Table Of Contents
I.
Sets of Numbers in the Real Number System
II.
Venn Diagrams
III. Subset Definition & Notation of Subsets
IV. Primes and Composites
V.
Tests For Divisibility
VI. Work with Fractions
VII. Operations with Signed Numbers
VIII. Absolute Value
IX. Order of Operations
X.
The Laws of the Real Number System
XI. Using The Laws
XII. Evaluating and Simplifying Algebraic Expression
XIII. The Distributive Law of Multiplication over Addition (Law
5)
XIV. More Laws (6-11)
XV. Using the Laws to Solve Equations

Table Of Contents, continued
XVI. Solving Formulas Using the Laws
XVII. Percent problems
XVIII. A Taste of Geometry
XIX. Solving Inequalities
XX. Chains of Inequalities
XXI. Graphing Two Variable Linear Equations
XXII. Graphing Single Variable Equations in Two Dimensions
XXIII. Graphing Two Variable Equations using Slope Method
XXIV. Nuances
XXV. Find Equations Given Facts
XXVI. Graphing Two Variable 2-D Inequalities
XXVII. Function Notation Introduced
XXVIII. Systems
XXIX. Two Methods of Solving Systems
XXX. Use Systems to Solve Word Problems
XXXI. Geometry Word Problems

Table Of Contents, continued
XXXII.Mixture Word Problems
XXXIII.Distance Word Problems
XXXIV.Graphing Systems of Linear Inequalities in Two
Dimensions
XXXV.Laws of Exponents
XXXVI.Exponent Examples
XXXVII.Expansions on the Schema
XXXVIII.Example of Zero & Negative Exponent Corollaries
XXXIX.Scientific Notation
XL. Operating With Polynomials
XLI. Review for Test VII
XLII.Multiplication of Polynomials
XLIII.Division of Polynomials
XLIV.Five Types of Factoring
XLV.Review of Types
XLVI.Uses of Algebraic Factoring

I
Natural Numbers
• Verbal definition:
• Symbolic definition:
Whole Numbers
•
Verbal definition:
• Symbolic definition:
Integers
• Verbal definition :
• Symbolic definition:
“the positive counting numbers”
N = {1,2,3, …}
“the natural numbers together with 0”
W = {0,1,2,3, …}
“ the whole numbers and their opposites”
J = {…-3,-2,-1,0,1,2,3…}

I
Rational Numbers
• Verbal definition:
• Symbolic definition:
Real Numbers
• Verbal definition:
• Symbolic definition:
“any number that can be expressed as a fraction with an integer numerator, and a nonzero integer denominator”
Q = {p/q  p ∈ J, q ∈ J, q ≠≠≠≠
0 } Read: the set of all p/q such that p is in J, q is in J, and q is not equal to 0.
Irrational Numbers
• Verbal definition:
• Symbolic definition:
“non-terminating, non-repeating decimal. It cannot be written as a fraction with an integer numerator, & denominator.”
H = {x  x ∈ Q} Read: the set of all x such that x is not a member of Q.
“any numbers that are rational or irrational”
R = {x  x ∈ Q U H } Read: the set of all x such that x is a member of “Q union H”

II
Venn diagrams or set diagrams are diagrams that show all hypothetically possible logical relations between a finite collection of sets (groups of things). Venn diagrams were invented around 1880 by John Venn. They are used in many fields, including set theory, probability, logic, statistics, and computer science
A Venn diagram is a diagram constructed with a collection of simple closed curves drawn in the plane. The principle of these diagrams is that classes be represented by regions in such relation to one another that all the possible logical relations of these classes can be indicated in the same diagram. That is, the diagram initially leaves room for any possible relation of the classes, and the actual or given relation, can then be specified by indicating that some particular region is null or is not null.
Source Wikipedia

II
R
2/3
-7
Q
J
W
N
Non-terminating, non-repeating decimals like
0 1
7 -23
π and e
3¼
-5/6

III
The definition of a subset could be stated: A is a subset of B
(Denoted A
⊆
B) if & only if (iff) all elements (numbers) of A are found in B. ( ⊆ is analogous to ≤ )
Ex1) J
⊆
Q
Ex2) J ⊆ N
Ex3) N ⊆ J
True
False
True

