9 __UNKNOWN__

d)

Rule:

and

For all x³0

Examples

a) ()² =

7

b) =

Addition of Radicals

Radical terms (any term with a radical) are said to be like terms if they have the same radicand. We can use the distributive property to combine like radical terms.

Examples

a) 7 + 5= (7 + 5) = 12

b) 3- 5 = (3 – 5) = -2

Products and Quotients of Radicals

Radical terms can be simplified using the product and quotient rules:

= (a,b ³ 0) and

= (a ³0,b>0)

Examples

a) = = 4

b) = = 5

c) = =

d) = = = 5

When simplifying or breaking down radicals you must factor the radicand as a perfect square and a remainder factor.For this reason you should know the perfect squares!

Perfect Squares

because 2² = 4

= 3 because 3² = 9 (because = b/c)

= 4 b/c 4² = 16

= 5 b/c 5² = 25

6 b/c 6² = 36

=7 b/c 7² = 49

=8 b/c 8² = 64

= 9 b/c 9² = 81

= 10 b/c 10² = 100

= 11 b/c 11² = 121

= 12 b/c 12² = 144

Example

SIMPLIFY :

a)

=

b)

=

c)

=

d)

=

=

e)

=

=

Exercises:

Exercise:

35) Simplify the following:

36) Simplify the following:

Key Focus:Rationalizing the denominator.

Overview of Concepts:

Here the students learn to Rationalize the denominator in a complex fraction.

Contents:

Caution!!! Never leave a radical in the denominator!!! It’s an impolite way of expressing a mathematical solution. Whenever we see a radical sign in the denominator of an expression we need to do away with the radical sign, the process of eliminating the radical sign(s) is called rationalizing. In this course we will focus on the following three flavors :

Recipe l: Whenever n = m we simply multiply the numerator and the denominator by

Example: Simplify

Recipe ll: Whenever n < m we simply multiply the numerator and the denominator by

Example: Simplify

Recipe lll: Whenever the denominator is of the form simply multiply the numerator and the denominator by its conjugate and utilize the difference of squares technique.

Example: Simplify

Complex Conjugate

Exercise: Simplify the following expressions:

Exercise:

1) Evaluate the following problems:

Evaluate each problem:

Key focus:Grasp the Concept of Function Notation

Overview of Concepts:

Here the students learn to grasp the concept of Function Notation. We’ll develop the necessary skill required make substitution and evaluate its output based on an input value.

Contents:

Function Notation (one of the first applications of signed numbers on the CEAFE exam is evaluation of functions

A function can be considered a rule that assigns to each input value a single output value. The concept of a function is often viewed as a box. This function box is a rule f(x) which assigns to each input value x an output value y.

x

y=f(x)

Thus, f(x) = x + 2 assigns to each value x the output value x+2. In this case, the output value is always two more than the input value x.

For Example:

f(12) = 12 + 2 = 14f(-7) = -7 + 2 = -5f(0) = 2f(57) = 59f(-99) = -97

Functions can be assigned any letter but the most common are f, g & h. Let g(x) = 2x + 3 in this function the output value is always 3 more than twice the input value x.

For example:

g(8) = 2(8) + 3 = 19g(-5) = 2(-5) + 3 = -10+3 = -7g(0) = 2(0) + 3 = 3

Example

If f(x) = x² + 5x + 10 to find the value of f(-3) note that – 3 is where x was, this indicates x = -3

Step #1: Substitute the value of x = -3, remember to use parentheses (-3)² + 5(-3) + 10

Step #2: Evaluate or simplify the expression, = 9 + (-15 ) +10

=-6 + 10

= 4

Thus, f(-3) = 4

Examples:

Chapter 1 Review

Chapter 2Proportions & Applications

Key focus:Use Ratio to compare two quantities

Overview of Concepts:

Here the students learn to develop the concept of a ratio by using ratio to compare two quantities.

Give many practical examples to the students in this lesson, don’t be complicated, stay simple. For instance, talk about the ratio of male to female in the class, or the number of students that travel by train to the number of students that drive to school, etc…

Contents:

Ratio: is a comparison of two things.

Let’s look at an example: what is the ratio of X to Y? There are many ways to do this I would show you all the possible ways I can think of:

X to Y

X : Y

Example:

What is the ratio of squares to trapeziums ?

9 to 5

9 : 5

And so on…
Exercise:

1) What is the ratio of squares to trapeziums from the representation below?

2) What is the ratio of trapeziums to squares from the representation above?

3) Write the following ratios in simplest form:

Key Focus:Understand the difference between Ratio vs. Proportion

Overview of Concepts:

Here the students learn the difference between Ratio vs. Proportion

Contents:

Proportion: Basically represent the equality of ratios.

Cross Product: Often refers to as cross multiplication (cross multiply), is the procedure used to:

Check/Compare if ratios are equal.

Solve proportions.

Follow the flow chart below to solve proportions:

Write a Proportion, using the letter N for the number you need to findSet up the cross product equationDivide tofind thenumber for N

Example 1: Richard ate 4 candies in 10 minutes; at the same rate how many candies

can he eat in 2 hours (120 minutes)?

Let the number of candies Richard eat in 120 minutes = N

Hence Richard could eat 48 candies in 2 hours.

Example #2 On a map, 2 centimeters represent 300 miles, how many miles do 5 centimeters represent?

Let x represent the unknown number of miles.

cross−multiplication

2x = 1500

x = 750 miles

Example #3 At the Dunkin Donuts, you paid $5.10 for a dozen donuts. At this rate, how much would you need to pay for 18 donuts?

Solution: since 1 dozen = 12 donuts

Rate of $5.10 for 12 donuts is and rate of $ x for 18 donuts is

Hence solve the proportion = cross multiplication

x = $7.65

Exercise:

1) Identify which of the following ratios are proportional:

Fill in the unknown value that makes the ratios a proportion:

3) Samantha heart beats 28 times in 20 seconds. How many times does her heart beat in one minute?

4) Ricky serves 7 aces in a tennis match. How many aces can he achieve in at the same rate in 9 matches?

5) Errol made 17 points in a basket ball game. How many points can he score at the same rate after 12 game tournaments?

6) One day Kevin read 2 books. How many books can he read at the same rate after a week?

Key focus:Convert fractions and decimals into percent.

Overview of Concepts: Here the students learn Convert fractions and decimals into percent. Previous knowledge of fractions, decimals and place value will be very important for continuation.

Contents:

The flow chart below illustrates the way in which one goes about and Convert fractions and decimals into percent.

