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Beginning Algebra Lesson 1: Section 1

by Viken Kiledjian Section 1: This section is about writing

Sets and the Absolute Value and Opposite of a Number.

A set is a grouping of objects. The objects in the set are called its elements. The set that contains no elements is called the null set.

A) SET NOTATION:

Roster Method: We enclose the elements of the set within braces.

ex. A = {1, 2, 4, 6}

The name of this set is A. To indicate that 2 is an element of this set, we write


by Viken Kiledjian 2 {1, 2, 4, 6} or 2 A

And to indicate that another number like 5 is not a member of this set, we would write

5 {1, 2, 4, 6} or 5 A

SETS OF NUMBERS:

a) Natural Numbers include the numbers starting from 1 then 2, 3, 4, 5, all the way to infinity.

b) Whole Numbers include the numbers starting from 0 then 1, 2, 3, 4, all the way to infinity.

c) Integers include numbers such as

-4, -3, -2, -1, 0, 1, 2, 3, 4 going all the way to negative and positive infinity.

(The Natural Numbers are a subset of Whole Numbers and Integers. The Whole Numbers are a subset of Integers)


by Viken Kiledjian d) Prime Numbers are the natural numbers greater than 1

that are divisible by themselves and 1.

Examples: 2,3,5,7,11,13,17 … are all prime numbers

ORDERING INTEGERS: To order integer numbers, imagine

putting them on a number line such as

| | | | | | | | | | | | | | |

-5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9

The more to the right a number is, the bigger it is. That’s the rule!!

Examples: 4 > 1, 5>-1, 0> -2 because it is to the right of -2

-4 > -7 because -4 is to the right of -7

-6 < 2 because -6 is to the left of 2


by Viken Kiledjian B) ASOLUTE VALUE OF A NUMBER: measures how far the

number is from the origin (which is the number 0 on the number line). The symbol for absolute value is | |

Ex: |4| = 4, |0|= 0, |-3| = 3, |-(-4)| = 4 because the two negatives inside the absolute value cancel each other and absolute value of 4 becomes 4

The OPPOSITE OR ADDITIVE INVERSE of a number is the Number which added to the original number yields a Zero.

Ex: The additive inverse of 5 is -5

The additive inverse of -6 is 6. The additive inverse of 0 is 0.

Example: Put the correct inequality sign between these two numbers: Additive Inverse of 6 Absolute Value of -4.

Answer: The 1st one is -6 and the 2nd one is 4. The answer is <


by Viken Kiledjian Section 2: This section concerns the Properties of real numbers and

how to Add and Subtract them.

ADDING and SUBTRACTING REAL NUMBERS:

Ex 1: 5 – 7 Start out at the point 5 on the number line and go 7

units to the left. It will take you to -2. Another way

you can think of this, is that 5 – 7 is the opposite to 7 – 5. Since, 7 – 5 gives you 2, then 5 – 7 must yield -2!!

Ex 2: -4 – 9 Start out at the point -4 on the number line and go 9

units to the left. It will take you to -13. Another way

you can think of this, is that -4 – 9 is the opposite to 4 + 9.

Since, 4 + 9 gives you 13, then -4 – 9 must give -13!!

Ex 3: -3 + 11 Start out at -3 and go 11 units to the right. The answer is 8. You can also say that -3 + 11 = 11 – 3 = 8!!


by Viken Kiledjian Section 4: In this section, we learn how to simplify fraction and to

transform a fraction into a decimal and percentage and vice versa. We also learn how to add, subtract, divide and multiply fractions.

A) Writing fractions in simplest form

When reducing a fraction down, the key is to look for the greatest Factor of the Numerator and Denominator.

Ex : 16/60 4 is the Greatest Common Factor (GCF) of 16 and 60.

Dividing 16 by 4 yields 4 and 60 by 4 yields 15. The answer is 4/15.

Writing fractions as Decimals.

The traditional way of doing this is to divide the denominator into the numerator, adding zeros after the numerator’s decimal when necessary. However, in the present age of calculators, you can just punch in the numerator and divide it by the denominator.

Ex: In the above example, 16/60 yields .2666666… repeating 6’s.


by Viken Kiledjian B) Transforming from a Percentage to a Fraction or Decimal

and Vice Versa.

When going from a Decimal to a Percentage, multiply the decimal by

100 and add a Percent Sign. When doing the opposite procedure,

divide by a 100 and remove the Percent sign.

When going from a Fraction to a Percentage, first express the fraction

as a decimal and then multiply by a 100 and add the percent sign.

When doing the opposite, divide the percent by 100 and remove

the percent sign. Then change the decimal to fraction!!

Ex 1: In the above example, 16/60 = .2666666… = 26.666..%

Ex 2: Change 15.5% into decimal and a fraction.

To change to decimal, divide by 100 (move the decimal point 2 places

to the left) and you will get .155


by Viken Kiledjian To change this to a fraction, place the .155 over a thousand (since the

1st decimal place is the 1/10th, the 2nd is the 1/100th, and the 3rd is

the 1/1000th place) and reduce it!!

.155 = 155/1000 Divide the numerator and denominator by 5!!

= 31/200 This is the answer, since it can’t be reduced further.

(You can check this with your calculator now; divide 31 by 200 and

you should get .155!!!)

C) Adding and Subtracting Fractions: The key to adding and

subtracting fractions is to make sure that they all have the same

denominator. If they don’t, then you have to change them so that

they share the same denominator (Least Common Denominator)

Ex : 2/5 – 3/8 + 1/10 The LCD for 5, 8, and 10 is 40. Now, we have

to write all of the fractions over a denominator of 40.


by Viken Kiledjian 2/5 = 16/40, In order to transform a denominator of 5 into a 40, we

multiply both the top and the bottom by 8, which yields 16/40.

3/8 = 15/40, In order to transform a denominator of 8 into a 40, we

multiply both the top and the bottom by 5, which yield 15/40.

1/10 = 4/40. Multiply both the top and bottom by 4!!

The answer is: 16/40 – 15/40 + 4/40 = (16 – 15 + 4)/40 = 5/40 = 1/8

D) Multiplying and Dividing Fractions: When multiplying

fractions, just multiply their numerators and denominators and

then simplify. Sometimes it saves some time to simplify the

denominator of one fraction with the numerator of the other one

and vice versa BEFORE multiplying them. When dividing

fractions, multiply the 1st fraction by the Reciprocal of the 2nd.

Beginning Algebra Lecture 1: Section 3

and Section 4 by Viken Kiledjian Ex 1:

Notice, that I divided the 3 into the 6 on top before multiplying them!

Ex 2:

Section 4: This section is about simplifying Exponents and long expressions using the correct Order of Operations.

A) Exponents are a short way of expressing many multiples of a number with itself. When you Multiply a Negative number by itself an Even number of times, you will get a Positive answer.

Ex 1: (-5)4 = (-5)(-5)(-5)(-5) = 625!!

Ex 2: (-2/3)3 = (-2/3)(-2/3)(-2/3) = -8/27!! You can also think of this as (-2)3/(3)3 = ((-2)(-2)(-2))/((3)(3)(3)) = -8/27!!

7

4

7

4

7

2

1

2

7

6

3

2

61

3

1

2

1

15

5

2

2

15

5

4

15

2

5

4


by Viken Kiledjian B) ORDER OF OPERATIONS: In an expression which contains

some addition, subtraction, multiplication and division,

multiplication and division always get the priority unless the

addition and subtraction are put in parentheses!!!!!

Ex. 2 – 3.5 = 2 – 15 = -13 unless the problem is (2-3).5 = -5

4 + 2/6 = 4 + 1/3 = 41/3 unless the problem is (4+2)/6 = 1

notice that I divided the 4 by 4 first since the order of precedence

between a division and multiplication sign is from left to right!!

423213132134432

7

32

7

2322

29

22162

243

22422

2


by Viken Kiledjian Section 1: This section is about evaluating variable expressions.

EVALUATING VARIABLE EXPRESSIONS: This Section

essentially reinforces the same skills of the previous Chapter,

except now you will work with variables.

Ex. Assume a = 2, b = -3, c = -2, then evaluate these expressions

1) a + b x c = 2 + (-3) x (-2) = 2 + 6 = 8 You just plug in the

variables and follow the order of operations rules. Practice the

order of operations before you move on to these!!!!!!

2)

4

74

7

4

310

)1(3

3)2()5(

2)2(3

)3()2()3(2

acb

bcba


by Viken Kiledjian Section 2: This section is about Simplifying variable expressions

using the Properties of Multiplication and Division that we

learned in Chapter 1. I will concentrate on Objective D.

D) Simplifying General Variable Expressions.

Ex 1: 3(a – b) – (a + b) Distribute the 3 into the 1st parenthesis

and the -1 into the 2nd parenthesis, which will change their sign

= 3a – 3b – a – b = 3a – a – 3b – b = 2a – 4b!!

Ex 2: -2[3x + 2(4 – x)] Distribute the 2 into the (4 – x) and

simplify the term in the brackets. Then, distribute the -2!!

= -2[3x + 8 – 2x] = -2[x + 8] = -2x – 16!!

Ex 3: -5x – 2[2x – 4(x + 7)] – 6 = -5x – 2[2x – 4x – 28] – 6 =

-5x – 2[-2x – 28] – 6 = -5x + 4x + 56 – 6 = -x + 50!!!


by Viken Kiledjian Section 3: In this section, we learn how to Translate a Verbal

Expression and then simplify it, if need be.

The expressions that are trickier tend to be the ones involving

Subtraction and Division. Addition and Multiplication are easier.

For example: “8 less than a number” means x – 8 (put the # first)

“A number decreased by 8” also means x – 8

“The difference between a number and 8” also means x – 8

“The quotient of a number and 8” means x/8

“The ratio of a number and 8” also means x/8

Ex 1: s increased by the quotient of 4 and s = s + 4/s!!

Ex 2: A number decreased by the difference between 8 and the

number = x – (8 – x) = x – 8 + x = 2x – 8 !!!!


by Viken Kiledjian Section 1: is about solving algebraic equations of the form

ax + b = c using the properties of equality.

A) SOLVING EQUATIONS using the properties of equality.

a) If a = b, then ca = cb (you can multiply both sides of the equation by the same number)

b) If a = b, then c + a = c + b (you can add the same number to both sides of the equation)

Given these two properties, you can solve any linear equation by combining like terms. Like terms are any terms that look similar and have the same power of the x.

Ex 1: Solve for x: 2x/5 – 3 = -2 Multiply everything by 5!!

(5)2x/5 – (5)(3) = (5)(-2) = 2x – 15 = -10 Add 15 to both sides

2x – 15 + 15 = -10 + 15 -----> 2x = 5 -------> x = 5/2!!


by Viken Kiledjian Ex 2: Solve for x: 10 = -3x + 6 Subtract 6 from both sides

10 – 6 = -3x + 6 – 6 -----> 4 = -3x Divide both sides by -3!!

4/(-3) = -3x/(-3) -------> x = -4/3

B) Application Problems involving Markup and Discount.

Here are the equation for each: S = C + rC, S = R – rR

S = selling price, C = regular cost of item, r = markup rate in decimals or discount rate in decimals, R = regular price of item

Ex 1: A sofa costing $320 is sold for $479. Find the markup rate.

Answer: C = $320, S = $479, solve for r --> 479 = 320 + r320

Subtract 320 from both sides, 479 – 320 = r320 -----> 159 = r320

Now, divide both sides by 320, 159/320 = r320/320 ----> .497 = r

Multiply by 100 to change to percentage, ----> r = 49.7% markup


by Viken Kiledjian Ex 2: A markup rate of 25% is used on a computer that has a

selling price of $2187.50. Find the cost of the computer.

Answer: Here C = unknown, S = 2187.5, r = 25% = .25

2187.5 = C + .25C -------> 2187.50 = 1.25C ---->2187.5/1.25 = C

C = $1750 is the regular cost of the computer

Ex 3: A battery with a discount price of $65 is on sale for 22% off the regular price. Find the regular price.