IV
The first hundred natural numbers and their designation
A Prime Number is a natural number greater than 1 that has only itself and 1 as factors
A Composite number is a natural number greater than 1 that is not prime
Factors Numbers that are being multiplied together
(See section VI for two methods of prime factorization)

V
There are three easy ways to test the divisibility of numbers
1) To test if a number is divisible by 2, look at the last digit.
If it’s even, then the number is divisible by 2.
2) Test for divisibility by 3 by taking the sum of the digits. If the sum is divisible by 3, then the original number is also divisible by 3.
Ex 1. 5280
⇒
5
+
2
+
8
+
0
=
15
3) Test for divisibility by 5, by looking at the last digit. If it’s a 5 or a 0, then the number is divisible by 5.

a) Mixed numbers are fractions
Ex:
2
3
4
=
11
4 b) When reducing fractions, use prime factorization. You should use divisibility tests. Prime factorize then cancel like factors.
c) Multiplying fractions is a lot easier if you use the prime factorization method.
• First you prime factorize the tops and bottoms.
• Next you will cancel like factors
• Last but not least, you will multiply straight across
Ex of PF method for multiplication
20
18
⋅
30
16
=
2
⋅
2
3
⋅
⋅
2
3
⋅
⋅
5
2
⋅
⋅
2
2
⋅
⋅
3
2
⋅
5
⋅
2
=
25
12 d) There are two methods of prime factorization in arithmetic
1.The Tree Method (Use for numbers < 100) Ex:
Primes in yellow

2.
2 ) 3760
=
2
⋅
5
⋅
47 (exponential form) for prime factorization
2 ) 1880
2 ) 940
2 ) 470
5 ) 235
47 e) Division of fractions: (two steps)
1. Invert the back or bottom fraction depending on how problem is written.
2. Next you must multiply using the multiplication rules. (Subheading C)
Ex:
7
8
÷
9
16
=
7
8
⋅
16
9
=
14
9 f) Addition and Subtraction of Fractions
To add or subtract, you must
1. get the least common denominator (LCD). In order to do this, you must first prime factorize (PF) the denominators only.
2. the next step is to Get Jealous.
This means you take any prime factors from one denominator that aren’t in the other, and you multiply the entire fraction by the missing primes.

Example of Addition and Subtraction of
Fractions:

a) When multiplying and dividing with signed numbers, if the numbers have like signs the the result will be positive, whereas unlike signs will result in negative answers.
−
3 (
−
4 )
=
12
(
7 )
7
+
(
−
18 )
= −
18 b) When doing addition and subtraction with signed numbers, think as if you’re working with money.
Ex1) .
9
+
.
5
+
(
−
.
2 )
+
(
−
.
9 )
=
.
3 or
90 cents
50 cents
20 cents
90 cents
30 cents
Ex2)
−
(
−
3 )
−
14
=
3
−
14
= −
11

(Da Wimpy Way, i.e. this will be explained in a more complicated way in higher math.)
The Absolute Value of a number is it’s distance away from zero on a number line. The symbol to indicate absolute value is
 n

.
Another way to think of it is the number with a positive sign.
3
Ex2) 5
−
7
= −
2
=
2
Ex3)
−
4 2
−
7
= −
4
−
5
= −
4 ( 5 )
= −
20

IX
Order of Operations
1. Do Single Term Exponents first
2. Do all operations within the parentheses or absolute value bars
3. If the result is a single term exponent, go back to rule 1.
4. Do Multiplication before Addition and Subtraction
Ex. 1 (note that terms are separated by + or -)
(
−
3 )
2 −
4 ( 5
−
7 )
3
9
−
4 (
−
2 )
3
9
−
4 (
−
8 )
9
+
32
Notes:
41
We do not do all the exponents first, only those that are single term.
Good algebraic form is done “step under step”.

Subtraction can be defined as addition with a negative number
Division can be defined as multiplication with a fraction
Law 1
The Commutative Law of Addition
(com
⊕
)
∀ a , b
∈
R a
b
b
a
Changing the order when adding does not affect the answer
If you add two natural numbers the sum will be a natural number
Law 2
The Commutative Law Of Multiplication
(com
⊗
)
∀ a , b
∈
R a
b
b
a or ab
= ba
Changing the order when multiplying does not affect the answer
A Law can also be called a property or axiom (self evident truth).