Equivalent fractionWith common denominator of 100

Given a Fractionor a Decimal

Write the percent byMultiply by 100(Numerator = %)

Convert to

Equivalent decimal fraction

Exercise:

Complete the following table, round all answer to the nearest thousandths, when necessary:

2) The 85% of the field were planted with rice. Represent the portion covered with rice as a fraction.

3) 7% of My English class was absent. What decimal part of the class was present?

4) The team scored 89 runs, 22 of them was made by Ronald. Represent Ronald’s score as a percent.

5) The doctor recommend 8 hours of sleep per day, what percent of the day is the doctor telling you to sleep?

Key Focus:Finding the percent of a number.

Overview of Concepts:

Here the students learn to find the percent of a number. A few methods could be used, such as, decimals or fractions, or by setting up a proportion.

Contents:

The flow charts below illustrates the way in which one go about and find the percent of a number.

Multiply
Problemstatement
Write the decimal for the percent

Example: Evaluate 15% of 45

15% of 45 = 6.75

ProblemStatement
Multiply
Write the fraction for the percent

Example: Evaluate 15% of 45

Write the proportion:P = PercentM = Part è ‘is’N = Total è ‘of’
15% of 45 = 6.75

Cross multiply and solve
Problemstatement

Example: Evaluate 15% of 45

15% of 45 = 6.75

Exercise:

1) Read each problem thoroughly before you attempt to solve each of the following:

What is 25% of 400?

What is 5% of 60?

25 is What percent of 400?

80% of 120 is what number?

95% of 50

35% of 175 is what number?

14% of 62

What is 6% of 56?

Clinton got 17 out of 20 question on his science test, what was his score expressed as a percent?

Errol made 14 out of 21 shots in a basketball game. What was his scoring percentage?

Chris ate 5 of his 12 chocolate. What percent of chocolate he has remaining?

Dolly made 25 cookies and give 21 to Linda. What percent of cookies did Dolly give away?

If Selena took a general knowledge test and missed 10 questions. She achieved a score of 92%. How many questions were given on the test?

Key Focus:Determine the percent of increase.

Overview of Concepts: Here the students learn to determine the percent of increase or decrease.

Contents:

* To find the Percent increase we use the following formula:

Increase = new amount – original amount

Example: five years ago one gallon of gas was $2.00 today gas sells for $3.00. Calculate the percent increase in the cost of gas.

Increase = new amount – original amount

Increase = ($3.00) – ($2.00)

Increase = $1.00

The percent of increase from $2.00 to $3.00 = 50%

Key Focus:Determine the percent of decrease.

Overview of Concepts: Here the students learn to determine the percent of decrease.

Contents:

* On the other hand to find the Percent decrease we use the following formula:

Decrease = original amount – new amount

Example: airline ticket for a trip to the virgin islands cost $450.00 last January, in May of the same year tickets cost $375.00. Calculate the percent decrease in the cost of a ticket.

Decrease = original amount – new amount

Increase = ($450.00) – ($375.00)

Increase = $75.00

The percent of decrease, rounded to the nearest whole percent

from $450.00 to $375.00 = 17%

Key Focus:Finding the simple interest

Overview of Concepts: Here the student learn to find the simple interest

Contents:

* To find the simple interest we use the following formula:

Where:

I = Interest P = Principal R = Rate T = Time

Example: Calculate the simple interest on a principal of $500.00 for 3 years, on a rate of 5%.

Givens:

I = ?

P = $500.00

R = 5% = 0.05

T = 3 years

The simple interest = $75.00

Exercise:

Given P = $450.00, R = 2% , and T = 3 years, calculate the simple interest.

Given P = $670.00, R = 2.5% , and T = 2.5 years, calculate the simple interest.

Given I = $40.00, R = 1.25% , and T = 1years, calculate the principal.

Determine the percent increase in the level of water in the rectangular tub below

BeforeAfter

5) Find the percent of increase or decrease to the nearest tenth of a percent.

11 to 16

23 to 14

Solving Percent Problems Using Proportions

Percent n% =

an amount out of a base of 100

To solve a percent problem, we translate as follows as;

Example #1 What is 15% of $600? is translated into an equation as n = 15% $600

Percent = 15%

Base = 600, this is the quantity we take 15% of.

Amount = n, we are looking for the quantity that is 15% of the base

Solution:

15100=n600

Cross multiply or translate ‘% of’ into multiplication to obtain:

n =

. $600 Multiply   n = $90 Therefore, $90 is 15% of $600Example #2 30 is what percent of 300?Percent = nBase = 300, we take n% of 300Amount = 30, this is the result of n% of 300 Solution: n100=30300 Cross multiply 100×30300=n  10 = n Therefore 10% of 300 is 30. Example #3 50 is 10% of what number? Percent = 10%Base = n, we take 10% of an unknown baseAmount = 50, the result of taking 10% of the unknown base is 50 Solution: 10100=50ncross multiply to solve 50×10010=nreduce  500 = n Therefore, 50 is 10% of 500. Percent Examples using proportion lines Example #4 What is 15% of $600? Solution: Quantity Percent

$600 100% [Whole/Base]    
$ x 15% [Part/Amount]
$0 0% The proportion is: = cross multiply 100x = 600(15)  100x = 9000 divide both sides by 100  x = 90   Therefore, $90 is 15% of $600.   Example #5 30 is what percent of 3000? Solution: QuantityPercent


3000 100% [Whole/Base]    

30 x% [Part/Amount]

0 0%  The proportion is: = Cross multiply  3000x = 30(100)  3000x = 3000 Divide both sides by 3000  x = 1 Therefore 30 is 1% of 3000.     Example #6 50 is 1% of what number? Solution: Quantity Percent


x100% [Whole/Base]      

50 1% [Part/Amount] 

0 0% The proportion is: = Cross multiply1x = 50(100) 1x = 5000 Divide both sides by 1 x = 5000  Therefore, 50 is 1% of 5000  Example #7 150% of $80 is what amount? Solution: Quantity Percent


x 150% [Part/Amount] 
$80 100% [Whole/Base]   

$0 0%   The proportion is: = Cross multiply 100x = 80(150) 100x = 12000 Divide both sides by 100 x = 120    Therefore, 150% of $80 is $120.   
Percent IncreaseExample #8 The marked price of an item is $200 and is on sale at 20% off. What is the new sale price? 
Solution:

$200 100% [Base] 

X 80% [20% off] [Amount]   $0 0%

The proportion is: = Cross multiply  100x = 16000 Divide both sides by 100  x = 160  Answer: Therefore, $160 is the new sale price.   Alternate Method using Percent Formula Percent = 20% Base = Original $200 Amount = The discount or the quantity off given by the 20% discount 20100=n200 Cross Multiply   (20)(200) = 100n  4000 = 100n   4000/100 = n  40 = n  This represents the discount to find the new sales price we subtract $200 -$40 = $160 sale price    Percent Decrease Example #9 The price of a skirt was increased from $20 to $30. The increase was what percent of the original price?Solution:
Method 1(Use of Proportion