Answer: R = unknown, S = $65, r = 22% = .22

65 = R - .22R -------> 65 = .78R ----> 65/.78 = R ----> R = $83.33

Ex 4: A luggage set with a regular price of $178 is on sale for

$103.24. Find the discount rate. 103.24 = 178 – r178 ------->

103.24 – 178 = -r178 -----> -74.22 = -r178 ----> -74.22/(-178) = r

-------> .417 = r -----------> r = 41.7% discount rate


by Viken Kiledjian Section 2: In section 2, we do some more equations, but this time

the x is found on both sides of the equal sign and sometimes in a parenthesis. We will also do some Application problems.

Ex 1: 4(3 – x) +2 = x - 5 first Distribute the 4 into the 3 – x

12 – 4x + 2 = x – 5 then combine the 12 and 2

10 – 4x = x – 5 then add 4x to both sides (usually it is good practice to collect the x’s to the side which makes their coefficient positive!!)

10 = 5x – 5 now add 5 to both sides!!!

15 = 5x finally divide both side by 5

x = 3 this is the answer!!!!

Ex 2: -4[x – 2(2x – 3)] + 1 = 2x – 3 First, Distribute the -2 into the parenthesis, ----> -4[x – 4x + 6] + 1 = 2x – 3 ------->

-4[-3x + 6] + 1 = 2x – 3 Now, distribute the -4 into the brackets


by Viken Kiledjian 12x – 24 + 1 = 2x – 3 --------> 12x – 23 = 2x – 3 Now, add 23 to

both sides ------> 12x = 2x – 3 + 23 ---> 12x = 2x + 20

Next, subtract 2x from both sides: 12x – 2x = 20 ----> 10x = 20

Finally, divide both sides by 10 ------> x = 2!!!

Application Problems: The Lever Problem The equation for the lever problem is F1x = F2(d – x)

Ex 1: An adult and a child are on a seesaw 14 ft long. The adult weighs 175 lb and the child weighs 70 lb. How many feet from

the child must the fulcrum be placed so that the seesaw balances?

Answer: Let x = the distance of the child to the fulcrum, d = 14ft

F1 = 70 lbs, since x refers to the distance of the child, F2 = 175 lbs

The equation becomes ----> 70x = 175(14 – x) = 2450 – 175x ---->

Add 175x to both sides ---> 245x = 2450 -------> x = 10 feet


and Section 3 by Viken Kiledjian Ex 2: A screwdriver 9 in. long is used as a lever to open a can of

paint. The tip of the screwdriver is placed under the lip of the can with the fulcrum 0.15 in. from the lip. A force of 30 lb is applied to the other end. Find the force on the lip of the can.

Answer: d = 9 inches, F1 = unknown, x = .15 inches, F2 = 30 lbs

The equation becomes: F1(.15) = 30(9 – .15) = 30(9) – 30(.15) --------> F1(.15) = 265.5 -------> F1 = 1770 lbs

Section 3: In this section, we will solve integer problems and general problems involving unknown variables.

Ex 1: Four times the sum of twice a number and three is twelve. Find the number.

Answer: Let x = the unknown number. The above sentence translates to the following equation: 4(2x + 3) = 12 --------->


by Viken Kiledjian 8x + 12 = 12 ---------> 8x = 0 --------> x = 0

Therefore, the number equals 0!!

Ex 2: The sum of two numbers is fourteen. The difference between

two times the smaller and the larger is one. Find the numbers.

Answer: Let x = the smaller number, then 14 – x = the larger

numbers, since their sum has to be equal to 14. The sentence

translates to the following equation: 2x – (14 – x) = 1 Now,

distribute the –1 into the parenthesis ---> 2x – 14 + x = 1 ---->

3x – 14 = 1 -------> 3x = 15 ------> x = 5, Therefore, 14 –x = 9

The smaller number equals 5 and the larger number equals 9!!

Ex 3: Find two consecutive even integers such that three times the

first equals twice the second.


by Viken Kiledjian Answer: Let x = the smaller even number, then x + 2 = the larger

even number, since there is always a difference of 2 between two consecutive even or odd numbers. ((There is No Difference to how you would set up this problem if the numbers were odd instead of even. Just set it up the same and the answer will reveal whether they are even or odd.)) The equation becomes ----

-----> 3x = 2(x + 2) ------> 3x = 2x + 4 -----> x = 4

The smaller even integer is 4 and the larger is 6!!!

Ex 4: The total cost to paint the inside of a house was $1346. This cost included $125 for materials and $33 per hour for labor. How many hours of labor were required to paint the house?

Answer: Let x = number of hours required to paint. The equation becomes: 1346 = 125 + 33x ------> 1221 = 33x ---> x = 37 hrs


by Viken Kiledjian Section 4: In this section, we will do examples which involve

Perimeter problems regarding Rectangles and Triangles.

Ex. 1 The width of a rectangular swimming pool is one-third its length. If its perimeter is 96 meters, find the dimensions of the pool.

Answer: Let w = width of rectangular in meters

Then L = 3w (the opposite of one-third is 3. I did this to avoid fractions. I’ll show you the setup for the other way also!!)

Now the Perimeter of a rectangle is the equivalent of Circumference for a circle. It is the distance all the way around the rectangle.

P = L + L + w + w = 2L + 2w therefore

96 = 2(3w) + 2w I substituted 3w in the place of L.

96 = 6w + 2w = 8w Therefore, w = 96/8 = 12 feet


by Viken Kiledjian And L = 3w = 3(12) = 36 feet. The final answer is

The width is 12 feet and the length is 36 feet.

((Now check your answer. Twice 36 equals 72 and twice 12 equals 24. If you add 72 to 24, you get 96 feet))

The other setup would be this way.

Let L = length of the rectangle

Then w = L/3 is the width, since it is one-third of the length.

P = 2L + 2w, so 96 = 2L + 2(L/3) = 2L + 2L/3

96 = 6L/3 + 2L/3 Get a common denominator so you can add them

96 = 8L/3 Now multiply both sides by 3/8

3/8(96) = L, and therefore L = 36 feet. Notice how much uglier it was. So, to avoid this, let your variable equal the smaller one!!


and Section 5 by Viken Kiledjian Ex 2: The perimeter of a triangle is 33 ft. One side of the triangle

is 1 foot longer than the second side. The third side is 2 feet

longer than the second side. Find the length of each side.

Answer: Let x = length of shortest side, which is the 2nd side.

Then, the 1st side = x + 1, and the third side = x + 2. The equation

becomes: x + (x + 1) + (x + 2) = 33 -------> 3x + 3 = 33 ---->

3x = 30 -----> x = 10 , so x + 1 = 11, and x + 2 = 12

The shortest side equals 10 ft, the next is 11 ft, the largest is 12 ft!

Section 5: In this section, we solve some more Application

Problems. In Part A, we solve Value Mixture problems, in

Part B, we solve Percent Mixture problems, and in Part C,

we solve Uniform Motion Problems.


by Viken Kiledjian A) Value Mixture Problems: I will illustrate each kind with an

example, explaining as I solve them.

Ex: A mixture of candy is made to sell 89 cents per pound. If 32 pounds of a cheaper candy, selling for 80 cents per pound, are used along with 12 pounds of a more expensive candy, find the price per pound of the better candy.

Answer: We start with 32 lbs of a candy selling at $.80/per pound, and mix with it 12 lbs of candy selling at x dollars per pound. We end up with 44 lbs of candy, right??? And the problem tells us that the price of the mixture = $.89/per pound. The idea here is:

Price of cheaper candy + Price of better candy = Price of Mixture

.80(32) + x(12) = .89(44) ---> 25.6 + 12x = 39.16 --->12x = 13.56

The better candy costs $1.13/per pound.


by Viken Kiledjian B) Percent Mixture Problems:

Ex: How much acid must be added to 60 grams of a solution that is 65% acid to obtain a new solution that is 75% acid?

This is like a chemistry problem. The idea here is that:

Original amount of acid + Amount of added acid = Total amount of acid.

Answer: Let x = amount of added acid.

The tricky part of this problem is to realize that when you add x grams of acid to 60 grams of a solution, the total mass of the mixture become ----> 60 + x!!!!

.65(60) + x = .75(60 + x) ---> 39 + x = 45 + .75x Subtract .75x from both sides and 39 from both sides ---> .25x = 6 -------->

x = 24 Therefore, the answer is 24 grams of acid must be added!!


by Viken Kiledjian Let me show you how to check this one.

There were 39 grams of acid in the original solution. If you add 24 grams of acid to it, you end up with 63 grams of acid. However, the total solution will now have a mass of 60 + 24 = 84 grams.

If you have 63 grams of acid in 84 grams of solution, what percentage is that??? 63/84 = .75 = 75% (Wow, it worked!!)

C) UNIFORM MOTION PROBLEMS: These tend to be the hardest for people because they are like physics problems.

Ex 1: A cyclist leaves Las Vegas riding at the rate of 18 mph. One hour later, a car leaves Las Vegas going 45 mph in the same direction. How long will it take the car to overtake the cyclist?

Answer: The Question is asking how long the car will take to reach the cyclist, So let t = time the car has been traveling


by Viken Kiledjian Then t + 1 = time the cyclist has been traveling, since she set out

1 hour before the car.

The distance traveled by cyclist = 18(t + 1)

The distance traveled by car = 45t Now, set them equal

18(t + 1) = 45t ----> 18t + 18 = 45 ---> 18t = 27 ---> t = 1.5

It will take the car 1.5 hours to catch up to the cyclist.

Ex. 2: Sarah walked north at the rate of 3mph and returned at the rate of 4mph. How many miles did she walk if the round trip took 3.5 hours?

Let t = time for the trip north

Then 3.5 – t = time for the return trip (since the total is 3.5 hours)

The distance traveled north = 3t (Remember that Distance

The distance for the return trip = 4(3.5 – t) Equals Rate x Time)


by Viken Kiledjian These two distances must be equal because she ended up where she

started, so set them equal to each other!!

3t = 4(3.5 – t) = 14 – 4t ---> 7t = 14 ---> t = 2 hours.

This means the trip north took 2 hours and the return trip took 1.5

hours. The problem wants to know the total round trip distance.

The distance north = 3t = 3(2) = 6 miles

The return distance = 4(3.5 – t) = 4(1.5) = 6 miles (Are you

surprised that it is the same??? You shouldn’t be!!)

Sarah walked a total distance of 12 miles!!!!!

The main formula here is::: Distance = rate x time

Beginning Algebra Lecture 4: Section 1 by Viken Kiledjian

Section 1: In this section, we define what a

Monomial is and how to Multiply

monomials and simplify their powers.

A Monomial is an expression such as

Axbyczd such that “b”, “c”, and “d” are All

positive integers or zero. They can

NOT be fractions or negative.

A) Multiplying Monomials by

Monomials: When you multiply two

monomials, you add the powers of the

variables that they share in common.

You also multiply their coefficients.

Ex: (2x2y4)(-4x3yz2) = (2)(-4)x2 + 3 y4 + 1 z2

= -8x5y5z2


by Viken Kiledjian B) Simplifying Powers of Monomials: When a Monomial

is raised to a certain power, it is like it is being multiplying by itself that many times.

Ex 1: (x2y)3 = (x2y)(x2y)(x2y) = x2+2+2y1+1+1 = x6y3

The quick way to do this is to multiply the power of each variable by the power that the whole monomial is being raised to. This is how it’s done ------------>

Ex 1: (x2y)3 = x(2)(3)y(1)(3) = x6y3 That’s much quicker!

Ex 2: (-2x3y2z4)3 = (-2)3x(3)(3)y(2)(3)z(4)(3) = -8x9y6z12

Now, let’s see an example that includes simplifying a power of a monomial and multiplying two monomials.

Ex 3: (-3x3)(-2xy3)2 = (-3x3)[(-2)2x(1)(2)y(3)(2)] =

(-3x3)(4)x2y6 = (-3)(4)x3+2y6 = -12x5y6

Section 2: In this Section, we learn how to multiply two monomials, a monomial and a polynomial, two binomials, and two binomials with special products.

A) Multiplying a Monomial by a Polynomial: Multiply the Monomial by each term of the Polynomial. In other words, Distribute the Monomial into the Polynomial.