(continued)
∀ a , b , c
∈
R
Law 3
The Associative Law of Addition
( a
+ b )
(assoc
⊕
)
+ c
= a
+
( b
+ c )
= a
+ b
+ c
Changing the grouping when adding does not affect the answer
Law 4:
a , b ,
The Associative Law Of Multiplication
(assoc
⊗
) c
R ( ab ) c
a ( bc )
abc
Changing the grouping when multiplying does not affect the answer
More Laws to come…

Ex1)
( 2
2
(
2
2 x x
−
−
3 )
3
+
+ x x
8 x
−
8 x )
+
−
6 x
+
3
( 6
6
−
−
8
8 x x
6
3
)
( 6
−
3 )
Given
Assoc +
Comm +
Assoc +
Combine Like Terms (CLT)
When using the associative law, you must be careful of how you write it. The
+ must be outside the parentheses.
Keep the negative sign closely attached to a negative number:
It is okay to write (2x – 8x) + (6 – 3)
It is okay to write (2x – 8x) + (-3 + 6)
It is not okay to write (2x – 8x) – (3 + 6)

Ex2) 3 x
−
4
=
5
3 x
3 x
+
+
4
( 4
−
−
4
4 )
=
=
4
9
+
3 x
+
0
=
3 x
=
9
9
5
1
3


1
3
⋅
3 x
3

 x
=
1 x
=
=
3
1
3
1
3
⋅
9
⋅
9 x
=
3
Given
⊕
⊕
⊕
⊗
⊗
⊗ l l of
Inv
Id
Id of
Inv
=
=
(law applied to Given to get this line, etc.) assoc
Note: LHS = left hand side reason, RHS = right hand side reason .

y
= −
3 y
= −
3
Evaluate 2 x
−
3 y
2
2 (
−
2 )
−
3 (
−
3 )
2
−
4
−
3 ( 9 )
= −
4
−
27
= −
31
Evaluate −
5 ( 1
+ x
3
)
1
+ x
−
5 ( 1
+
(
−
2 )
3
)
1
−
2
−
5 ( 1
−
8 )
−
−
5 (
1
−
7 )
−
1
35
−
1
−
35

Law 5
This law can be stated symbolically as follows:
a, b, c,
R a(b + c) = ab + ac
It can also be stated verbally: The result of first adding several numbers and then multiplying the sum by some number is the same as first multiplying each separately by the number and then adding the products.
Source Britannica concise dictionary
The abbreviation of this Law is:
⊗ ⁄ ⊕

(continued)
Ex1) −
3 ( 6
−
4 x )
−
3 ( 6 )
+ −
3 (
−
4 x )
−
18
+
12 x
Ex2)
−
2
3


5
6
3 x


+
1
2


2
3
−
10
18
+
6 x
3
+
2
6
−
4 x
2
4 x


−
5


−
9
5
9
+
+
2 x
1
3


+
+
1
3
−
2 x
( 2 x
−
2 x )


−
5
9
+
3
9


+
0
−
2
9

(Continued)
Law 6
The Additive Identity Law
∀ a
∈
R
(
⊕
I.D.) a
+
0
= a
Ex 1) 3
+
0
=
3
Ex 1)
Law 7
The Multiplicative Identity Law
(
⊗
I.D.)
∀ a
∈
R 1 a
= a



2 x
2 x
−
3
−
3



( 3 x
−
5 )
=
( 1 )( 3 x
−
5 )
=
3 x
−
5

(Continued)
Law 8
The Additive Inverse Law (
⊕
Inv)
∀ a
∈
R a
+
(
− a )
=
0 a
− a
=
0
+ − =
0
Ex 2)
3 x
− y
+ y
−
3 x
=
0
∀ a
Law 9
The Multiplicative Inverse Law (
⊗
Inv)
∈
R a



1 a



=
1 a
1


1 a


=
1 a a
=
1
Ex 1)
2
3


3
2


=
1
Ex 2)
−
3


1
3


=
1

∀ a , b , c
∈
R
(Continued)
Law 10
The Addition Law of Equality (
⊕
Law of =) a
= a
+ c
= b
+ c
Ex 1) if 2
+
9
=
5
+
6 then 2
+
9
+
8
=
5
+
6
+
8
Law 11
The Multiplication Law of Equality (
⊗
Law of =)
a , b
Ex 1) if
, c
R if a
= b
2
+
3
=
7
−
2 then then ac
= bc
3 ( 2
+
3 )
=
3 ( 7
−
2 )