) $30 x [Amount]  

$20 100% [Whole/Base]   

$0 0% The proportion is: = Cross multiply  20x = 3000 Divide both sides by 20  x = 150 Percent Increase = 150%−100% = 50%  Answer: Therefore; the increase was 50% of the original price. Alternate Method Percent Formula The base is the original price = $20 The amount is the increase in sale price: $30−$20 = $10 n100=1020 Cross Multiply (100)(10) = 20n  1000 = 20n  1000/20 = n  50 = n  Answer: Therefore, the increase was 50% of the original price.
Review Exercises:#1) An item that costs $720 is discounted 20%. Find the reduced sale price.  a) $144b) $700c) $576d) $676#2) A refrigerator that usually costs $120 is on sale for $90. Find the percent discount?  30%33%75%25%#3) An e−Pad that usually costs $800 is on sale for $560. Find the percent decrease.  240%30%24%70%#4) A bus travels 240 miles in 4 hours how far in miles will it go at this same rate in 9 hours?  a) 600b) 480c) 720d) 540#5) The price of a small house increased from $400,000 to $480,000. What is the percent increase?  a) 40%b) 80%c) 20%d) 16%#6) If 1” on a map represents 75 miles then how far apart in miles are two towns 2 ¼ inches apart on the map?  a) 150 b) 150 ¼ c) 162 ½ d) 168 ¾  #7) If 6 lb of cheese costs $24 then how much will 10 lb cost?   a) $60b) $30c) $96d) $40#8) If 9 mg. of a cancer drug are mixed with 21 mg of an enzyme inhibitor for a dose to be used in an IV solution. Then how many mg of enzyme inhibitor are required to for 12 mg of the cancer drug? a) 24b) 33c) 25d) 28 #9) A bookstore observes that 2 out of 9 books sold are returned, if 360 books are sold how many will be returned?   40 80180162 #10) Females need 44g of protein a day. If every 2 tablespoons of peanut butter provides 8g of protein, how many tablespoons of peanut butter would a female need to eat in order to get all of her protein from peanut butter?  5 ½ 181011  #11) A car that originally sold for $20,000 is reduced to $12,000 find the percent discount.   8%80%4%40%#12) If 6 oz of butter and 2 eggs are required for a cake. Then how many eggs are required to make a larger cake with 15 oz?  45 30 56#13) An item that costs $200 is discounted 35% find the reduced sales price.  a) $165b) $199.65c) $70d) $130 #14) The number of students in the Fall semester was 5,000 and in the following Spring was 5500. Find the percent increase a) 5%b) 50%c) 11%d) 10% #15) How many small containers are needed if 1623 lbs. of sugar is divided into small 123 lb. container? a) 7b) 16c) 70d) 10#16) A camera that sells regularly for $280.00 is discounted by $89 in a sale. what is the sale price? a) $369b) $201c) $191d) $89  Chapter 3Algebraic Expressions & Applications Key Focus:Explain and identify the difference between an Algebraic expression and an equation. Overview of Concepts: Here the students learn to explain and identify the difference between an algebraic expression and an equation. Contents: VariableCoefficientTerm
Variable: A variable is basically a symbol used to represent an unknown numeral. Coefficient: A coefficient is basically another word that means number, or number part of a term. Term: A term is a combination of a variable and a number/coefficient. A variable or a number are also considered as a term whenever they stand independently.  Expression: The putting together of one or more terms forms what we call an algebraic expression. It does not have an equal signs separating any term.  Equation: The direct comparison of term is called algebraic equation, i.e. two expressions separated by an equal sign.  
Yes
By inspection: check if the expression has an equal sign
* Follow the following flow chart below to guide you through the process of identifying Equations vs. Expressions: Equation
Algebraic Problem
No


   
 Expression
    Example: By inspection: Algebraic ExpressionNumber of TermsExplanation   1 A number multiplied by a variable   2 Subtraction signs separate two terms   3Addition and subtraction signs separate three terms   1A number multiplied by a variable raised to the third power   1A number multiplied by three variable divided by another variable 
Exercise:  1) Identify the following as an equation or as an expression.              2) Fill in the blank:           3) Evaluate the following by combining like terms:                  4) State the variable, the exponent and the coefficient of each of the following: ExpressionCoefficientVariableExponent                    
 Key Focus:Substituting values into algebraic expressions   Overview of Concepts: Learn to substituting values into algebraic expressions for the purpose of evaluation and simplification.  Contents: A method used in algebra to simplifying algebraic expressions, where you’re to replace the variable(s) with an assigned value. Let’s look at the example below. Example 1: Evaluate the expression:: when X = 3 and Y = −2. Always Use Parentheses whenever you’re substituting numbers in an algebraic expression
  Example 2: Evaluate the expression based on the given formula: the classical Pythagorean’s theorem is given by use the formula to calculate the exact value of C when A = 3 and B = 4Always Use Parentheses whenever you’re substituting numbers in an algebraic expression