Ex: -2x2yz3(x4 – 3yz2 + 2x2y3) = -2x2+4yz3 – (2)(-3)x2y1+1z3+2 – (2)(2)x2+2y1+3z3 = -2x6yz3 + 6x2y2z5 – 4x4y4z3

B) Multiplying Polynomials by Polynomials: Multiply each term of one Polynomial by each term of the other Polynomial. If both Polynomials are Binomials, there will be 4 multiplications. If one of them is a Binomial and the other is a Trinomial, there will be 6 multiplications overall. Let’s do some examples:

Ex 1: (2x – 3y)(3x + 6y) = (2x)(3x) + (2x)(6y) – (3y)(3x) – (3y)(6y) = 6x2 + 12xy – 9yx – 18y2 = 6x2 + 3xy – 18y2


by Viken Kiledjian


by Viken Kiledjian Ex 2: (x – 3y)(x2 + 3xy + 9y2) This is a binomial times a trinomial

= (x)(x2 + 3xy + 9y2) – (3y)(x2 + 3xy + 9y2) Distribute each one now

= x3 + 3x2y + 9xy2 – 3x2y – 9xy2 – 27y3 There are 6 terms. Now,

= x3 – 27y3 combine like terms, and the middle ones cancel out!!

C) Special Products: are short ways of doing a few specific kind of

multiplications. Although you can still do these multiplications

the regular way as in Ex 1, it is recommended that you learn

these short cuts to speed up your work.

There are 3 of them. Here are the formulas:

(a + b)2 = a2 + 2ab + b2 “a” and “b” can stand for any 2 expressions

(a – b)2 = a2 – 2ab + b2 For example: a = 4x and b = 3y

(a + b)(a – b) = a2 – b2 Let’s do a few examples to illustrate this.


by Viken Kiledjian Ex 1: (4x – 3y)2 Here a = 4x and b = 4y Use formula 2 above

= (4x)2 – 2(4x)(3y) + (3y)2 = 16x2 – 24xy + 9y2 Here is the regular way of doing this:

(4x – 3y)2 = (4x – 3y)(4x – 3y) = (4x)2 – (4x)(3y) – (3y)(4x) + (3y)2

= 16x2 – 12xy – 12xy + 9y2 = Gives us the same answer!!!!!!!!!!!

Ex 2: (5x – 2y)(5x + 2y) Here a = 5x and b = 2y Use Formula 3

= (5x)2 – (2y)2 = 25x2 – 4y2 That’s pretty quick, wouldn’t you say!!

Here is the regular way of doing it:

(5x – 2y)(5x + 2y) = (5x)2 + (5x)(2y) – (2y)(5x) – (2y)2 = 25x2 +10xy

– 10xy – 4y2---> the middle terms cancel and we get the same answer


by Viken Kiledjian Section 3 is about dividing Monomials and expressing numbers in

Scientific Notation.

A) When dividing Monomials or raising them to some power,

we’ve got to follow the Law of Exponents. Here they are:

PROPERTIES OF EXPONENTS:

a) Product Rule of Exponents: xmxn = xm+n

b) Power Rule of Exponents: (xm)n = xmn and (xy)n = xn yn

and if y equals not zero.

c) Law of Zero Exponents: if x is not 0, then x0 = 1

d) Law of Negative Exponents: if x is not 0, then

and

n

nn

y

x

y

x

n

n

xx

1 n

nx

x

1


by Viken Kiledjian e) Quotient Rule of Exponents: if x is not 0, then

f) Theorem of Reciprocals: if x and y are not 0, then

Here are examples of each one. It is very important to practice these until you know them by heart!!!!

a) 2324 = 23+4 = 27 = 128, 4(-5)47 = 4(-5 + 7) = 42 = 16

b) (42)3 = 46 = 4096 (notice that you don’t add the 2 and 3!!!)

c) (4.2)3 = 43 23 = 64.8 = 512 (here you don’t have to use the law;

you can just multiply 4 by 2 and get 8 and 8 cubed is 512)

nm

n

m

xx

x

nn

x

y

y

x


by Viken Kiledjian here is another example of

law B: -------------------->

d) and

e)

Here are some more complex examples which combine the Rules!!

a)

b)

27

8

3

2

3

23

33

9

1

3

13

2

2 644

4

1 3

3

81333

3 426

2

6

4

6

42

6642232

4222

x

y

x

yyxyx

34

3312430)1(213

31

023

zx

yzyxzyx

zxy

zyx


by Viken Kiledjian c)

B) Scientific Notation: Scientific Notation is used to express large or small numbers and to make calculations easier.

Ex .000000456 = 4.56 x 10-7, .0000389 = 3.89 x 10-5

345,023,000,000 = 3.45023 x 1011, 20,000,000,000,000 = 2 x 1013

Notice that you always move the decimal to the right of the 1st digit!

Now, we can also go backwards.

Ex 2.4 x 10-4 = .00024 (you move the decimal point 4 places to the left)

3.5 x 100 = 3.5 since 10 to the power 0 is equal to 1

4.62 x 105 = 462,000 (move decimal point 5 places to the right!)

846

864

)2)(4()2)(2()2)(3(

)2)(4()2)(3()2)(2(2

423

432

2

2

2

2

2

2

yx

yx

yx

yx

yx

yx

162164610)8(8)4(6)6(4 102422 yxyxyx


by Viken Kiledjian

Section 4: In this section, we learn how to divide a Polynomial by a Monomial and a Polynomial by another Polynomial.

A) When we divide a Polynomial by a Monomial, we divide each term of the Polynomial by the Monomial.

Ex:

B) When we divide a Polynomial by another Polynomial, we use a process similar to how Long Division is done with numbers.

Suppose we wanted to divide 4056 by 72. The 4056 is called the Dividend and the 72 the Divisor. The answer of the division is called the Quotient. If the 72 does not divide perfectly into the 4056, then there will be a Remainder.

The general equation is: Dividend = Quotient + Remainder

Divisor Divisor

xyy

xy

y

xy

x

xy

xy

xy

yxxy 3

2

5

2

6

2

5

2

2

2

652 233


by Viken Kiledjian You put the 72 to the left of the 4056 and a bar between them.

Then you ask yourself “How many times does the 72 go into the

405?” The answer is 5 times. Then you put the 5 over the 4056

and multiply it by 72 which gives 360. Now, write 360 below the

405 of the 4056 number. Subtract 405-360 and you get 45. Now,

bring down the 6 and attach it to the 45 and you get 456. Now,

repeat the process and you get a remainder of 24. This is

illustrated in the following picture:

In light of the general equations, what this

means is: 4056 = 56 + 24

72 72

24/72 = .333333 and indeed, when you

do this division in the calculator, you get

56.33333333

56

0024

04320456

36004056

72


by Viken Kiledjian

Now, Let’s do an actual Example:

Ex: -------------->

Step 1: “x” goes into (4x2) how many times? Answer: 4x

Step 2: Put the (4x) on top and multiply it by (x + 1).

Step 3: Write this answer below the 4x2 -3x and Subtract.

Step 4: Bring down the (+7) from the original Polynomial.

Step 5: “x” goes into (-7x) how many times? Answer: -7

Step 6: Put the (-7) on top and multiply it by (x + 1).Answer: -7x - 7

1

734 2

x

xx 7341 2 xxx

xx

xx

xxxx

x

04

70

4434

12

2

2


by Viken Kiledjian Step 7: Write this answer below the -7x + 7 and Subtract.

Here is the picture of what happens so far:

Notice that when you subtract (-7x)

from (-7x), you should get Zero, but

when you subtract (-7) from (7), the

answer will be (14). This is because

(7) – (-7) = 14. A common mistake

here is to Add the two equations and

obtain a Remainder of zero. Our Remainder is 14!

Therefore, the answer looks like the following:

74

1400

770770

044734

1

2

2

2

2

2

x

xx

xxxx

xxxx

x

1

1474

1

734 2

xx

x

xx


Section 1 is about Factoring Monomials from Polynomials and to

factor Polynomials by Grouping.

A) Factoring out the Greatest Common Factor (GCF) The first

step is to always look for the GCF. This includes the greatest

number and greatest powers of the variables that ALL the terms

of the polynomial share in common.

Ex 1: 13ab2c3 – 26a3b2c The GCF = 13ab2c because both terms

have at least a 13 in them and “a” to the power of 1 and “b” to

the power of 2 and “c” to the power of 1. Factor out this GCF!

= 13ab2c(c2 – 2a2) You leave in the parenthesis whatever is left over

Ex 2: 18y2z2 + 12y2z3 – 24y4z3 The GCF = 6y2z2 Factor this out!!

6y2z2(3 + 2z – 4y2z) Leave in the parenthesis whatever is left over


by Viken Kiledjian B) Factoring by Grouping: is the second major technique used in

simplifying and factoring expressions. You group together like

terms and factor the GCF from those and then you might be able

to factor out one more time from the entire expression.

Ex : x2 + 4y – xy – 4x Group the 1st and 3rd terms and the 2nd and

x2 – xy + 4y – 4x 4th terms. Now factor out the GCF of each pair.

x(x – y) + 4(y – x) Now, change the sign of (y – x) = -(x – y)!!!

x(x – y) – 4(x – y) Finally, factor out the expression (x – y)

(x – y)(x – 4) You can check your answer this way: Multiply out

these 2 binomials and you should get the original expression!!!

(x – y)(x – 4) = x2 – 4x – yx + 4y It Checks out!!!


by Viken Kiledjian

Section 2: This section is about factoring Trinomials of the form

x2 + bx + c where b, c may be any numbers and to Factor a

Polynomial completely.

A) Let’s First concentrate on factoring trinomials.

Ex 1: x2 + 4x + 3 Here b = 4 and c = 3. This is the easiest case

because both “b” and “c” are positives. Therefore, the 2

parenthesis will be BOTH positives --------> = (x + 3)(x + 1)

Notice that the factors of 3 are 3 and 1 and they add up to 4!!!

Ex 2: x2 – 7x + 12 This is a little harder but not much. Here “b” is

a negative and “c” is a positive. In these cases, BOTH

parenthesis will have a Negative sign. You have to ask yourself

the question, “What are the factors of 12 which add up to 7?”

It is 4 and 3, right? Therefore the answer is (x – 4)(x – 3)


by Viken Kiledjian Ex 3: x2 + 3x – 28 This is even a little harder. Here “b” is positive

and “c” is negative. In these cases, One of the parenthesis will be positive and the other one a negative. The question now should be “What are the factors of 28 whose difference is 3?”

It is 7 and 4, right? Therefore the answer is (x + 7)(x – 4)

Notice that the 7 has the + sign because the “b” = +3.

If the question was ------> x2 – 3x – 28, the answer is (x – 7)(x + 4)

If the question was ------> x2 + 4x – 28, the answer would be Prime

or NOT Factorable because no factors of 28 have a difference of 4.

B) When you have a General Polynomial, first factor out the GCF from every term of the Polynomial, and then apply the techniques of either Grouping or Factoring Trinomials.

Ex 1: 18 + 7x – x2 = -x2 + 7x + 18 Now, factor out a Negative 1!!

-(x2 – 7x – 18) = - (x – 9)(x + 2) The product of 9 and 2 is 18 and

their difference is 7. Since the 7 is negative, the 9 gets the minus sign


and Section 3 by Viken Kiledjian

Ex 2: 4x2y + 20xy – 56y First, factor out a 4y from each term!!

= 4y(x2 + 5x – 14) = 4y(x + 7)(x – 2) since the product of 7 and 2 is 14 and their difference is 5. 7 gets the positive this time, since the 5 in the middle has a plus sign.

Section 3: In this section, we are going to cover a very useful topic which is about factoring Special Forms.

a2 – b2 = (a – b)(a + b) This is called the Difference of Two Squares

a2 + b2 = NOT Factorable. Let’s do some examples.