Example 1)
3 x
=
29
1
3
⋅
3 x
=
29
⋅
1
3


1
3
3


 x
=
29
3
1 x
=
29
3 x
=
29
3
Given
(
⊗
Law of =)
(assoc
⊗
) and Arithmetic
(
⊗
Inv)
(
⊗
I.D.)

XV Using the Laws to Solve Equations, continued
Ex 2)
Given
(
⊕
Law of =)
(assoc
⊕
) & Arithmetic
(
⊕
Inv)
(
⊕
I.D.)
(
⊗
Law of =)
(assoc
⊗
) & Arithmetic
(
⊗
Inv)
(
⊗
I.D.)

Ex 1)
3 x
−
2
3 x
−
3 x
−
3
2
3 y y
=
= x
+ z x
−
3 x
+ z
( 3 x
−
3 x )
−
2 y
= −
2 x
+ z
3
2
( 0 )
− y
= −
2 x
+ z
3
−


−
3
2
3
2


 −
⋅ −
−
2
3
2
3 y

 y
=
=
−
−
2 x
3
2
+ z
(
−
2 x
+
2
3

 y
=
3 x
−
3
2 z z )
( 1 ) y y
=
=
3 x
−
3 x
−
2
3
3
2 z z
Given
Addition Law of Equality
Associative Law of Addition
Additive Inverse
Additive Identity
Multiplication Law of Equality
Associative Law and Distributive Law of of Multiplication multiplication over Addition
Multiplicative Inverse
Multiplicative Identity

XVII Percent problems
(Translation method)
English
Is
Of
Percent
What
Find
Change “What is” to
100 x
=
Math
=
or
100 x
Ex 1) What is 2% of 18 x
=
2
100
⋅
18
1
=
1
50
⋅
18
1
=
1
25
⋅
9
1
=
9
25
When doing a problem like this, it is best to prime factorize these numbers for easier reduction.

To find the area of a square, the proper formula is A = s 2
The formula to find the area of a rectangle is
A = LW
Therefore
4’(2’) = 8ft 2
2’
4’
L is the length and
W is the width
To find the perimeter of a square, the formula is P = 4s
The formula to find the perimeter of a rectangle is
P = 2L + 2W
Therefore
4’(2) + 2’(2) = 12’

Ex 1)
2’
(Continued)
4’
The formula for the area of a triangle can be expressed:
A
= bh
2
A
=
1
2 bh b = base of the triangle h = height of the triangle
1
2
( 4
′ ⋅
2
′
)
=
4 ft
2
2’
4’
There is no formula to find the perimeter of a triangle, except for an equilateral triangle. The formula for that would be:
P
= b
⋅
3
To find the perimeter of any other triangle, you would follow this:
P
= s
1
+ s
2
+ s
3

(Continued)
Ex 1) r
=
3 "
The area of a circle can be expressed as:
A
= π r
2
Where r = the radius
To find the circumference of this circle we will use the formula:
P
= π d
Where d = the diameter
C
Therefore the circumference of the circle is
= π
( 6 )
≈
3 .
14 ( 6 )
=
18 .
84 in
π is one of the most commonly used irrational numbers. It is equivalent to the number of times a radius goes around a semi-circle, and is roughly equivalent to
3 .
14 or
22
7
Therefore the above circle will have an area of:
A
= π
( 3 )
2
≈
22
7
( 9 )
=
2
28
7 in
2
When talking about circles, the word circumference replaces the word perimeter

Law 12
The Addition Law of Inequality (
⊕
Law of ineq)
∀ a, b, c,
∈
R
If a < b then a + c < b + c
Law 13
The Multiplication Law of Inequality (
⊗
Law of ineq)
∀ a, b, c,
∈
R
If a < b then ac < bc iff c > 0 but
If a < b then ac > bc iff c < 0