Exercise: 1) Given that X = −2, Y = 2, and Z = −1. Evaluate the following algebraic expression:                2) The formula for computing the area of a triangle is given by. Find the area of a triangle if the base, b = 6 inches and the height h = 13 inches.     3) The simple interest I on principal P dollars at an interest rate R, for time T in years, is given by the formula I = PRT. Find the simple interest on a principal of $12,000 at 6% for 1 years (Hint: 6% = 0.06)      Temperature conversion: The formula that relates Celsius and Fahrenheit temperature is , if the temperature of the day is −35^oC, what is the Fahrenheit temperature?        5) Find the value of where X = 2 6) Find the value of where X = 2   Key Focus:Interpreting/translating word problems into algebraic expressionsOverview of Concepts: Here the students learn identifying, interpreting/translating word problems into algebraic expressions. I will develop a list of to help you comprehend and order the way in which you translate a phrase/expression into an algebraic expression.Contents: One of the first steps in solving word problems is translating the conditions of the problem into algebraic symbols. The following tables illustrate some key word/phrase we need to learn:   Common Phrases Describing Addition Algebraic Expression  Eight more than some number The sum of a number and eightA number increased by eightEight is added to a number Eight greater than a numberA number plus eightThe total of a number and eight       X+ 8 Common Phrases Describing Subtraction Algebraic Expression  A number decreased by 2Two less than some number Two is subtracted from a number Two smaller than a number A number diminished by 2 A number minus 2The difference between a number and 2       X − 2 
 Common Phrases Describing Multiplication Algebraic Expression  Double a number Twice a number The product of a number and 2Two of a number Two times a number    2X   Common Phrases Describing Division Algebraic Expression  A number divided by 5The ratio of some number and fiveThe quotient of a number and five      
Key Focus:Interpreting/translating word problems into algebraic expressions involving more than one operation.   Overview of Concepts: Here the students learn to interpret/translate word problems into algebraic expressions involving more than one operation.  Contents: The key to this is reading the question carefully and apply critical thinking. Use the previous information collectively and summarize the words into symbolic form. The following table illustrates some key concepts we need to learn:
QuestionAlgebraic ExpressionSummarize  6 more than the quotient ofa certain number and 2.    6 more than something è something + 6quotient è means to  8 less than 2 times a number  8 less than something è something − 8Times è means to    Seven Times the sum of a number and 4   Times è means to Sum of è means to + some number and 4 collectively  The square of the difference between a number and 4    Difference è something −Squared è means to use exponent/power   
Exercise:1) Make appropriate translation to convert each phrase into an algebraic expression:  A certain number plus 6   The quotient of 4 and a number   The product of a number a 9   A number multiplied by 5   45% of a certain number   The difference between a number and 7   6 subtracted from the quotient of 3 and a number   Twice the difference between a number and 7   the ratio of 7 and some number   3 more than the product of a number and 4   3 more than twice a number   Twice the sum of a number and 4   The quotient of a number and 8   5 decreased by a number    2) Make appropriate translation to convert each algebraic expression into a phrase                    Key Focus:Identifying and setting up problems involving consecutive integers, consecutive odd integers and consecutive even integers  Overview of Concepts: Here the students learn to identify and set up problems involving consecutive integers, consecutive odd integers and consecutive even integers. Contents:Consecutive Numbers: Consecutive numbers are integers that are arranged one after the other from least to greatest: i.e. 1, 2, 3, 4, 5, 6, 7, 8 …Hence, if we consider the first of the consecutive integer to be X then the second would be X+1, X+2, X+3, and so on… Consecutive Odd Numbers: Consecutive odd numbers are integers that are not multiples of two, arranged one after the other from least to greatest: i.e. 1, 3, 5, 7 …Hence, if we consider the first of the consecutive integer to be X then the second would be X+2, third would be X+4, and so on… Consecutive Even Numbers: Consecutive even numbers are integers that are multiples of two, arranged one after the other from least to greatest: i.e. 2, 4, 6, 8 …Hence, if we consider the first of the consecutive integer to be X then the second would be X+2, third would be X+4, and so on… ProblemApproach   Consecutive Numbers  1^st è X2^nd è X+13^rd è X+24^th è X+35^th è X+4 And So on …    Consecutive Odd Numbers   1^st è X2^nd è X+23^rd è X+44^th è X+65^th è X+8 And So on …    Consecutive Even Numbers   1^st è X2^nd è X+23^rd è X+44^th è X+65^th è X+8 And So on …  
Example 1: Give an algebraic expression for the sum of three consecutive integers. Consecutive Numbers:
1^st è X2^nd è X+13^rd è X+24^th è X+35^th è X+4 And So on …Hence the algebraic expression:    Example 2: Give an algebraic expression for the sum of four consecutive odd integers. Consecutive Odd Numbers:
1^st è X2^nd è X+23^rd è X+44^th è X+65^th è X+8 And So on …Hence the algebraic expression:    Example 3: Give an algebraic expression for the sum of five consecutive even integers. Consecutive Even Numbers:
1^st è X2^nd è X+23^rd è X+44^th è X+65^th è X+8 And So on …Hence the algebraic expression: 
Key focus: Define and understand the meaning of algebraic term(s) as related to polynomials. Overview of Concepts: Here the students learn the definition and understand the meaning of algebraic term(s) as related to polynomials Contents: Monomial: One term.Example:  Binomial: Two termsExample:  Trinomial: Three termsExample:  Polynomials: Polynomial meaning many terms.Example:    Like Terms: Like terms are terms that have the same variable raised to the same powers. Only the like terms can be combined  Example: Combine like terms in the given expression:  

LIKE TERMS  è We usually write the final answer from largest to smallest (descending order) 
Exercise: Simplify the following polynomials by combining like terms   2) Evaluate the following:   a) Subtract from the sum of and    b) A rectangle has sides of 8x + 9 and 6x − 7. Find the polynomial that represents its perimeter.      3) Simplify:     4) The length of a rectangle is given by yards and the width is given by yards. Express the area of the rectangle in terms of x.    Suppose an orchard is planted with trees in straight rows. If there are rows with tress in each row, how many trees are there in the orchard?  6) Perform the following operations:    Key Focus:Adding polynomials. Overview of Concepts: Here the students learn to add polynomials. Contents: Reinforce to the students that polynomials are like numbers. If we see two polynomials in a question we can always add them to get another polynomial. Like we previously discussed before, we need to focus our attention on the idea of combining like terms. The following example will illustrate what I mean:  Example 1: 1^st re−write the expression without parentheses è 2^nd combine like terms è     Example 2: 1^st re−write the expression without parentheses è 2^nd combine like terms è     And so on…
Exercise:  1) Add the following polynomials:                            
Key Focus:Subtracting polynomials. Overview of Concepts: Here the students learn to subtract polynomials. Contents: Reinforce to the students that polynomials are like numbers. If we see two polynomials in a question we can always subtract them to get another polynomial. Like we previously discussed before, we need to focus our attention on the idea of combining like terms. Remembering many times in subtracting polynomials we are required to change signs for obvious reasons. The following example will illustrate what I mean:  Example 1: 1^st re−write the expression without parentheses è 2^nd combine like terms è     Example 2: 1^st re−write the expression without parentheses è 2^nd combine like terms è     And so on…
Exercise:  1) Simplify the following polynomials:                              
Key Focus:Multiplying Polynomials Overview of Concepts: Here the students learn to multiplying Polynomials.  Contents: Reinforce to the students that polynomials are like numbers. If we see two polynomials in a question we can always multiply them to get another polynomial. Like we previously discussed before, we need to focus our attention on the concepts behind the laws of exponents. The following example will illustrate what I mean:  

Example 1: 1^st re−write the expression without parentheses è 2^nd Multiply to get è    