Ex 1: 16x4 – 81y2 We can put this in the a2 – b2 form if we write it like this -------------> (4x2) – (9y)2 where a = 4x2 and b = 9y

Therefore the solution is (4x2 – 9y)(4x2 + 9y)

Ex 2: 64r6 – 121s2 = (8r3)2 – (11s)2 = (8r3 – 11s)(8r3 + 11s)

Ex 3: x2y2 – 4 = (xy)2 – 22 = (xy – 2)(xy + 2)


by Viken Kiledjian Then, there are Perfect-Square Trinomials of the form:

(a – b)2 = a2 – 2ab + b2 and (a + b)2 = a2 + 2ab + b2

Ex 1: y2 + 14y + 49 Notice that the last term (49) is a perfect

square whose square root is (7). Also, notice that 2 times (7) is

14, which gives the middle term. Therefore, the trinomial can

be expressed as -------> y2 + 2(7)y + (7)2 This fits into the form

(a + b)2 = a2 + 2ab + b2 where a = 1, b = 7. Therefore, the final

factored form is y2 + 14y + 49 = (y + 7)2

Ex 2: 64x2 – 48xy + 9y2 (64) is a perfect square whose square

root is (8) and (9) is a perfect square whose square root is (3).

So, it can be rewritten as -------------> (8x)2 – 2(8x)(3y) + (3y)2

Since 8 times 3 gives you 24 and 2 x 24 = 48, this works out

well. This fits into the form (a – b)2 = a2 – 2ab + b2 where

a = 8x, b = 3y. Therefore, the final answer is (8x – 3y)2!!!


by Viken Kiledjian Section 4: In this section, we use the techniques of factoring to solve

equations. To do this, we use the Zero-Factor Theorem which states that ------> If ab = 0, then a = 0 or b = 0. We will also solve some Application Problems using this technique.

Ex 1: x2 – 9 = 0 First factor the expression x2 – 9

(x – 3)(x + 3) = 0 Now, treat (x – 3) as “a” and (x + 3) as “b”

Therefore the answer is: x – 3 = 0 OR x + 3 = 0 Now, solve each individual equation for x and you get ---------> x = 3 or x = -3

Ex 2: x2 – 3x = 0 Factor this expression first.

x(x – 3) = 0 Treat x as “a” and (x – 3) as “b” Use the Zero-Factor theorem and you get x = 0 OR x – 3 = 0 Now, solve for x in each equation and you get. x = 0 OR x = 3

Ex 3: x2 = 24 + 5x First gather everything to the left side and factor it out. x2 - 5x – 24 = 0 -------------> (x – 8)(x + 3) = 0

x – 8 = 0 OR x + 3 = 0 Therefore, the answer is x = 8 OR -3


by Viken Kiledjian Ex 4: (x + 4)(x – 1) = 14 Here, you can’t equate x + 4 = 14 and

x – 1 = 14. You have to first multiply out the (x + 4)(x – 1) expression and gather all the terms to the left side and then factor again and then finally solve the equation. Here it is:

(x + 4)(x – 1) = x2 – x + 4x – 4 = 14 -----------> x2 + 3x – 4 – 14 = 0 -------------> x2 + 3x – 18 = 0 ----------> (x + 6)(x – 3) = 0 ---------------> x = -6, x = 3

Now, let’s solve some Application Problems with these techniques.

Ex 1: The sum of the squares of two consecutive positive integers is 85. Find the integers.

Let x = the smaller integer, then x + 1 = the larger integer.

Therefore, x2 + (x + 1)2 = 85 Now, we have to solve for x.

Expand the (x + 1)2 = x2 + 2x + 1 and add this to the x2. You get --------> 2x2 + 2x + 1 = 85 Subtract 85 from both sides.


by Viken Kiledjian 2x2 + 2x – 84 = 0 Divide everything by 2.

x2 + x – 42 = 0 Factor this trinomial (x + 7)(x – 6) = 0

Therefore, x = -7 or x = 6. Since, x has to be a positive integer,

discard the 1st answer. Therefore, x = 6 and x + 1 = 7

The smaller integer is 6 and the larger integer is 7.

Ex 2: One side of a rectangle is three times longer than another. If

its area is 147 square centimeters, find its dimensions.

Let w = the width, then l = the length = 3w (3 times longer)

The Area = lw = (3w)w = 3w2 = 147 Take the 147 to the left side.

3w2 – 147 = 0 Divide everything by 3 --------------> w2 – 49 = 0

(w – 7)(w + 7) = 0 Therefore, the width = 7 or -7. Discard the

negative answer. Width = 7 cm , Length = 3w = 21 cm


by Viken Kiledjian

Ex 3: A woman plans to use one-fourth of her 48 foot by 100 foot

backyard to plant a garden. Find the perimeter of the garden if

the length is to be 40 feet longer than the width.

Let w = width of her garden, then l = 40 + w.

The Area of her garden will be ¼ of the Area of her backyard =

(48)(100) = 4,800 square feet. ¼ of this equals 1,200 square feet.

Therefore, A = lw = (40 + w)w = 1200 ---------> w2 + 40w =

1200

w2 + 40w – 1200 = 0 ---------> (w + 60)(w – 20) = 0 ------------>

w = 20 or w = -60 Discard the negative answer again. Therefore, the w = 20 and l = 40 + w = 60

The Perimeter = 2l + 2w = 2(60) + 2(20) = 120 + 40 = 160 feet


by Viken Kiledjian Ex 4: After how many seconds will an object hit the ground if it was

thrown straight up with an initial velocity of 208 feet per second?

Solution: The equation that you have to use here is h = vt – 16t2

h = the height above the ground that a projectile reaches

v = the initial velocity of the projectile, t = time of flight

In our problem, h = 0 (when it hits the ground, its height above the

ground will be zero) and v = 208 ft/sec Therefore, the equation

becomes ----------> 0 = 208t – 16t2 Factor 16t out of this.

16t(13 – t) = 0 Therefore, t = 0 or t = 13 Discard the 1st answer

Because it represents the original time when the projectile was

launched. The projectile will hit the ground in 13 seconds.


Section 1: is about simplifying Rational

Expressions. A Rational Expression is any

expression which is the Quotient of two

Polynomials. In this section, we will learn

to Simplify, Multiply, and Divide Rational

Expressions.

A) Simplifying Rational Expressions

Ex 1: Simplify Each Rational Expression

Ex 2:

2

5

2

345

23

45

7

5

7

5

21

15

c

ba

c

ba

cb

ba

2)2()2(

)2(233

2

x

x

x

x

x

xx


Ex 3: In Ex 3 & 4, we are going to utilize a technique of

Factoring out a (-1) -----------> (a – b) = -(b – a)

Ex 4:

Ex 5:

Notice that we are utilizing the same

Techniques of Factoring that we learned

In Chapter 5. Take out the GCF first and

Then factor out the remaining Parenthesis.

1)(

)(

))((

))((

pr

pr

qppr

rpqp

xxx

x

x

xx

3)3(

)3(

)3(

3

96 22

22

2

23

3

)2(2

)2)(2(

)44(6

)4(3

24246

123

x

xx

xxx

xx

xxx

xx

)2(2

2

x

x

B) Multiplying Rational Expressions

Ex:

The x2 on the top cross-cancels with

the x3 on the bottom and similar with the y’s.

Also, the (x +1) cross-cancels with the (x + 1)2

C) Dividing Rational Expressions: Just multiply the first

expression by the Reciprocal of the Second.

Ex:


by Viken Kiledjian

22

32

23

2

)1(

)1)(4(

12

45

x

y

x

xx

xx

yx

yx

xx

)1(

)4(

xx

xy

2811

42

)2)(5(

)82(

42

2811

107

282

22

2

2

2

2

xx

xx

xx

xx

xx

xx

xx

xx

5

6

)5(

)6(

)4)(7(

)6)(7(

)2)(5(

)2)(4(

x

x

x

x

xx

xx

xx

xx


by Viken Kiledjian Section 2: This section involves Adding and Subtracting Rational

Expressions, which is a totally Different process than simplifying,

multiplying, or dividing them.

When you add 2 or more Fractions, they must ALL have the same

Denominator which is called the LCD (least common

denominator). If they don’t, then you need to multiply each

denominator by an appropriate number or expression so that it

becomes equal to the LCD. When you do this, you have to also

multiply the numerator by the same expression or number so you

won’t change the Fraction.

Ex 1: Here, the LCD = 120. You need to multiply

8 by (15) in order to achieve this LCD. You

need to multiply 6 by (20) and 10 by (12). However, you have to

also multiply their numerators by these numbers so that the

values of the Fractions are not altered.

10

3

6

1

8

5

59/120 is NOT simplifiable any more so it is the Final Answer.

Ex 2: Here, they already have the Same

Denominator, so just combine the

Numerators, Factor the 9 and cancel.

Ex 3: Here, the LCD = 10ab. Multiply the 5a by

(2b) and the 2b by (5a) to achieve this LCD

Ex 4: Here, the LCD = (x +3)(x + 6).


120

59

120

362075

120

36

120

20

120

75

10)12(

3)12(

6)20(

1)20(

8)15(

5)15(

yx

y

yx

x

99

9)(999

yx

yx

yx

yx

ba 2

3

5

2

ab

ab

ab

a

ab

b

ba

a

ab

b

10

154

10

15

10

4

2)5(

3)5(

5)2(

2)2(

6

4

3

7

x

x

x


Since the Numerator is not Factorable

any further, this is the Final Answer.

Ex 5:

Here, we had to do some work on the original problem in order to

determine what the LCD is. Now, it is evident that the LCD =

(x – 2)(x + 2). The 2nd fraction already has this LCD so nothing

needs to be done with it. You just need to multiply the top and

bottom of the 1st fraction by (x + 2).

)6)(3(

124427

)6)(3(

4)3(

)3)(6(

7)6( 2

xx

xxx

xx

xx

xx

x

)6)(3(

42194 2

xx

xx

)2)(2(

5

2

4

4

5

2

4

4

5

2

422

xxxxxxx

)2)(2(

34

)2)(2(

584

)2)(2(

5

)2)(2(

4)2(

xx

x

xx

x

xxxx

x


Ex 6: Whenever an expression does not have

a Denominator, you can put a (1) for its

Denominator and the LCD = x – 1

Ex 7:

Here, the LCD

is (x + 3)(x – 2)(x – 5). Multiply the top and bottom of each

fraction by whatever it needs to achieve this LCD!!!!!!!

xx

41

24

1

)2)(3(4

1

)6(4

1

2444

1

4424 222

x

xx

x

xx

x

xx

x

xx

1)1(

)4)(1(

1

24

1

4

1

24

x

xx

x

x

x

107

32

152

53

6

12222

xx

x

xx

x

xx

x

)2)(5(

32

)3)(5(

53

)2)(3(

12

xx

x

xx

x

xx

x

Section 3: In this section, we solve equations which include Rational

Expressions. The technique to solve them is to multiply both sides

of the equal sign by the LCD.

Ex 1: Here the LCD is n, so multiply everything

by n, gather like terms, and solve



)2)(5)(3(

)32)(3(

)3)(5)(2(

)53)(2(

)2)(3)(5(

)12)(5(

xxx

xx

xxx

xx

xxx

xx

)2)(3)(5(

)932()10113()5112( 222

xxx

xxxxxx

)5)(2)(3(

143

)2)(3)(5(

910531111232 2222

xxx

xx

xxx

xxxxxx

58

3 n

Notice that, when we multiplied by the LCD, all the Denominators

were cancelled. That’s what we want!!!

Ex 2: Here, the LCD is (3x – 4)(1 – 2x).

Multiply everything by this LCD!!

Ex 3: Here, the LCD is just (x – 3)


428583)(58

)(3)( nnnnnn

nn

xx 21

3

43

5

)21(

3)43)(21(

)43(

5)43)(21(

xxx

xxx

xxxxx 7129105)3)(43(5)21(

3

2

3

6

x

x

xx



Section 4: is about Ratios and Proportions. We will also do some

Application Problems that involve proportions and similar

Triangles. Here are their definitions:

A Ratio is the quotient of two numbers.

A Proportion is an expression which indicates that 2 ratios are

equal to each other.

To solve a proportion, we can cross multiply the Denominator of

one ratio with the Numerator of the other ratio and vice versa. In

other words, ad = bc

d

c

b

a

xxxx

xx

xxxx 263

)3(

2)3(

)3(

6)3()3( 2

1,60)1)(6(0652 xxxxx


by Viken Kiledjian Here are some examples that utilize this technique:

Ex 1: After we Cross-Multiply, we end up with

4(x + 1) = 6(x – 1) 4x + 4 = 6x – 6

10 = 2x and x = 5 (That’s pretty easy, isn’t it???)