(continued)
Ex 1) 1
3
+ c
5
> −
3
2


1
3
1
3
−
1
3
+ c
5
1
3


+ c
5
> −
3
2
−
1
3
> −
3
2


3
3


−
1
3


2
2


0
+
0
+ c
5 c
5
>
>
−
−
9
6
9
6
−
−
2
2
6 c
5
5
1
⋅ c
5
> −
11
6
> −
11
⋅
6
5
1
1
⋅ c
> −
55
6 c
> −
9
1
6
Given
⊕
Law of ineq
⊕
Assoc |
⊗
I.D. Twice
⊕
Inv | Arith
Arith
⊕
I.D. | Arith
⊗
Law of ineq
⊗
Inv | Arith
⊗
I.D. | Arith
Now that we have isolated the variable, we must show the solution for the inequality in one of 3 ways. We can either graph it, or write it in one of two types of solution sets.
Solution sets can either be written in
Set Builder Notation
(SBN), or Interval
Notation (IN)
-10 -5 0 5 10
(
Set Builder Notation c | c
> −
9
1
6
Interval Notation
−
9
1
6
,
∞

Whenever using Set
Builder notation, you must put braces { } on the outside of the set to indicate that this is the solution set.
When solving an inequality, it is important to note that when multiplying by a negative number, the direction of the inequality sign must be reversed.
XIX
Solving Inequalities,
(Continued)
When graphing or writing the solution set in interval notation, if the end point is known, you must use a bracket.
Use a parenthesis to indicate an unspecifiable end point.
When solving inequalities, it is also important to take note that 2
3 y
=
2
3 y
2

(continued)
Ex 2)
−
3
4 x
+ 

−
3
4 x
5
6
+
5
6
5



6
≤
≤
−
−
1
2
1
2
−
5
6
Note that we multiplied by a negative number so the inequality changed direction.
(Mult. Law of inequality)
−
4
3
⋅ −
3
4 x
≥ −
4
3


−
1
2
−
5
6

 x
≥ −
4
3


−
1
2
−
5
6

 x
≥
4
6
+
20
18
Here we use the distributive law of multiplication over addition
In order to add these two fractions together, we find an LCD using the
Jealousy Method (VI) x
≥
2
3
+
10
9 x x
≥
≥


3
3


2
3
6
+
10
+
9
10
3
⋅
3 x
≥
1
7
9
1 7
9
[
-2 -1 0 1 2
Solution Sets
IN SBN

 1
7
9
,
∞

 x | x
≥
1
7
9

(Compound inequalities)
Ex 1)
When two inequalities are combined to show that an expression lies between two fixed values, this is known as a
Compound Inequality.
−
2
+
3
≤
5 x
<
1
2
+
3
1
≤
5 x
<
3
1
2
1
5
⋅
1
≤
1
5
⋅
5 x
<
1
5
⋅
7
2
Compound
Inequalities must be solved from the middle
1
5
≤ x
<
7
10
The Solution can be verbalized as all of the x’s between one fifth and seven tenths.
IN


1
5
7
,
10


SBN x |
1
5
≤ x
<
7
10
0 ¼ ½ ¾ 1
[ )
1
5
7
10

a) Descartes (Inventor of the Cartesian Coordinate System)
All real Numbers can be graphed using the Cartesian Coordinate System
When plotting on these graphs, it is imperative to remember that x always comes first when writing the points
It is good practice to plot three points when you are first starting, simply to check your work, however, it is only necessary to plot two points.

XXI Graphing Two Variable Linear Equations
(Continued)

XXI Graphing Two Variable Linear Equations
(Continued) b) Go to Work
W hen trying to graph an equation, then the
Point Plotting Method is a good method to use. This method is a quick and easy way to find where the x and yintercepts are located.
First a table, as shown in the previous slide, must be created, only instead of the complicated table shown earlier, the only table necessary to graph an equation like
2 x
+
3 y
=
1 looks like this:
When substituting a number for x or y , always use 0, as it will save endless time and energy.
x
0 y
?
?
0
So if we substitute zero for x to find y , our equation will look something like this
2 ( 0 )
+
3 y
=
1
3
3 y y
=
=
1
1
3
3
So now we can fill in the ? For y in the above table
And if we substitute zero for y to find x , our equation will look something like this
2 x
+
3 ( 0 )
=
1
2 x
2 x
=
1
=
2
1
2