Example 2:
1^st Apply the distributive property the è
2^nd Multiply to get è   3^rd combine like terms è    And so on…
Exercise:  1) Find the product of the following polynomials:                                                                                  
Key Focus:Dividing Polynomials Overview of Concepts: Here the students learn to divide Polynomials.   Contents:Reinforce to the students that polynomials are like numbers. If we see two polynomials in a question we can always divide them to get another polynomial. Like we previously discussed before, we need to focus our attention on the concepts behind the laws of exponents. The following example will illustrate what I mean:  Example 1: 1^st Re−write the expression without parentheses è 2^nd Divide to get è     Example 2: 1^st Apply the distributive property the è   2^nd Divide to get è    And so on…
Exercise:  1) Divide the following polynomials:                                                               
Key Focus:Factoring Prime polynomials  Overview of Concepts: Here the students learn to identify polynomials that are Prime   Contents: Prime polynomials are algebraic expressions that are not factorable.  Example 1: Factor:   è Prime   Example 2: Factor:   è Prime Note: Prime! meaning the expression cannot be factored.    And so on…
Key Focus:Factoring polynomials using Greatest Common Factor Overview of Concepts: Here the students learn to factor polynomials using the Greatest Common Factor.  Contents: Factor: Is defined as a number or an expression that is multiplied by another to produce a product. Greatest Common Factor (GCF): Is the greatest: whole number or a variable or both a whole number and a variable that is a factor of each term in an algebraic expression. Example 1: Factor out the greatest common factor: 1^st Identify the GCF of è 2X 2^nd Factor the expression to get è  Example 2: Factor out the greatest common factor: 1^st Identify the GCF of è 3Z 2^nd Factor the expression to get è  Example 3: Factor: -28m³n⁴– 14m²n³ Step 1: Find the GCF of numbers and the common variablesGCF(28,14)=14 because 14 is the greatest number that divides evenly into both numbers GCF {m³ , m²}b= m² as m² is the smaller of the two terms it divides evenly into both. GCF {n⁴ ,n³} = n³ as n³ being the smaller of the two terms divides into both. Hence the GCF for -28m³n⁴, – 14m²n³) = – 14 m²n³  Step 2: Divide each term by the GCF.-28m3n4-14m2n3 = 2nm and -14m2n3-14m2n3 = 1  Step 3: Write the GCF outside the parentheses and write the quotients from step−2 inside the parentheses Hence -28m³n⁴– 14m²n³ = -14m²n³( 2nm + 1)
Example 4: Factor: 8x²y⁹+12x⁵y⁷ (note there are two terms)  Step 1: Find the GCF of numbers and the common variables4 is the GCF of 8 and 12.x² is the GCF of x²and x⁵. <sup>y7 is the GCF of y7and y The GCF = 4x2y7 Step 2: Write 8x2y9+12x5y7 as product of the GCF with (two terms−in this example) parenthesis carry down the addition operation between the terms:  8x2y9+12x5y7 = 4x2y7( ____ + ____ )  Step 3: to find the terms inside the parentheses divide each term by the GCF=2y2 and =3x3. Step 4: Write the answer 8x2y9+12x5y7 =4x2y7(2y2+3x3)     Example 5: Factor: 27a5 + 18a4– 36a6 Step 1: Find the GCF of numbers and the common variablesGCF ( 27, 18, 36) = 9GCF( a5, a4 , a6) = a4GCF = 9a4 Step 2: Write the gcf product with parentheses leaving 3 terms and carry down the two operations: 27a5 + 18a4– 36a6 = 9a4 ( __ + __ – __ )  Step 3: Divide by GCF to find the terms inside the parentheses27a59a4=3a, 18a49a4=2, -36a69a4=-4a2 Step 4: Write the answer: 27a5 + 18a4– 36a6 = 9a4( 3a + 2 – 4a2)          Exercise: Find the GCF ( Greatest common factor) for each pair of terms.GCF of { x6, x10 } = ?GCF of {a5b3 , a2b6 }= ?GCF of { 16, 20 }= ?GCF of { 15, 14 }= ?GCF of { (x+2)(x+7) , (x+2)(x-5) }= ?GCF of { 40x3y5, 16x8y4 }= ?GCF of { 8m7n3 , 10m2 }= ? Factor each of the following expression completely: #1) Factor Completely: 30a2x2– 18a4x9 6a2x26a2x(5x – 3a2x8)6a2x2(5 – 3a2x7)6a2x2(5 – 3ax7) #2) Factor Completely 3x2y – 6xy2 3(x2y – 2xy2)3xy(x – 2y)3x2y(1 – 2y)3xy2(x – 2) #5) Factor Completely 32xy – 18x2  a) 2(16xy – 9x)b) 2x(16y -9)c) 2x(16y -9x)d) 2x(4y – 3x)(4y + 3x )#6) Factor Completely 5n3m – 15n2m + 10nm2  a) 5nm(n3m – 3n2m + 2nm2)b) 5nm(n2m – 3nm + 2nm)c) 5nm(n2– 3m + 2m)d) 5nm(n2– 3n + 2m)#7) Factor Completely 14r7s6 + 21r6s7– 28r5s8  a) 7rs(2r6s5 + 3r5s6– 4r4s7)b) 7r2s2(2r5s4 + 3r4s5– 4r3s6)c) 7r7s6(2 + 3s – 4s2)d) 7r5s6(2r2 + 3rs – 4s2)#8) Factor Completely 6a2b3+8a3b2+12a3b3  2ab(3ab2 + 4a2b + 6a2b2)2a2b(3b2 + 4ab + 6ab2)2a2b2(3b + 4a + 6ab)2a2b3(3 + 4a + 6ab) #9) Factor Completely 4x2y+12x3y#10) Factor Completely: 28x3y5 -21x7y3 + 35x9y2    
Key Focus:Factoring polynomials of the form , where A = 1A = 1
 Overview of Concepts: Here the students learn to factor polynomials of the form , where A, B, and C are constants. First we’ll look at it when A = 1, then later on we’ll look at it when A > 1.   Contents: Polynomials of the form are refer to as trinomials. We factor them by asking what two numbers have a product of C and a sum of B. Pay attention, if each term of the trinomial has a GCF, we’ll need to factor it out as our first step and then work with the parentheses if it is factorable.30 * 115 * 210 * 3 6 * 5