Ex 2: After we Cross-Multiply, we get

10 = -2x(x + 6) 10 = -2x2 – 12x

2x2 + 12x + 10 = 0 x2 + 6x + 5 = 0 (x + 1)(x + 5) = 0

Therefore, x = -5 or -1

Application Problems & Similar Triangles:

Ex 1: In a city of 25,000 homes, a survey was taken to determine the

number with cable television. Of the 300 homes surveyed, 210

had cable television. Estimate the number of homes in the city

that have cable television.

4

6

1

1

x

x

5

2

6

2 x

x


Answer: This is a proportion problem. We will use the proportion

equation: Let a = 25,000 homes = total

number of homes in the city, b = total number that have cable, which

is unknown, c = total number of homes surveyed = 300, and d =

the number of homes from the survey, which had cable. Then, the

equation becomes:

There are 17,500 homes in the city which have cable.

Ex 2: As part of a conservation effort for a lake, 40 fish are caught,

tagged, and then released. Later, 80 fish are caught. Four of the

80 fish are found to have tags. Estimate the number of fish in the

lake.

d

c

b

a

500,17300

)210)(000,25(300)210)(000,25(

210

300000,25 bb

b


Answer: This is a proportion problem again. Let a = total number

of fish in the lake, which is unknown, b = the total number which

are caught and tagged = 40, c = the number of fish that are

caught later = 80, d = the number of fish out of the 80, which are

found with a tag = 4. The equation is:

There are a total of 800 fish in the lake.

Ex 3: Find the area of the triangle ABC, given that it is similar

C F to triangle DEF

15 cm

A 12 cm B D 22.5 cm E

Answer: First, we need to find the height of the triangle ABC.

8004

)80)(40()80)(40(4

4

80

40 aa

a

Beginning Algebra Lecture 6: Section 4 and

Section 5 by Viken Kiledjian Let a = height of the ABC triangle, b = base of the ABC triangle =

12 cm, c = height of the DEF triangle = 15 cm, d = base of the DEF triangle = 22.5 cm. The equation becomes

Now, we use the formula for the Area of a Triangle for the ABC triangle: ½ Base x Height -------> ½ (12)(8) = 48 cm2

Section 5: In this section, we will do some more Application Problems involving Work and Uniform Motion.

Ex 1: A proofreader can read 50 pages in 3 hours, and a 2nd proofreader can read 50 pages in 1 hour. If they both work on a 250 page book, can they meet a six-hour deadline?

Answer: When people work together, their rates add up.

The rate for person A = 50/3 = 16.6666667 pages/hr

The rate for person B = 50/1 = 50 pages/hr

The combined rate = 250/t (We want to see if t is less than 6)

cmaaa

85.22

)15)(12()15)(12(5.22

5.22

15

12


Now, let’s add up their rates and solve for “t”

The LCD = 3t so multiply everything by this.

50t + 150t = 750 t = 3.75

Since t < 6 hours, they can meet this deadline.

Ex 2: Sally can clean the house in 6 hours, and her father can clean the house in 4 hours. Sally’s younger brother, Dennis, can completely mess up the house in 8 hours. If Sally and her father clean and Dennis plays, how long will it take to clean the house?

Sally’s rate for cleaning = 1/6 house/hour

The Father’s rate for cleaning = ¼ house/hour

Dennis’s rate of messing up the house = 1/8 house/hour

Combined rate for cleaning house = 1/t house/hr

t

250

1

50

3

50

)250

(3)1

50

3

50(3

ttt


When we add up their rates, we have to put a Minus sign before Dennis’ rate because he is messing up the house.

It will take 24/7 = 3 and 3/7 hours to clean the house.

Ex 3: Two trains made the same 315 mile run. Since one train traveled 10 mph faster than the other, it arrived 2 hours earlier. Find the speed of each train.

Solution: The basic equation is Distance = rate x time

Let x = rate of slower train. Then x + 10 = rate of faster train

Time for slower train = 315/x,

Time for faster train = 315/(x + 10)

We also know that the faster train’s time is 2 hrs less than the slower trains time. Therefore, the equation becomes

t

1

8

1

4

1

6

1 )

1(24)

8

1

4

1

6

1(24

ttt 24364 ttt


by Viken Kiledjian

Therefore, x = -45 or 35. We discard the negative answer.

The slower train has a speed of 35 mph and the faster train has a speed of 45 mph.

Ex 4: A commercial jet can fly 550 mph in calm air. Traveling with the jet stream, the plane flew 2400 mi in the same amount of time it takes to fly 2000 mi against the jet stream. Find the rate of the jet stream.

Answer: Let r = rate of the jet stream. Then 550 + r = rate of the plane with the stream, 550 – r = rate of plane against the stream. Using the equation t = Distance/Rate, we get

2315

10

315

xx]2

315)[10()

10

315)(10(

xxx

xxx

)10(2)10(315315 xxxx xxxx 2023150315315 2

03150202 2 xx 01575102 xx 0)35)(45( xx


Cancel 2 zeros

from both sides

The rate of the jet stream is 50 miles per hour.

Practice these problems. The more you practice, the better you will

get!!!!!!!!!!

)550(20)550(24550

2000

550

2400rr

rrt

5044220020000,1124200,13 rrrr


Section 1: covers the topic of the Cartesian

or Rectangular Coordinate System.

Origin . Point P

Quadrant 2 Quadrant 1

x-axis

Quadrant 3 Quadrant 4

y-axis

Every point on this System is given by a

pair of points (x, y). The origin is the

point (0, 0). The point P is roughly

about (3, 5). The 3 is the x-coordinate

of point P and 5 is the y-coordinate.


by Viken Kiledjian In the ordered pair (x, y), the 1st number (x) is called the abscissa,

and the 2nd number (y) is called the ordinate. The x values are usually also called the Domain, and the y values are called the Range.

Objective C: This Objective is about Functions and Relations.

Function Definition: A function is a correspondence between the elements of one set (called the domain) and the elements of another set (called the range), where exactly ONE ELEMENT IN THE RANGE CORRESPONDS TO EACH ELEMENT IN THE DOMAIN.

Relation Definition: A relation is a correspondence between the elements of one set (called the domain), and the elements of another set (called the range), where ONE OR MORE ELEMENTS IN THE RANGE CORRESPONDS TO EACH ELEMENT IN THE DOMAIN.

NOTE: A function is ALWAYS a relation, but a relation is not always a function.


by Viken Kiledjian

Ex. 1 Give the domain and range of each relation and tell whether

the relation is a function

Golfer Tournament Champion

Hale Irwin United States Open

Jack Nicklaus The Masters

Greg Norman The British Open

Solution: The Domain are the Golfers: Irwin, Nicklaus, Norman

The Range are the Championships they have won: the US Open, The

Masters, the British Open.

This Relation is NOT a function because one member in the Domain

corresponds with more than one member in the range.

It would be a function if Jack Nicklaus had ONLY won one of the 3

championships. Suppose, the situation was like this ------>


by Viken Kiledjian

Golfer Tournament Champion

Hale Irwin United States Open

Jack Nicklaus The Masters

Greg Norman The British Open

Now, the Range would consist of Only 2 members: the US Open and

the British Open. But this relation WOULD BE A FUNCTION.

This is because 2 members in the Domain are getting mapped

into 1 member in the Range. This is acceptable for a function.

Ex. 2 Tell whether the following equations are functions:

a) y = 4x – 1 b) y2 = x + 1 c) y = x2 + 1 d) y = |x| e) x = |y|

Usually, “x” takes the role of the domain and the variable “y” takes

the role of the range. In this example, (a), (c ), and (d) are all

functions but (b) and (e) are not.


by Viken Kiledjian Let’s take each one separately,

a) y = 4x – 1 for each value of x, there is Only one value of y

b) y2 = x + 1 for each value of x, there are Two values of y. For Example, suppose x = 3. Then y2 = 4 and y = +2 or –2. One member in the domain (the x-value) cannot correspond with 2 values in the range (the y-value); therefore, it’s not a function.

c) y = x2 + 1 for each value of x, there is Only 1 value of y. For example, suppose x = 3 again. Then y = 32 + 1 = 10. Therefore, it is a Function. (It is true that two different values of x will yield the same y, but that is Acceptable for a function. Remember the second case in Example 1??

d) y = |x| This is similar to example (c ). For each value of x, there is only 1 value of y. Therefore, it is a Function

e) x =|y| This is similar to example (b). Each value of x corresponds with 2 values of y. For example, if x = 3, then y can equal 3 or –3. Therefore, it is NOT A FUNCTION.



Function Notation We usually write a function as:

y = f(x) which states that “y is a function of x”

Note: It does not mean that “y equals f times x”

In the notation y = f(x), x is the domain of the function f, and y is the range of the function f. Another way of saying it is: x is the independent variable of the function f, and y is the dependent variable because it depends on x via the function f.

Ex: If f(x) = x2 – 3x, find f(-1)

f(-1) means “f of –1” NOT “f times –1”. Therefore, wherever you see an “x” in the equation, you substitute a “–1” for it!!

f(-1) = (-1)2 – 3(-1) = 1 + 3 = 4

Section 2: This section is about Graphing Linear Equations in Two Variables. There are two main ways of writing a linear equation: The 1st way is the slope-intercept method y = mx + b and the 2nd way is the general method Ax + By = C.


by Viken Kiledjian

Graphing Linear Equations: A linear equation is an equation that relates a variable y with another variable x, such that both of their powers are 1. Ex. y = 2x – 5 (notice that the powers of the variables y and x are both 1) To graph this equation one only needs two pairs of (x, y) points since a straight line can be constructed from just 2 points.

We can arbitrarily choose 2 values for x, and solve for the corresponding values of y.

If x = 0 -----> y = 2(0) – 5 = -5 -------> this yields the point (0, -5)

If x = 1 -----> y = 2(1) – 5 = -3 -------> this yields the point (1, -3)

Therefore, the graph will look something like:

(0, -5) means you don’t move to the right,

but you go 5 down.

(1, -3) means you move 1 to the right,

and you go 3 down.


by Viken Kiledjian

Graphing Lines Parallel to the X- and Y- Axis:

Ex. 1 The line y = 5 means that y is always equal to 5 no matter

what the value of x is. This means that you go up 5 and draw a

straight horizontal line parallel to the x-axis such as:

Ex. 2 The line x = -3 means that x is always equal to –3 no matter

what the value of y is. Therefore, go to the left 3 and draw a

straight vertical line parallel to the y-axis such as:


by Viken Kiledjian

The General Form of the equation of a line is written as:

Ax + By = C. We can also graph the general form as we did the

Slope-Intercept form: Just pick two values for x and find the

corresponding y values. Use those 2 points to draw a straight line.

Ex: Graph the Equation 4x – 3y = 12.

Answer: Let x = 0, then y = -4 and let y = 0, x = 3. (Sometimes, it is

easy to set both the x and y = 0 and solve for the other variable!!)

Now, we have the following points ----------> (3, 0) and (0, -4)

We go to the right 3 units from the origin and put a point there!!!

We go down 4 units from the origin and put a point there!!


by Viken Kiledjian Section 3 deals with the idea of Slope and the X and Y- intercept of

a line. The slope of a line measures its steepness. You can compare the idea of Slope to the idea of Grade on a road. Often when you drive on a road, there might be a sign that says, “7% Downhill Grade: Trucks Use Low Gear”

The y-intercept of a line is the point where it crosses the y-axis.

In order to find it, set the value of x equal to zero, and solve for y.

The x-intercept of a line is the point where it crosses the x-axis.

In order to find it, set the value of y equal to zero, and solve for x.

In the previous example, 4x – 3y = 12,

If we set x = 0, then y = -4 ----> this means the y-intercept = -4

If we set y = 0, then x = 3 ----> this means the x-intercept = 3

This means that the line given by 4x – 3y = 12 will definitely go through these 2 points --------> (0, -4) and (3, 0)

This is the Same technique that we used in the previous example, without giving them a name. Now, we know what they are!!!


by Viken Kiledjian

Now, let’s go to slope again. It is common to define slope as “Rise over Run” and use “m” as its symbol. The formal equation is:

m = change in y = Dy = y2 – y1

change in x Dx x2 – x1

Ex 1: Find the slope between the points (-2, 6) and (1, 4).