Single
Ex 1) Graph
When x is the variable, the line is vertical.
x y
3½ 0
3½ 1
3½ 2
3½ 3
3½ 4
3½ 5
3½ 6

In order to use the Slope Method we must first change the equation from General Form, into Slope Intercept Form y = mx + b
Where m = slope b = y -intercept
To make this change, we must get y alone in the equation.
Ex 1)
Now that we have the slope intercept form, we know that the y -intercept is located at and we know that the slope is positive so now we can graph it as shown:

(Continued)
There are three key steps in graphing equations with the Slope Method.
1.
Get y alone, or change the equation from General Form to Slope Intercept Form
2.
Pick off, and spot the yintercept. It is important to remember to do this step before the next step.
3.
Plot the rise over run, or the slope. As shown in the previous slide, the slope of the equation is m 3
7 of three parallel to the y -axis, we will run positive seven parallel to the x -axis.
The slope of a line passing through points ( x
1 is
, y
1
) and ( x
2
, y
2
) m =
Vertical Change
Horizontal Change
=
Rise
Run
=
Change in y
Change in x
= y x
1
1
−
− y x
2
2 if x
1
≠ x
2

There are several facts that must be observed when working with single variable equations. It must be remembered that a horizontal line has a slope of zero, while a vertical line has an undefined (infinite) slope. A good way to remember this is to remember that when solving for y in a single variable equation (like Ex 2 in XXII), the slope will be zero because y = mx + b becomes y = (0) x + b which reduces to y = b .
However, when solving for x in a single variable equation, the slope will be undefined
(infinity). This is because in any vertical line, no matter what the rise is, the run is always 0. Division by 0 is undefined (rise/0 = ∞ ), hence infinite, since there are an infinite number of 0’s in any number.
Parallel lines have the same slope, and can be symbolized

Perpendicular lines will always have slopes that are “negative reciprocals” of each other.
The symbol for perpendicular lines is ⊥ . All horizontal lines are ⊥ to all vertical lines x
− y Find the slope of a line
⊥ to this line
−
4 y
−
4
=
−
3 x
−
4
+ y
=
3
4 x
−
7
4
7
−
∴ is a symbol that means therefore
4
−
4
∴ a line that is ⊥ to this line would have a slope of
3

XXV
Find Equations Given Facts
(Analytical Geometry)
Step 1
Draw a picture (DAP)
Ex 1) If a line goes thru (-2,4) and has a slope of -3, find the equation of the line in both GF & SIF
Step 1) Draw A Picture
Step 2
Write the SIF formula: y = mx +b
(-2,4)
(-1,1)
Step 3
Plug in known “ m ” value and any known ( x,y ) point.
Step 4
Solve for “ b ”
Step 5
Insert m and b values into y = mx + b to get answer in SIF. For GF, clear any fractions and move the x term to the far left side of the equation.
The x term must also be positive.

XXV
Find Equations Given Facts
(Continued)
Step 2) Write y = mx + b
Step 3) Plug in Known Facts
This is the most critical part of the process, because extreme care must be taken to insure that all facts are plugged in at the proper place, or the equation will be wrong. We know slope = -3 and it has the point (-2,4) y = -3 x + b
4 = -3(-2) + b
Step 4) Solve for b.
4 = 6 + b
-2 = b
Step 5) Insert m and b values into y = mx + b
Now that the y -intercept has been found, all the information is known to write the equation in SIF y = -3 x – 2 which can be changed to 3 x + y = -2 to obtain General Form (GF)

XXV
Find Equations Given Facts
(Continued)
Ex 2) Find the equation of a line parallel to 3 x – 4 y = -1 that goes thru (-2,-1)
Step 1)
DAP

XXV
Find Equations Given Facts
(Continued)
Step 2)
Write y
= mx
+ b
Step 3)
Plug in known facts (slope and point (-2,-1))
−
1
=
3
4
(
−
2 )
+ b
Step 4) Solve for b
−
1
= −
3
2
+ b
3
2
−
1
= b
1
2
= b
Step 5) Now we have both slope and y-intercept.
∴ y
=
3
4 x
+
1
2

XXV
Find Equations Given Facts
(Continued)
Step 5) continued
SIF is: y
=
3
4 x
+
1
2
To get the GF, we must put the x on the other side in order to arrange the equation into general form ax
+ by
= c
3
Or
− x
+ y
=
1
4 2
But we cannot have any fractions or a negative in front of x , so we must multiply the equation by the LCD of all the denominators and by -1 in order to simplify the equation into GF
Multiply by (4)(-1) = -4
−
4


−
3
4 x
+ y


= −
4
⋅
1
2
3 x
−
4 y
= −
2
Voila!