 Example 1: Factor the Trinomial:  1st Identify if a GCF existsè No GCF 2nd Factors of 30è  3rd Factors to get è  18 * 1 9 * 2 6 * 3
Example 2: Factor the Trinomial:  1st Identify if a GCF existsè 2nd Factors of 18è  3rd Factors to get è  And so on…
Exercise:  1) Factor the following polynomials completely:                                                                      
Key Focus:Factoring polynomials of the form, where   Overview of Concepts: A > 1
Here the students learn to factor polynomials of the form , where A, B, and C are constants. We’ll look at it when A > 1.   Contents: Polynomials of the form are usually referred to as trinomials. We factor them by the trial and error method or by the grouping method. We factor them by asking what two numbers have a product of (A * C) and a sum of B. Pay attention, if each term of the trinomial has a GCF, we’ll need to factor it out as our first step and then work with the parentheses if it is factorable.60Y * 1Y30Y * 2Y20Y * 3Y15Y * 4Y12Y * 5Y10Y *6Y
 Example 1: Factor the Trinomial:  1st Identify if a GCF existsè No GCF, Other than 1 2nd A * C = è  3rd Re−write the polynomial as è  4th Factor the polynomial by Grouping è   5th Factor the polynomial by Grouping è    And so on…
Exercise:  1) Factor the following polynomials completely:                                                                               Key Focus:Factoring polynomials using Special Products Overview of Concepts: Here the students learn to factor polynomials using Special Products Contents: Pay attention, if each term of the trinomial has a GCF, we’ll need to factor it out as our first step and then work with the parentheses if it is factorable.  Difference of squares: if by inspection the expression fits into your special case formula, then use it. It can be a very effective and efficient method of factoring.  Example : Factor:  1st Identify if a GCF existsè GCF = 1 Don’t bother using it! 2nd è  3rd Re−write the polynomial as è  4th Factor the polynomialè      And so on…
Perfect squares trinomial: if by inspection the expression fits into your special case formula, then use it. It can be a very effective and efficient method of factoring. Example 1: Factor:  1st Identify if a GCF exists è GCF = 1 Don’t bother using it! 
2nd è  3rd Re−write the polynomial as è  4th Factor the polynomialè    Example 2: Factor:  1st Identify if a GCF exists è GCF = 1 Don’t bother using it!  
2nd è  3rd Re−write the polynomial as è  4th Factor the polynomial è   And so on…
Exercise:  1) Factor the following polynomials completely:  a)           b)           c)                                                     
Exercise: Apply appropriate technique to factor completely.  #1) Factor completely 5a3b3– 20ab3   a) 5ab3(a -2)2b) 5ab3(a2-6a) c) 5ab3(a+2)(a+b) d) 5ab3(a+2)(a-2)  #2) Factor Completely: 5x3 + 10x2– 175x    a) 5x(x + 2x -35) b) 5x(x + 5)(x – 7) c) 5x(x2 + 4x – 35) d) 5x(x – 5)(x + 7) #3) Factor Completely: 32x3– 50xy2   2x(16x2-25y2)2x(4x -5y)22(16x2-25y2)2x(4x + 5y)(4x -5y)#4) Which of the following is a factor of: 12ax + 15bx -20ay -25by   (3x + 5y)(4a – 5b)(4x -5y)(4a + 5b) #5) Factor completely: 9xy3 − 36xy    a) 9xy(y2+4) b) 9xy(x2−4)c) 9xy(y−2)(y+2) d) 9x(x−2y)(x+2y) #6) Which of the following is the complete factorization of 2x2 − 13x − 24 ?   (2x − 6)(x + 4)(2x − 3)(x − 8)(2x + 3)(x − 8)2(x + 3)(x − 4) #7) Factor completely: 32x2y2– 16x3y   a) 16xy(2xy -x2)b) 16x2y2(2 -x)c) 8x2y(4y – 2x)d) 16x2y(2y – x)#8) Factor completely: 8x6-24x5 + 16x4-8x3   a) 8( x6– 6x5 + 4x4– 2x3 )b) 8x3( x3-3x2 + 2x )c) 8x3( x3-3x2 + 2x – 1)d) 4x3(2x3– 6x2 + 4x – 2)  #9) Factor Completely 54x3– 6x2   a) 3x(18x2-2x)b) 6x(9x2-x)c) 6x2(9x -1)d) 6x2(9x)#10) Factor Completely 2x4– 4x3 + x2   a) 2(x4– 2x3 + x2)b) 2x(x3– 2x2 + x)c) 2x2(x2– 2x + 1)d) x2(2x2– 4x + 1)#11) Factor Completely 128x2– 162y2   2(64x2– 81y2)2(8x – 9)(8x + 9)2(8x – 81y)(8x + y)2(8x – 9y)(8x + 9y)#12) Factor Completely: 50x2– 98   2(25x2– 49)2(5x – 7)(5x + 7)2(5x + 7)(5x + 7)2(5x – 7)(5x – 7)#13) Factor Completely: 6x5y3– 18x4y2 + 36x3y3   6(x5y3– 3x4y2 + 12x3y3)6xy(x4y2– 3x3y + 12x2y2)6x2y2(x3y – 3x2 + 12xy)6x3y2(x2y – 3x + 12y) #14) Factor Completely:12x3– 27x    3(4x3– 9x)3x(4x2– 9)3x(2x – 3)(2x + 3)3x(2x + 3)(2x + 3) #15) Factor Completely: 32x2 + 50  2(16x2 + 25)2(4x + 5)(4x + 5)2(2x + 5)(4x – 5)2(2x – 5)(4x – 5)#16) Factor completely 4x3 + 4x2 − 8x   a) (4x2 − 4x)(x + 2) b) 4(x + 2)(x − 1)c)4x(x − 1)(x + 2) d) x(4x + 4)(x − 1)#17) Factor completely 4x2−4  a) (2x − 2)(2x + 2) b) 4(x2− 1)c) 4(x + 1)(x − 1) d)2 (2x+2)(x −1)#18) Factor completely y2 + 49  y( y + 49)(y + 7)(y+7)(y + 1)(y + 49)Prime Key Focus:Complex Fraction.  Overview of Concepts: Here the students learn to simplify complex fractions.  Contents:As in regular fractions, we appreciate the effort of having our answers in its lowest term, i.e… Reduced form. Polynomials can also be written in fractional form, to simplify such expressions we’ll need to factor the numerator and the denominator individually, and then make the appropriate cancellation(s). All the tactics we have learned in factoring thus far would be beneficial in simplifying these kinds of problems. The following example will illustrate this concept:  Example : simplify:  1st Factorè