Solution: Let x1 = -2, then y1 = 6. Therefore, x2 = 1, y2 = 4

m = (4 – 6)/(1 - -2) = -2/3 (The slope is negative!!!)

You can also let x1 = 1 and then y1 = 4, x2 = -2, y2 = 6. So the slope,

m = (6 – 4)/(-2 – 1) = 2/(-3) = -2/3 (The answer is the same!!!)

Ex 2: Find the slope of the line in the first example of Section 2:

y = 2x – 5 (You get any two points on this line and apply the slope equation to it. Remember in Section 2, we already found 2 points? They were (0, -5) and (1, -3). Let x1 = 0, so


by Viken Kiledjian

y1 = -5, x2 = 1, y2 = -3. Therefore, m = (-3 - -5)/(1 – 0) = 2/1

m = 2 (Notice that the slope is the same as the coefficient of the x

in the equation. This is because the equation is already written in the slope intercept method ------------> y = mx + b

Ex. 3 Find the slope of the equation 4x – 3y = 12 This is the same equation as we have already done. Using the 2 points (0, -4),(3, 0)

m = (0 - -4)/(3 – 0) = (4)/(3) = 4/3!!! We can also do this by writing this equation in the slope-intercept method, by solving for y ---> -3y = 12 – 4x ---> y = -4 + (4/3)x, ---> m = 4/3 and b = -4.

Graphical Interpretation of Slope: Lines which have Positive slope are “rising” and look like:


by Viken Kiledjian

Lines which have Negative slope are descending and look like:

Lines which have Zero slope are horizontal such as the line y = 5:

Vertical lines such as x = -3 have Infinite or Undefined slopes:


by Viken Kiledjian

Besides helping us to draw a line, the slope of a line helps us to ascertain if it is parallel to another line. Here is the rule:

The slopes of parallel lines are equal.

Ex. 1 Tell whether the lines with the following slopes are parallel

m1 = 3, m2 = -1 Since they have different slopes, they are not parallel. If m2 = 3 or m1 = -1, then they would have been parallel

Ex. 2 Tell whether the line PQ is parallel to a line of slope –2.

P(6, 4) Q(8, 5) -----> m = (5 – 4)/(8 – 6) = ½ They are not

parallel to each other

Ex 3: Tell whether the line through P1 and P2 is parallel to the line through Q1 and Q2. P1 (4, -5), P2 (6, -9), Q1 (5, -4), Q2 (1, 4)

m1 = (-9 - -5)/(6 – 4) = -4/2 = -2 (The slope of a line through P1P2)

m2 = (4 - -4)/(1 – 5) = 8/(-4) = -2 (The slope of a line through Q1Q2)

Since these two slopes are the same, the two lines are parallel!!!


by Viken Kiledjian

Section 4: In this section, we learn how to find the equation of a line given a Point and the Slope of the line OR 2 points on the line.

The Point-Slope form of the equation of a line is written as:

y – y1 = m(x – x1)

The idea is that if you know the slope of a straight line and you know one point that it goes through, then you can write the general equation of the line.

Ex. 1 If the slope of a line = -2 and it goes through the point (1, -4), what is its equation??

Solution: x1 = 1, y1 = -4, and m = -2 Therefore,

y – (-4) = -2(x – 1) Now, distribute the –2 into the parenthesis

y + 4 = -2x + 2 ---------> y = -2x – 2 is the answer!!

2 Points of a Line: To find the equation of a line when 2 points are given, 1st calculate its slope and then use the Point-Slope form.


by Viken Kiledjian

Ex. 2 Find the equation of the line that passes through the two points

(-1, 4) and (3, -6).

Solution: First find the slope m! Let x1 = -1, then y1 = 4, x2 = 3,

y2 = -6

Therefore, m = (-6 – 4)/(3 – -1) = -10/4 = -5/2

Then use the point slope form, y – y1 = m(x – x1)

You can let the point (-1, 4) or the point (3, -6) represent the x1 , y1

Since we let the point (-1, 4) represent the x1 and y1 while we were computing the slope, I’ll stick to the same practice.

So x1 = -1 and y1 = 4 ---------> y – 4 = -5/2(x - -1) = -5/2(x + 1)

Now, distribute the –5/2 into the parenthesis, y – 4 = (-5/2)x – 5/2

And, add 4 to both sides -----> y = (-5/2)x + 3/2 This is the Slope-Intercept Form where the slope = -5/2 and the y-intercept = 3/2


Section 1 In this section, we will solve a system

of 2 equations with 2 unknowns using the

Substitution Method and we will solve some

Application Problems involving Investments.

There are 2 special situations to be aware of :

An Inconsistent System results in an impossible

situation and therefore there are NO

SOLUTIONS FOR IT. Graphically, this

means that the two Lines Never Cross.

A System with Infinitely Many Solutions

results when the 2 equations are a redundant

repetition of each other, which is also called a

Dependent system. Graphically, this means

that the 2 Lines overlap and are the same line.


by Viken Kiledjian Ex. 1 Solve this system of 2 equations by the Substitution Method:

3x – 2y = -10 In this method, we have to Solve for one of

6x + 5y = 25 the variables and substitute it in the other

equation. It is usually easiest to solve for the variable that DOES

NOT HAVE A COEFFICIENT IN FRONT OF IT. In this

example, since all of the variables have a coefficient in front of it,

it does not matter which one we solve for or isolate. Let’s solve

for the x in the upper equation:

3x = 2y – 10 --------> x = (2y – 10)/3 Now, plug this in for the

6(2y – 10) + 5y = 25 x in the bottom equation!!!!

3 The 3 divides into the 6!!!

2(2y – 10) + 5y = 25 Distribute the 2 into the parenthesis!!

4y – 20 + 5y = 25 --------> 9y = 45 --------> y = 5

Now, plug this in the equation for the x: x = [2(5) – 10]/3 = 0


by Viken Kiledjian Ex. 2: Solve this system of 2 equations by the Substitution Method:

x = 4 – 2y

y = 2x – 13 In this example, since Either of the variables

has a coefficient of 1 in front of it, we can substitute it in the other

equation. Let’s solve with both ways and get the same answer:

Method 1: Substitute the x from the 1st eq. into the 2nd equation.

x = 4 – 2y --------> y = 2(4 – 2y) – 13 = 8 – 4y – 13 = -5 – 4y ----

------> y + 4y = -5 ----------> 5y = -5 Therefore, y = -1

Plug this y into the 1st equation for the x: x = 4 – 2(-1) = 6

Method 2: Substitute the y from the 2nd eq. into the 1st equation.

y = 2x - 13 --------> x = 4 – 2(2x – 13) = 4 – 4x + 26 = 30 – 4x

--------> x + 4x = 30 --------> 5x = 30 Therefore, x = 6

Now, plug this x in the equation for the y: y = 2(6) – 13 = -1


by Viken Kiledjian Now, let’s do some Investment Problems with 2 equations:

Ex 1: A mortgage broker purchased two trust deeds for a total of $250,000. One trust deed earns 7% simple annual interest, and the second one earns 8% simple annual interest. If the total annual interest from the two trust deeds is $18,500, what was the purchase price of each trust deed?

Answer: Let x = purchase price of deed at 7%, y = purchase price of deed at 8%. Then, the amount that you would earn at simple annual interest would be .07x and .08y for the two trust deeds.

The two equations become: x + y = 250,000 since that’s the total value purchased and .07x + .08y = 18,500 since that’s the total interest earned. Now, substitute for the x from the top equation:

x = 250,000 – y -------> .07(250,000 – y) + .08y = 18,500 -------->

17,500 - .07y + .08y = 18,500 ----> .01y = 1,000 ----> y = $100,000 at 8% and x = 250,000 – 100,000 = $150,000 at 7%


by Viken Kiledjian Ex 2: A mortgage broker purchased two trust deeds for a total of

$200,000. One trust deed earns 12% simple annual interest, and

the second one earns 9.5% simple annual interest. If the annual

interest from the 9.5% trust deed is twice the annual interest from

the 12%, what was the purchase price of each trust deed?

Answer: Let x = purchase price of deed at 9.5%, y = purchase price

of deed at 12%. Then, the amount that you would earn at simple

annual interest would be .095x and .12y for each trust deed.

The two equations become: x + y = 200,000 since that’s the total

value purchased and .095x = 2(.12y) since the interest earned at

9.5% is twice the interest earned at 12%. Substitute for the x:

x = 200,000 – y -------> .095(200,000 – y) = 2(.12)y -------->

19,000 - .095y = .24y -----> .335y = 19,000 ----> y = $56,716.42 at

12% and x = 200,000 – 56,716.42 = $143,283.58 at 9.5%


by Viken Kiledjian Section 2: In this section, we use the Addition Method to solve a

system of 2 equations and 2 unknowns.

Ex. 1: Solve the following 2 equations by the Addition Method

a) 2y – 3x = -13 1) First, rearrange the 17 and the 4y in the

b) 3x – 17 = 4y bottom equation to group the variables.

b) 3x – 4y = 17 2) Now, rewrite the top equation so that the

a) –3x + 2y = -13 x comes first and then the y.

3) Now, just add the equations, and the x’s

-4y + 2y = 17 + (-13) will cancel, because 3x + (-3x) = 0

-2y = 4 ---------> y = -2 4) Now, plug this value of y into either

equation “a” or “b” to solve for x.

3x – 17 = 4(-2) = -8 I’ll choose equation “b” because the 3x = 17 – 8 = 9 coefficient of the x is positive in that x = 3 equation!!


by Viken Kiledjian

Ex. 2: Solve the following 2 equations by the Addition Method:

a) 8x – 3y = -96 It looks like the “y” variable is the easiest one

b) 4x – 3y = -72 to get rid of because its coefficient is the same in both equations. However, we have to multiply one of the equations by (-1) so that the coefficients of the “y” will be opposite of each other and the “y” will cancel when we add the 2 equations. So, let’s multiply the Top equation by (-1)!!

-8x + 3y = 96 I simply changed the sign of each member!!

4x – 3y = -72 Now, add the 2 equations, and the y’s cancel

-8x + 4x = 96 –72 --------> -4x = 24 --------> x = -6

Now, plug this in to either equation and solve for y. Let’s plug it into the changed equation above because the coefficient of the y is positive: -8x + 3y = 96 -------> -8(-6) + 3y = 96 ------->

48 + 3y = 96 ----------> 3y = 48 ---------> y = 16 Great!!


by Viken Kiledjian Section 3: In this section, we solve some more Application

Problems about Rate of Wind or Current and General Problems involving 2 equations and 2 unknowns. You may either use the Substitution or the Addition Method to solve these.

Ex 1: A sporting goods salesperson sells 2 fishing reels and 5 rods for $270. The next day, the salesperson sells 4 reels and 2 rods for $220. How much does each cost?

Solution: Let “x” be the cost of each fishing reel and “y” the cost of each rod. Then the 2 equations become:

2x + 5y = 270 It is best to use the addition method here and

4x + 2y = 220 eliminate the x variable. Multiply the top equation by (-2) to make the coefficients of the x’s opposite to each other, so that the x’s will cancel when we add them.

(-2)(2x) + (-2)(5y) = -2(270) -------------> -4x – 10y = -540 Now, add this to the bottom equation: -10y + 2y = -540 + 220


by Viken Kiledjian -8y = -320 -------> y = 40 Now, plug this into either equation to

solve for “x”. Let’s plug it into the top equation:

2x + 5(40) = 270 -------> 2x + 200 = 270 ---------> x = 35

Therefore, the reels cost $35 each and the rods cost $40 each.

Ex. 2: A Hard Candy costs $2/lb and a Soft Candy costs $4/lb.

How many pounds of each must be mixed to obtain 60 pounds of

candy that is worth $3 per pound.