XXV
Find Equations Given Facts
(Continued)
Ex 3) Find the equation of a line that passes thru the two given points (-3,0) & (3,1)
Step 1) DAP

XXV
Find Equations Given Facts
(Continued)
Write
Step 2) y
= mx
+ b
Step 3)
Plug in known facts, 1/6 & point (3,1) y
=
1
6 x
+ b
Step 4)
Simplify and return “ b ” to SIF
1
=
1
6
⋅
3
+ b
1
=
1
2
+ b
1
−
1
2
= b
1
2
= b y
=
1
6 x
+
1
2 Now see if you can convert to GF!!

XXVI Graphing Two Variable 2-D Inequalities
Step 1: Graph the equation as if it is an equality. Use either
Slope Method, or Point plotting method.
Step 2: Use a solid line if it is ≤ , or ≥ , and use a dotted line if <, or >.
Step 3: Test a point. The easiest and best point is (0,0). If the point is false, shade the other side of the line.
SLOPE METHOD
Ex1) 2x – 3y < -1
POINT PLOTTING METHOD
-3y < -2x – 1
For X
2x – 3(0) < -1
For Y
2(0) – 3y < -1
-3y < -2x – 1
-3 -3 y < 2x + 1
3
2x <-1
2 2 x < -½
-3y < -1
3 3 y < ⅓ x
0
-½ y
⅓ o

XXVI Graphing Two Variable 2-D Inequalities
(Continued)

XXVI
Graphing Two Variable 2-D Inequalities
(Continued)
Ex2) -2x + 8 > 5
-2x > -3
2 2
-x > -3
2 x < 3
2

XXVII
Function Notation Introduced f(x) means a function (an algebraic expression) evaluated at whatever x is
Or
A statement where the only variable is x
Or
Y is a statement in terms
A function is a set of ordered pairs (a relation) in which to each first component, there corresponds exactly one second component. Therefore, if to each value of x in a given domain, there is assigned exactly one value of y in the range, then y is said to be a function of x.
The notation y = f(x) denotes that the variable y is a function of x.
F(x) = f of x not f ✖ x

XXVII
Function Notation Introduced
(Continued)
Ex1) f(x) = 2x – 3 find f(-1) f(-1) = 2(-1) – 3 f(-1) = -2 – 3 f(-1) = -5
Find f(2) f(2) = 2(2) – 3 f(2) = 4 – 3 f(2) = 1

Ex1) Graph a) 2x – 3y = 8 b) -x + y = -7 a)
X
0
4
Y
-8
3
0 b)
X
0
7
Y
-7
0
XXVIII
Systems
(Bunches of Equations)

XXIX
Two Methods of Solving Systems a) Substitution Method
1) Solve either equation for either variable.
2) Force result (a.k.a. THE BLOB) into the other equation
3) Solve the resulting single variable equation.
4) Put the result into the BLOB to get the other variable value.
Ex1) a) 2x – 3y = 8 b) –x + y = -7
Step 1: b) y = x – 7
Step 2: a) 2x – 3(x – 7) = 8
As equation b) is already solved for y, that will be the BLOB.

Step 3
XXIX
Two Methods of Solving Systems
(Continued) a) 2x – 3(x – 7) = 8
2x – 3x + 21 = 8
The substitution method is best used when
∃∃∃∃ positive 1 on at least one variable.
( ∃∃∃∃ means there exists)
-x + 21 – 21 = 8 – 21
-x = -13 x = 13
Step 4) Now that we have solved for x, we can force it into the BLOB
∴ y = 13 – 7 y = 6
This point (13,6) is is actually where equations a & b intersect, and is the
Solution Set
(13,6)
.