2nd Reduceè  3rd Final resultè     And so on…
Exercise:  1) Simplify the following completely:                                                                   Mixed Exercise:  Fill in the blanks: a) List all the factors of 12 in increasing order. _____, ______, ______, _____, ______, ______ .b) 2 and 6 are called the ___________ of 12.c) 12 = 4 × 3 is ______ of 12 but not complete factor of 12.d) Prime factors of 12 = _____ × _____ × ______e) 35 = ____ × 7f) x2– 9 = ( ____ )( x – 3 )g) one factor of y2 + 6y + 8 if y + 2, what is the other factor ___________h) ( m + 5 )( ____ ) = m2 + 10m + 25i) GCF of 3, 6, 9 is ___________j) GCF of (x+2)(x +7) and (x – 4)( x + 2) is _____________k) ( 2x + 1 ) ( x +__ ) = 2x2 + 5x + 3l) ( 5x- 3) ( x + ___ ) = 5x2 + 7x – 6m) ( x + 5) ( ___ + ___ ) = 2x2 + 13x + 15   #1) Factor Completely: 2x2 + 11x + 15    (2x + 15)(x + 1)(2x + 5)(x + 3)(2x + 1)(x + 15) (2x + 3)(x + 5)#2) Factor Completely: 2x2 + 13x + 15(2x + 15)(x + 1)(2x + 5)(x + 3)(2x + 1)(x + 15) (2x + 3)(x + 5)#3) Factor Completely: 2x2 + 31x + 15    (2x + 15)(x + 1)(2x + 5)(x + 3)(2x + 1)(x + 15) (2x + 3)(x + 5)#4) Factor Completely: 2x2 + 17x + 15     (2x + 15)(x + 1)(2x + 5)(x + 3)(2x + 1)(x + 15) (2x + 3)(x + 5)#5) Factor Completely: 2x2– 15x + 21    (2x – 3)(x – 7)(2x – 21)(x – 1)(2x – 7)(x – 3) (2x – 1)(x – 21)#6) Factor Completely: 2x2– 9x + 21     (2x – 3)(x – 7)(2x – 21)(x – 1)(2x – 7)(x – 3) (2x – 1)(x – 21) #7) Factor Completely: 2x2 – 43x + 21    (2x – 3)(x – 7)(2x – 21)(x – 1)(2x – 7)(x – 3) (2x – 1)(x – 21) #8) Factor Completely: 2x2 – 23x + 21     (2x – 3)(x – 7)(2x – 21)(x – 1)(2x – 7)(x – 3) (2x – 1)(x – 21)#9) Factor Completely: 2x2 – 7x -15   a) (2x – 3)(x + 5)b) (2x +3)(x -5)c) (2x -5)(x + 3)d) (2x + 5)(x -3)#10) Factor Completely:2x2 – 29x -15   a) (2x – 1)(x + 15) b) (2x + 1)(x – 15)c) (2x + 15)(x – 1)d) (2x – 15)(x + 1) #11) Which of the following is a factor of: 2x2 + 13x -15   a) (2x – 1)b) (2x + 1)c) (2x + 15)d) (2x – 15)#12) Which of the following is a factor of: 2x2 – x -15   a) (x – 3)b) (2x – 5)c) (x + 5)d) (2x – 3) #13) Which of the following is a factor of: 2x2 -13x – 7    2x -1x + 72x + 7x – 7 #14) Factor Completely: 5x3 + 10x2 – 175x      a) 5x(x + 2x -35) b) 5x(x + 5)(x – 7) c) 5x(x2 + 4x – 35) d) 5x(x – 5)(x + 7)#15) Which is a factor of : 3x2 + 11x + 10      a) (3x + 2) b) (x + 5) c) (3x + 5) d) (x + 10)#16) Which of the following is a factor of: 10ax + 15bx – 6ay – 9by   2a – 3b5x + 3y5x -3y2ax -3by#17) Factor completely: 9xy3 − 36xy  a) 9xy(y2+4) b) 9xy(x2−4)c) 9xy(y−2)(y+2) d) 9x(x−2y)(x+2y)#18) Which of the following is the complete factorization of 2x2 -13x – 24?    a) (2x – 6)(x + 4) b) (x – 6)(2x + 4) c) (2x – 3)(x- 8) d) (2x + 3)(x – 8) #19) Which of following is factor of the polynomial: 3x2 + 13x -10  x – 23x + 2x – 53x – 2#20) Which of following is factor of the polynomial: 3x2 – 13x -10  x – 23x + 2x + 53x – 2#21) Which of following is factor of the polynomial: 3x2 + x – 10    x – 23x – 5x + 53x + 5 #22) Which of following is factor of the polynomial: 3x2 – x -10    3x + 23x – 5x + 53x + 5 #23) Factor completely: 3x2 + 17x + 8     (3x + 2)(x + 4)(3x + 4)(x + 1)(3x + 1)(x + 8)(3x + 8)(x + 1)#24) Which is a factor of the polynomial: 15nx + 20mx – 6ny – 8my    5x + 2y5x – 2y3n – 2m3x + 4m #25) Factor: 3x2 – 16x – 35    #26) Factor Completely using grouping: 28rx + 21sx – 20ry -15sy    #27) Factor: 9xy3 – 64x3y         #28) Factor completely:16a2b-100b3   Key Focus:Identifying and solving quadratic equations. Overview of Concepts: Here the students learn to identifying and solving quadratic equations of the form , where A, B, and C are constants. Factoring polynomials are a prerequisite for this topic. Contents: A quadratic equation of the form , sometimes considered as a second degree polynomial equation. We will utilize the knowledge of factoring to solving quadratic equations in most cases 30 * 115 * 210 * 3 6 * 5
Example: Solve the equation:  1st Identify if a GCF existsè è GCF = 1 Don’t botherusing it! 2nd Factors of 30è  3rd Factors to get è  4th Solve the Equation è ;  ;  ;  ; 5th Hence the solution set:è   And so on…
Exercise:  1) Solve the following quadratic equations:                                                                    
 Key Focus:Application quadratic formula.  Overview of Concepts: Here the students learn to apply quadratic formula to solve problems  Contents:  What do we do with this formula? Simple! We do just the same thing we would do with any other formula, that is to say we use substitution to solve for an unknown.We have probably seen equations of the form if we have then that’s great! If we don’t we are surely seeing it now, it represents the standard form of the quadratic equation so we will use it to match the equation against it to evaluate the value(s) of x.  Example: Solve the equation: By observation we see that the equation is in standard form, so we identify some of the known things, that is:Notice that the variables a, b and c is related to: Substitute these values in the equation (it works like a recipe!):
Exercise:  1) Use the quadratic formula to solve the following equations, ALL ANSWERS MUST BE EXACT (DO NOT ROUND/APPROXIMATE):          c)         d)         2) Find the roots of the following equations, express your answer(s) to the nearest tenth, (when possible):                        3) Solve the equation:  
Chapter 4Coordinate Geometry & Applications Key Focus:Apply the Pythagorean’s theorem  Overview of Concepts: Here the students learn to apply the Pythagorean’s theorem to find the missing side of a right angle triangle. Contents:  Example 1: Consider the triangle below where B = 3 ft and A = 4 ft. What is the dimension in feet for C?Example 2: Consider the triangle below where C = 13 ft and A = 5 ft. What is the dimension in feet for B?  Exercise:  Find the length of the indicated side of the right triangle.  2. In the rectangle below the length of the diagonal D is If a 10 ft ladder leans against a wall, and it is given that the distance from the foot of the ladder to the wall is 8 ft. What is the total vertical distance the ladder meet up the wall?
Prove that the hypotenuse AC of the triangle shown on the plane below is