Solution: Let x be the pounds of Hard Candy and y the pounds of

Soft Candy. Here are the 2 equations:

x + y = 60 (Since there has to be a total of 60 pounds of candy)

2x + 4y = 3(60) (Since the cost of the total candy is $3 x 60 lbs)

Let’s use the Substitution method since the coefficient of the x or y

on the top equation is 1. Let’s solve for x in the top equation:

x = 60 – y Now, plug this into the bottom equation for x:


by Viken Kiledjian 2(60 – y) + 4y = 180 -------> 120 – 2y + 4y = 180 --------> 2y = 60

y = 30 Now, plug this into the equation for x ----> x = 60 – 30 = 30

Therefore, we should mix 30 lbs of Soft and 30 lbs of Hard Candy.

This makes sense since the total mixture is going to cost the exact average. The average of $2/lb and $4/lb is $3/lb.

Now, let’s do some Wind and Current Problems:

Ex 1: A plane flying with the jet stream flew from Los Angeles to Chicago, a distance of 2250 miles, in 5 hours. Flying against the stream, the plane could fly only 1750 miles in the same time. Find the rate of the plane in calm air and the rate of the wind.

Answer: Let r = rate of the plane in calm air, w = rate of the wind

Then, r + w = rate of the plane with the stream, r – w = rate of the plane against the wind. Using the equation D = rate x time, we get

2250 = (r + w)(5) ---------> r + w = 450 Now, add the 2 equations

1750 = (r – w)(5) ----------> r – w = 350 to eliminate the w variable


by Viken Kiledjian 2r = 800 ------> r = 400 Therefore, 400 + w = 450 ---> w = 50

The rate of the plane is 400mph and the rate of the wind is 50mph

Ex 2: A seaplane pilot flying with the wind flew from an ocean port to a lake, a distance of 240 miles, in 2 hours. Flying against the wind, the pilot flew from the lake to the ocean port in 2 hours, 40 minutes. Find the rate of the plane and the rate of the wind.

Answer: Let r = rate of the plane in calm air, w = rate of the wind

We have to change the time of 2 hrs, 40 min into hours alone ----->

2 + 40/60 = 2 + 2/3 = 6/3 + 2/3 = 8/3 hours

Then, r + w = rate of the plane with the stream, r – w = rate of the plane against the wind. We again use D = rate x time:

240 = (r + w)(2) ----------> r + w = 120 Again, add the 2

240 = (r – w)(8/3) --------> r – w = (3/8)240 = 90 equations:

2r = 210 -------> r = 105 Therefore, 105 + w = 120 ----> w = 15

The rate of the plane is 105mph and the rate of the wind is 15mph


by Viken Kiledjian Section 1: This section is about writing

Sets and Graphing Inequalities on the Number Line.

A set is a grouping of objects. The objects in the set are called its elements. The set that contains no elements is called the null set.

1) SET NOTATION: There are 2 main methods of writing sets.

a) Roster Method: We enclose the elements of the set within braces.

ex. A = {1, 2, 4, 6}

Ex: Write the set of integers between -10

and -4. [Note: between does NOT include

the end points] A = {-9,-8,-7,-6,-5}


by Viken Kiledjian

b) Set Builder Notation: We could also express the previous set A

like this. A = {x| x = 1, 2, 4, 6}

This reads “the set of all numbers x, such that x = 1, 2, 4, or 6.”

Another example would be:

B = {x| x is a positive integer less than 10}

This would represent the numbers 1, 2, 3, 4, 5, 6, 7, 8, 9.

Ex 1: Write the set of “The positive integers less than 5” using set

builder notation.

Answer: {x|x < 5, x positive integers} The symbol means

“member of”.

Ex 2: Write the set of “The odd integers less than -2” using the set

builder notation again.

Answer: {x|x < -2, x odd integers}


by Viken Kiledjian

2) UNIONS AND INTERSECTIONS:

a) The Union of two sets is the set of all of their elements

combined together. ( The symbol is )

b) The Intersection of two sets is the set of all elements that they

share in common. (The symbol is )

Ex 1: Find the Union between sets A and B, where

A = { -3, -2, -1}, B = {-1, 1, 2}

Answer: The union of the two sets is the combined elements of both

sets. However, since they have -1 in common, you don’t need to

write that twice!!! A B = {-3, -2, -1, 1, 2}

Ex 2: In the previous ex., find the intersection between sets A and B.

Answer: The intersection of the two sets are the elements that they

share in common. In this case, there is only one!! A B = {-1}


by Viken Kiledjian 3) Graphing Inequalities on the Number Line is an important topic

and it helps to visualize what inequalities mean. Here are the

different kinds of Inequalities:

< means “less than”

> means “greater than”

means “less than or equal to”

means “greater than or equal to”

Ex 1: Graph {x|x -1}

Answer: Since -1 is included in this case, we use a Bracket notation

to mean “equal to”, which includes the end point!!!!

[

. . . -3 -2 -1 0 1 2 3 . . .

Ex 2: Graph {x|x < 4} In this case, we use a Parenthesis to

indicate that the 4 is not included in the interval.


by Viken Kiledjian

Answer: )

. . . -3 -2 -1 0 1 2 3 4 . . .

Ex 3: Graph {x|x > 4} {x|x < -2}

Answer: Here, we want the Union between all the numbers greater

than 4 and less than -2. In other words, just take the combined

list of numbers included in both sets.

) (

. . . -3 -2 -1 0 1 2 3 4 . . .

Ex 4: Graph {x|x > -3} {x|x < 3}

Answer: Here, we want the Intersection between these two sets. In

other words, take the Common numbers greater than -3 and less

than 3. These will be the set of numbers between -3 and 3!!

( )

-3 -2 -1 0 1 2 3


by Viken Kiledjian Section 2 covers the topic of Solving Linear Inequalities. It is pretty

much the same as solving equations with ONE exception. When you divide/multiply by a negative number, you must change the sign of the inequality. We will also solve some Application Problems involving Inequalities.

Ex 1: Solve for x:

-2x + 6 16

First, subtract 6 from both sides:

-2x 10 Now, divide both sides by –2 and CHANGE THE

x -5 INEQUALITY SIGN!!

Ex 2: -9(h – 3) + 2h < 8(4 – h) First, Distribute the –9 and 8!!

-9h + 27 + 2h < 32 – 8h ---------> -7h + 27 < 32 – 8h

Now, add 8h to both sides and subtract 27 from both sides!!!

8h – 7h < 32 – 27 ------------> h < 5


by Viken Kiledjian Now, let’s do some Application Problems Involving Inequalities:

Ex 1: The cost of renting a truck is $29.95 for the first hour and

$8.95 for each additional hour. How long can a person have the

truck if the cost is to be less than $110? (Part of an hour is

considered a Full hour)

Solution: Let x = the number of additional hours beyond the 1st hour

The total cost of renting must be less than $110.Here is the Equation

29.95 + 8.95x < 110 -----------> 8.95x < 80.05 ------> x < 8.94 hrs

Since, part of an Hour is considered a Full hour, if the person drives

for 8.94 hours, beyond the original 1st hour, he/she will be

charged for 9 Additional hours. Therefore, they should only drive

it for 8 Additional hours!!!!! Adding 1 hour to this yields 9 hours.

The Total Number of Hours must be less than 9 hours


by Viken Kiledjian Ex 2: A student who can afford to spend up to $2000 sees the

following advertisement: Big Sale on Computer for $1695.95 and All CD-ROMs for $19.95. If she buys the computer, find the greatest number of CD-ROMs that she can buy.

Solution: let x = the number of CD-ROMs, then the total cost is:

1695.95 + 19.95x 2000 (Notice that the wording “up to” is NOT the same as the wording “less than”. “Up to” includes the number whereas “less than” does NOT include the number.)

19.95x 304.05 ----------> x 15.24 (However, you can’t buy fractional CDs, so you have to round this number down)

The student can buy a maximum of 15 CD-ROMs.

Ex 3: An excavating company charges $300 an hour for the use of a backhoe and $500 an hour for the use of a bulldozer. The company employs one operator for 40 hours per week. If the company wants to take in at least $18,500 each week, how many hours per week can it schedule the operator to use the backhoe?


by Viken Kiledjian In this problem, Part of an hour counts as a full hour.

Solution: Let x = the number of hours for the use of the backhoe,

then 40 – x = the number of hours for the use of the bulldozer

(Since the total # of hours must be 40 hrs) Therefore, the total cost is

300x + 500(40 – x) 18,500 (Note that the phrase “at least”

translates into a “greater or equal” sign.) Now, Distribute the 500!

300x + 20,000 – 500x 18,500 ------> -200x + 20,000 18,500

--------> -200x -1500 (Now divide by –200 and change the sign)

x 7.5 (Now, round this down to the nearest integer, since part

of an hour is considered an hour)

The backhoe can only be used for 7 hours Maximum. The

bulldozer should get 33 hours Minimum, since they have to

add up to 40 hours.


by Viken Kiledjian Section 3 is about Graphing 2-dimensional Linear Inequalities.

This section is pretty short and easy!!

Graphing Linear Inequalities: Let’s do a few examples.

Ex 1: Graph the inequality y < 2x –1

Solution: First, you graph this line on the x-y axis just like in

chapter 7. Since there is NO equal sign in this problem, you

make the graph a DOTTED LINE.

Now, the next thing you do is to see

which region you have to shade ----->

the one above or the one below this line??

The easiest way to do this is to test the point

(0,0) in the original equation ------> 0 < 2(0) – 1 ??? This leads to

0 < -1 which is Not True. Therefore, you have to shade the

region BELOW THE LINE since(0,0) did not satisfy the inequality.


by Viken Kiledjian Ex 2: Graph the Inequality 2x -3y - 12

Solution: Add 3y to both sides and you get -----> 2x + 3y -12

Now, graph this equation as in chapter 7. However, this time make

the line SOLID because the inequality has AN EQUAL SIGN

included in it.

Again, check the point (0,0)

in the inequality ------------>

2(0) + 3(0) -12 ????

0 < -12 ?? Since this is Not True,

you must shade the region below this line.

Another Way to look at this is to solve for y in the original equation:

y (-12 – 2x)/3 This means that you must shade the region

BELOW the line, since y is “less than or equal to” this equation!!


by Viken Kiledjian Section 1: is about Simplifying Numerical and

Variable Radical Expressions. We will focus

on 1 major kind of Radical Expression, the

square root.

Square Roots: if x2 = y, then x is the Square

Root of y denoted as

Actually, x can also equal the Negative of this

because when you square a negative, it

becomes a positive. The positive square root

of a number is called the Principle Square

Root.

What happens when you take the square root of

a square of a number?

If x > 0, then

If x < 0, then

yx

2x

xx 2

xx 2


by Viken Kiledjian In other words, If you square a number and then take its square root,

the result is always the Positive of the original number. This is

equivalent to the Absolute Value Function.

Therefore, . We will assume that Variables are always

Positive, unless otherwise stated.

Ex 1:

Ex 2:

Ex 3:

Ex 4:

In the previous examples, we are using the following Properties

of Square Roots:

We will now illustrate some examples where the Number inside the

square root is not a Perfect Square Root Number.

xx 2

xxxx 441616 22

2244 552525 xxxx

55)5(2510 22 xxxxx

xxxx 774949 22

yxxy


by Viken Kiledjian Ex 1:

Ex 2:

The idea is to use the Perfect Square Numbers to factor the original

number until it is Not factorable by any Perfect Square Number.

Here are the first few Perfect Squares:

4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144, 169, 196, 225, 256

Ex 3:

Here, we even broke up the variables into their Square Root

component and their non-square root component (i.e. x9 = x8x)

Ex 4:

6264)6)(4(24

)2)(36()2)(9(7249)72)(4(92889

21082)6)(18(2)36(18

xyyxxyyxxyyxyx 26236)2)(36(72 4282839

)2)(64(5)2)(64(51285 422422532 abbaababaabaa

abbaababaabbaa 2402)8)(5(2645 2322422


by Viken Kiledjian Section 2: In this section, we learn how to Add and Subtract Radical

Expressions involving Square Roots.

When we add radicals together, we first simplify each one as much as

possible and then we add Like Radicals.

Ex 1:

Ex 2:

Ex 3:

Notice that we can’t combine the with the

225229)2)(25(2)2)(9(50218

213210232)5(223

21624)2)(16()2)(4(328

262422

)2)(144()2)(64()5)(16(28812880

245421228542144264516

5 2


by Viken Kiledjian Section 3: is about Multiplying and Dividing Radicals and

Rationalizing Radicals involving Square Roots.

