC are equivalent.

Note: Multiplying or Dividing by a negative number requires the inequality sign be reversed.

• Example 1: Solve -2x < 10 x > -5

Solution set: (-5,)• Example 2: Solve 2x < -10

x < -5 Solution set: (-,-5)

Note: The first example requires a symbol change because both sides are multiplied by (-1/2). The second example does not because both sides are multiplied by (1/2)

Solving linear inequalities using the Multiplication Property:

• Example 3: Solve –9m < -81 and graph the solution setm > 9

Solution set: (9, )

Solving a Linear Inequality: Step 1: Simplify each side separately. If necessary, use the distributive property to clear parenthesis and combine like terms.

Step 2: Use the addition property to get all terms containing the specified variable on the left side of the inequality sign and all other terms on the right side.

Step 3: Isolate the variable. (Use the multiplication (division) property to get an inequality where the coefficient of the variable is 1.)

• Example 4: Solve 6(x-1) + 3x -x –3(x + 2) and graph the solution setStep 1: 6x-6 + 3x -x –3x – 6

9x - 6 –4x – 6Step 2: 13x 0Step 3: x 0

Solution set: [0, )

Solving applied problems using linear inequalities:

• Example 6: Solve 5 < 3x – 4 < 9 and graph the solution set

5 + 4 < 3x < 9 + 4

9 < 3x < 13

3 < x < (13/3)

Solution set: ( 3, )

2 )(3 13

Systems of Linear Equations in Two Variables

Solving Linear Systems by Graphing. One way to find the solution set of a linear system of equations is to graph each equation and find the point where the graphs intersect.

• Example 1: Solve the system of equations by graphing. A) x + y = 5 B) 2x + y = -5

2x - y = 4 -x + 3y = 6

Solution: {(3,2)} Solution: {(-3,1)}

Solving Linear Systems by Graphing. There are three possible solutions to a system of linear equations in two variables that have been graphed:

• 1) The two graphs intersect at a single point. The coordinates give the solution of the system. In this case, the solution is “consistent” and the equations are “independent”.

• 2) The graphs are parallel lines. (Slopes are equal) In this case the system is “inconsistent” and the solution set is 0 or null.

• 3) The graphs are the same line. (Slopes and y-intercepts are the same) In this case, the equations are “dependent” and the solution set is an infinite set of ordered pairs.

Graphing

What to know about graphing?

• How to find the slope of a line

• How to find the y-intercept

• How to graph a solution

Review Lecture on graphing (MS Word File)

End of Review

1 final exam review math 002 college algebra. 2 write sets using set notation a set is a collection...

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