Key Focus:Solving linear equation in two variables by the process of elimination and by the method of substitution.Overview of Concepts:Here the students learn to Solve linear equation in two variables by the process of elimination and by the method of substitution Contents:In this section we will only focus on system of linear equations in two variables. We will discuss two methods for solving such equations, later you will be introduced with a third method (Graphically). Practice all the methods they are equally important.  We’ll now explore system of equations algebraically: This Chapter is about solving the system of linear equations containing two equations in two variables using Addition ( Elimination ) method and Substitution method.  Fact: The solution of a system of equations is the point of intersection, or the points which are common to both graphs. In other words, the solution a point ( x, y) will satisfy both equations. Look at the three figures below: Figure #1 y=-x+4y=x-2 Or In Figure 1, The point at which these lines meet is called a solution to the system of equations. It appears that the lines meet at the point (3,1) in order to verify that (3,1) is this a solution to this system of equations we substitute these values into both equations we obtain:1=-3+41=3-2Since both these statements are true the point (3,1) lies on both lines and is a unique solution to this system of equations. Figure #2  y=2x+4y=2x+2 Or  Notice there is not solution or point of intersection of these parallel lines, which have the same slope.  This system of equation has no solution and the system is called Inconsistent.  Figure #3  y=x-22y=2x-4 Or  In Figure 3, both equations have same graph and hence each and every point is the point of intersection.  This system is called Consistent and has infinite solutions.  
Method I: Elimination of one Variable by Substitution  Procedure  Example This method is very useful especially when one variable is expressed in terms of the other in one of the two given equations.   Replace one variable in terms of the other; this will reduce the problem to one equation in one variable.   Solve for the remaining variable.   Substitute the now known value in an   Equation involving both variables, and solve for the second variable.   Check both values in each ‘original equation.’ Question:Solve the system of equations:  Solution:We would substitute (replace) y by 3x in the 2nd equation:  now we substitute (replace) x = 10 into one of the ‘original equations’ and solve for y:   check both values of x and y into each ‘original equations’:hence; x = 10, y = 30 or (10, 30) 
Method II: Elimination of one Variable by Addition or Substitution   Procedure  Example This method is also very useful; it requires that the coefficient of one of the variables (x or y) be equal in absolute value. Your job is to identify or use multiplier(s) to attain what is desired  By observation: decide on the variable that is to have equal or opposite coefficients. If necessary we will use multipliers.  We will now add or subtract to combine the equations; this reduces the pair of equations to one equation in one variable.  Solve for this variable.  Take this new value and use it to substitute it in one of the ‘original’ equation, the second variable can now be solve for.  As always we will check both values in each ‘original equation.’  Question:Solve the system of equations:   Solution: We will multiply the 1st equation by 2 and the 2nd by 3: è now we substitute (replace) x = 1 into one of the ‘original equations’ and solve for y:  check both values of x and y into each ‘original equations’:hence; x = 1, y = − 2 or (1, − 2) 
Exercise: Use an appropriate method to solve the following system of equations  1) Solve the system:        a) (2, 6) b) (10, -2) c) (6, 2) d) (4, 4)2) Solve the system:        a) x = 2, y =1 b) x = 1, y = 2c) x = -1, y = 3 d) x = -1, y = 23) Solve the system: For the y−value       a) 1b) 2c) -2d) 2.5 4) The value of y=?3x-2y=12x+5y=7        a) 0b) 1c) 2d) -15) Find the x−value of the solution to the system of equations: 2x + y = 3 5x + 2y = -3    a) 3b) -1c) 1d) -36) Find the y−value for the system of linear equations: x + 3y = 2 3x + 8y = -4   a) -2b) 10c) 6d) -107) Find the value of y  3x-2y=12x+5y=7     a) 0b) 1c) 2d) -18) Find the y value of the solution to the system of equations: x-3y=-3x+6y=15     a) -3b) 2c) 3d) -29) Find the x value to the solution of the system of equations:5x+3y=83x+y=4    a) 1b) 2c) -1d) -2 10) Find the value of x=? 3x-2y=12x+5y=7     a) 0b) 1c) 2d) -1
Key Focus:Understand the difference between equations and inequalities.  Overview of Concepts: Here the students learn to represent the solution of equations and inequalities. Also comprehend the inclusion of the absolute value symbol  Contents: Absolute Value Equations:An absolute value is defined as the nonnegative equivalent number that is equal numerically to a given real number. Mathematically this is represented as:       Example 1: If x = 5, then = 3, since 3 > 0    Example 2: If x = −5, then = − (−3) = 3 since 3 < 0 
Example 3: solve:  By definition absolute value leads us to two possible equations:  Then we check both values in the original equation:   Hence the solution set is   Solving and Graphing InequalitiesAs in the case of solving equations, there are certain manipulations of the inequality which do not change the inequality symbol. Here is a list of “permissible” manipulations: Rule 1. Adding/subtracting the same number on both sides.  Example: Solve the inequality x – 3 < 5 x – 3 < 5 (add 3 to both sides) x – 3 + 3 < 5 + 3 x < 8 Rule 2. Dividing or multiplying by a positive number  Example: Solve 4x > 24 4x > 24 (divide both sides by 4)4×4>244 Hence x > 6  Note: You can change the orientation of the sides of inequality as long it points and opens up to the same values. Example: x < 5 is all numbers smaller or less than 5 while 5 > x means 5 is bigger than or more than the unknown number x both these statements are equivalent.
 Rule 3However, when you divide or multiply by a negative number you change the sense of the inequality  Example: -x < 7 becomes -x-1>7-1  or x >-7 Note: To understand why to switch the inequality, consider this, while $50 < $70 however -$50 >-$70 Thus a debt of $50 is a higher value of money that a debt of $70.  Example : Solve the inequality 2x + 5 < 15 and graph it on the number line. The basic strategy for inequalities and equations is the same: isolate x on one side, and put the number terms on the other side.   2x + 5 < 15 Subtract 5 on both sides   2x + 5 -5 < 15 – 5  2x < 10 Divide by 2 both sides 2×2<102</sup>