We will use the Multiplication and Quotient property of radicals.

Let’s do some examples of multiplying and dividing radicals first,

then we’ll learn the concept of Rationalizing radicals.

Ex 1:

Another approach to this problem would be the following:

I am usually going to use the 1st approach where I Multiply or Divide

the radicals first, then I simplify the combined radical.

Ex 2:

57549)5)(49()5)(7)(7()35)(7(357

575495)7)(7(577)5)(7(7357

24

2

5

3

5

3 416

7

112

7

112

a

b

a

b

ba

ab

ba

ab


Ex 3:

Ex 4: We can use the Special Form formula

(a – b)2 = a2 – 2ab + b2

Rationalizing Fractions:

When the denominator of a fraction contains a non-reducible Radical

sign, we rationalize the fraction by multiplying BOTH the

Numerator and the Denominator by the Conjugate of the Radical

expression. After the process is done, all of the radical signs will

only be found in the Numerator!!!

103)2()5(7375327 yyyyyy

1029211063521 yyyyy

235 x

315253352522

xxxx


The conjugate of a Radical Expression is another Radical Expression

that when multiplied by the original will yield an integer and will

eliminate the Radical sign.

Ex: The conjugate of is itself, because

The conjugate of is also itself, because

The conjugate of a Binomial Radical expression such as

is because (a + b)(a – b) = a2 – b2

Let’s do some examples of Rationalizing.

Ex 1:

x xxx

x3 xxx 333

23 23

12323232322

abc

cba

abcabc

abccab

abc

cab

10

50

1010

105

10

5 23222

2

2

10

25

10

)2)(25( 222b

abc

babc

abc

bcba


Or we can do it another way even easier:

Ex 2:

Ex 3:

)2)(2(33

3233

2323

233

23

3

32343

323

yxyx

yxyx

yx

yx

yx

yxyx

yyxx

yyyxyxxx

2

2

2

22

2

2210

5

10

5 22 bbbb

abc

cab

abc

cab


Section 4: is about Solving Equations which contain Radical Expressions. The basic idea is to square both sides of the equal sign until you get rid of all Radicals. We will also solve some Application Problems involving Square Roots.

Ex 1: Add 2 to both sides and you’ll get.

Now, square both sides of the equal sign.

Ex 2:

Now, let’s do some Application Problems:

Ex 1: The infield of a softball diamond is a square. The distance between successive bases is 60 ft. The pitcher’s mound is on the diagonal between home plate and Second base at a distance of 46 ft from home plate. Is the pitcher’s mound more or less than halfway between home plate and

52136 x

7136 x

6366491367136 22

xxxx

22

326326 xxxx

133326 xxxx


Second Base??

Answer: Here we have to use the Pythagorean Theorem for right triangles, c2 = a2 + b2 where c = hypotenuse of the right triangle and a and b = the two legs of the triangle.

In our case, a = distance between home plate and 1st base = 60 ft, and b = distance between 1st base and 2nd base = 60 ft and c = distance between home plate and 2nd base.

Therefore, c2 = 602 + 602 = 3600 + 3600 = 7200 -------------->

And

Since the distance between home plate and the pitcher’s mound is 46 and twice 46 equals 92. Therefore, The

Distance between home plate and the pitcher’s mound is More than half way between home plate and 2nd base!

Ex 2: Find the length of a pendulum that makes one swing in 1.5 seconds. The equation for the time of one swing of a pendulum is given by the equation --------------------------->

85.847200 cc


where T is measured in seconds and L is the

length of the pendulum in feet. Round to the

nearest hundredth.

Answer: In our case, we know the value of the Period (T) of the pendulum, but we don’t know the Length (L). The equation becomes:

In order to solve this, we square both sides to eliminate the square root sign, and then we solve for L in the equation:

322

LT

3225.1

L

8324

322

3225.1

22

2

22

2

2 LLLL

ftL 82.118)5.1)(8(

22

2

Section 1: covers the topics of Solving Quadratic Equations by Factoring and by Taking Square Roots.

The Square Root Property:

If c > 0, then the equation x2 = c has two solutions given by and

We can use this property to solve problems like the following:

Ex 1: 5x2 – 49 = 0

5x2 = 49 Therefore, x2 = 49/5 and

Ex 2: (x + 3)2 – 7 = 0 (x + 3)2 = 7


cx cx

5

57

5

57

55

57

5

7

5

49 Orx

3773 xx


In situations where there are more than one power of x, we can

either try to Factor the expression like we did in Chapter 5 or

we can use the Quadratic Formula, which we will learn in

Section 2. Of course, whenever the expression is factorable, it

is easier and faster to Factor it. Here are some examples of

factorable equations where we don’t need the Quadratic Eq:

Ex 1: x2 + 6x + 5 = 0 When we factor it, we’ll get

(x + 5)(x + 1) = 0. Therefore, x = -5 or -1

Ex 2: x2 + 5x – 4 = (2x + 1)(x – 4) Here, we have to multiply out

the right side of the equation and gather all like terms on 1 side

x2 + 5x – 4 = 2x2 – 8x + x – 4 = 2x2 – 7x – 4 ------------------>

0 = 2x2 – 7x – 4 – (x2 + 5x – 4 ) = 2x2 – 7x – 4 – x2 – 5x + 4 ------->

x2 – 12x = 0 Now, factor out the x!! x(12 – x) = 0

Therefore, x = 12 or 0!!!


Section 2: is about the Quadratic Equation, as we previously mentioned. The general formula is known as the Quadratic Formula. For any quadratic equation of the form, ax2 + bx + c = 0, the solution is given by the equation:

The advantage of this equation is that it works for Factorable as well as non-factorable equations. Let’s illustrate with some examples:

Ex 1: 4w2 + 6w + 1 = 0 Here, a = 4, b = 6, c = 1. Therefore,

Since the Solution came out with a Radical, this equation is Non- Factorable ---> You have to use the Quadratic Formula for this one!!

a

acbbx

2

42

8

526

8

206

8

16366

)4(2

)1)(4(466 2

x

4

53

8

532

x


Ex 2: 3x = x2/2 + 6 6x = x2 + 12 x2 - 6x + 12 = 0

Here, a = 1, b = -6, c = 12 . Therefore, the solution is

No Real Number Solution can be found for this equation. The

solution is an imaginary #, which is covered in the next course.

Ex 3: 6y2 + 5y – 4 = 0 Here, a = 6, b = 5, c = -4.

This one is actually a factorable equation, but it’s not easy to factor

it. For this one, it’s probably easier to use the Quadratic Formula!

#Im2

126

2

48366

)1(2

)12)(1(4)6()6( 2

aginaryx

12

115

12

1215

12

96255

)6(2

)4)(6(4)5()5( 2

x

3

4

12

16

12

115

2

1

12

6

12

115

ORx


by Viken Kiledjian Section 3: In this Section, we will do further Application Problems

involving solving by Factoring or using the Quadratic Formula.

Ex 1: The length of a rectangle is 4 ft more than twice the width.

The area of the rectangle is 160 ft2. Find the length and width of the

rectangle.

Answer: Let w = the width of the rectangle, then l = 4 + 2w =

length of the rectangle, and A = lw = (4 + 2w)w = 160 ----------->

4w + 2w2 = 160 -----> 2w2 + 4w – 160 = 0 ------> w2 + 2w – 80 = 0

Now, factor this trinomial: (w + 10)(w – 8) = 0 So, w = 8 or -10

Since the width can’t be negative, discard the -10 answer.

Therefore, the Width is 8 ft, and the Length = 4 + 2(8) = 20 ft.

Ex 2: A square piece of cardboard is to be formed into a box to

transport pizzas. The box is formed by cutting 2-inch square

corners from the cardboard and folding them up. If the volume of


by Viken Kiledjian the box is 512 in3, what are the dimensions of the cardboard?

Answer: The picture looks like the following: 2 in

The corners are 2 inch squares and the box x

has width x all around. When the corners are

folded and turned over, the box will have a base which will also be a

square, and its sides will equal x – 4 inches long. This is because

each side is losing two corners which are 2 inches long each. The

box will also have a height of 2 inches because that’s how large the

corners are. Now, we use the formula for the volume of a box ---->

V = (Area of Base)(Height) = (x – 4)2(2) This will equal 512

(x – 4)2(2) = 512 ---------> (x2 – 8x + 16)(2) = 512 Divide both

sides by 2 to reduce the large number 512 -----> x2 – 8x + 16 = 256

-------> x2 – 8x – 240 = 0 --------> (x – 20)(x + 12) = 0 So, x = 20

or -12. We discard the negative answer again. x = 20 inches.


by Viken Kiledjian Ex 3: The hypotenuse of a right triangle is cm. One leg is 1 cm

shorter than twice the length of the other leg. Find the lengths of the

legs of the right triangle.

Answer: Here, we use the Pythagorean Theorem c2 = a2 + b2 where

c = hypotenuse of the triangle = in our problem. Let a = length

of 1st leg and b = length of the 2nd leg. Then, according to the

problem --------> a = 2b – 1 because it’s 1 cm shorter than twice

the other one!! Now, plug all these facts into the Pythagorean Th.

and we get

0 = 5b2 – 4b – 12 This Trinomial might be factorable, but it is

easier to use the Quadratic Equation on it rather than trying to see

how to factor it!!! a = 5, b = -4, c = -12 Plug these all in!!

13

13

22222

)144(131213 bbbbb

10

164

10

2564

10

240164

)5(2

)12)(5(4)4()4( 2

b


by Viken Kiledjian So, it was factorable and the solution for b = 20/10 or -12/10. We

discard the negative answer again because the Leg of a Triangle

can’t be negative. Therefore, one of the Legs of the Triangle

= 2 cm and the other Leg = 2b – 1 = 2(2) – 1 = 3 cm.

Ex 4: A tank has two drains. One drain takes 16 min longer to

empty the tank than does a second drain. With both drains open, the

tank is emptied in 6 min. How long would it take each drain,

working alone, to empty the tank?

Answer: Let t = time for faster drain to empty the tank, then

t + 16 = time for slower drain to empty the tank, since it takes 16

minutes longer. Therefore, the Rate for the fast drain = 1/t and the

Rate for the slower drain = 1/(t + 16). Remember that when two

drains work together, their rates add up. The Total rate = 1/6 ---->

1/6 = 1/t + 1/(t + 16) Multiply everything by the LCD = 6t(t + 16)


by Viken Kiledjian [6t(t + 16)]1/6 = [6t(t + 16)]1/t + [6t(t + 16)]1/(t + 16) --------->

t(t + 16) = 6(t + 16) + 6t ---------> t2 + 16t = 6t + 96 + 6t ------->

t2 + 4t – 96 = 0 --------> (t + 12)(t – 8) = 0 So, t = -12 or 8.

Discard the negative answer again because time can’t be negative.

The faster drain can drain in t = 8 minutes and the slower drain

can drain in t + 16 = 24 minutes!!!!

Ex 5: It took a motorboat 1 hour longer to travel 36 miles against

the current than to go 36 miles with the current. The rate of the

current was 3 mph. Find the rate of the boat in calm water.

Answer: Let r = rate of the boat in calm water, then r + 3 = rate

of the boat with the current, and r – 3 = rate of boat against current

We will again use the basic equation for rate problems: Distance =

Rate x Time. In this case, the Distance both ways = 36 miles. Also,

let t = time to travel with the current, then t + 1 = time to travel


by Viken Kiledjian against the current, since it took 1 hour longer. Here are the

equations we are left with:

Distance with the Current = 36 = (r + 3)t

Distance against the Current = 36 = (r – 3)(t + 1) Since the

question asks us to find r, we are really not interested in t.

Therefore, solve for t from the top equation and substitute it into the

bottom equation: t = 36/(r + 3) Plug this into the bottom equation:

Now, distribute the (r – 3) into the parenthesis

and multiply everything by the LCD = (r + 3)

1

3

36)3(36

rr

)3)(3()3(

)3(36)3()3(36)3(

)3(

)3(3636

rr

r

rrrr

r

r

mphrrrrrr 152259216)9(1083610836 222

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