XXIX
Two Methods of Solving Systems
(Continued)
Ex2) a) x + y = 1
4 4 b) y + 11 = x
2 20 10
Because the y is a positive 1, the substitution method should be used, with equation a used for the BLOB.
However, before this equations can be solved, all fractions must be cleared
Eq. a) 4 ( + y = )
1 4 4
Step 1) x + 4y – 4y = 1 – 4y x = -4y + 1 BLOB
Eq. b) 20 ( + = )
Step 2) 10y + 11 = 2(1 – 4y)
Step 3) 10y + 11 = 2 – 8y
10y +8y + 11 = 2 – 8y + 8y a) x + 4y = 1
18y + 11 – 11 = 2 - 11 b) 10y + 11 = 2x
18y = -9
18 18
Now these equations can be solved using the substitution method y = -½
Step 4) Put answer into
BLOB x = 1 – 4(-½) x = 3
The solution set for equations a & b is:
(3, -½)

XXIX
Two Methods of Solving Systems
(Continued) b) Addition Method
(a.k.a. Elimination Method)
Step 1) Be sure x’s, y’s, and constants
(no variable) are lined up.
it
Step 2) Multiply through equation a or b, or both, by whatever takes so that when added, either x or y will cancel.
Step 3) Solve the resulting single variable equation.
Step 4) Use that result in either equation to get the value of the other variable.

XXIX
Two Methods of Solving Systems
(Continued)
Ex1) Eq a) 2x – 3y = 8
Eq b) 5x + 2y = -7
Step 2: 2(2x – 3y = 8)
3(5x + 2y = -7)
2( ) – 3y = 8
19
_ 10 – 3y = 8
19
-3y = + 8 19
19
-3y = 10 + 152
19
_ 1
3
(-3y) = ( )
19 3 / y =
19
19x = -5
19 19 x = _ 5
19

Ex1) Alaska: Most of the 1,422 mile-long Alaskan Highway is actually in Canada.
Find the length of the highway that is in Canada if it is known that the difference in lengths is 1,020 miles.
a)x + y = 1,422 b) x – y = 1,020 b) x = y + 1,020 BLOB a)(y + 1,020) + y = 1,422
2y + 1,020 – 1,020 = 1422 – 1020
2y = 402
2 2
Y = 201 miles of the highway are in the U.S.
∴ x + 201 – 201 = 1422 - 201
Or
X = 1221 miles are in Canada

(Continued)
Ex 2) Pension Funds: A state employees’ pension fund invested a total of one million dollars into two accounts that earned 3.5% and 4.5% annual interest. At the end of the year, the total interest earned from the two investments was $39,000. How much was invested at each rate?
a) x + y = 1 million
( x +
100 y = 39,000)1000
Y = -x + 1 million BLOB
35x + 45(-x + 1 million) = 39 million a) 600,000 + y = 1,000,000
-10x = -6 million
10 10
600,000 – 600,000 + y = 1,000,000 – 600,000
-x = - $600,000 y = $400,000
X = $600,000

a) Angles of a triangle add up to 180 ° b) Right angles are 90 ° c) A whole circle is 360 ° d) Quadrantal angles are 90 ° , 180 ° , 270 ° , 360 ° , 450 ° , etc.
e) Supplementary angles add up to 180 ° f) Complimentary angles add up to 90 °
Ex1) Two angles are supplementary. The measure of one angle is 20 less than 19 times the measure of the other. Find the measure of each angle.
a) x + y = 180 b) x – 19y – 20
(19y – 20) + y = 180
20y = 200
20 20 y = 10 therefore x = 170

Ex1) Mouthwash a pharmacist has a mouthwash solution that is 6% ethanol acohol and another that is 18% ethanol alcohol. How many millimeters of each

Distance = Rate × Time
Ex1) The jet stream is a strong wind current that flows across the United States. Flying with the jet stream a plane flew 3000 miles in five hours. Flying against the same wind, the trip took six hours. Find the speed of the plane in still air, and the speed of the wind. a)3000 = r(5) b) 3000 = r(6)
R = 600 r = 500
This is not correct.
The different rates must be distinguished a) 3000 = (r p
+ r w
)5 b) 3000 = (r p
– r w
)6
5r p
+ 5r w
= 6r p
– 6r w
11r w
= 1r p
BLOB a) 3000 = (11r w
+ r w
)5
3000 = 60r w
60 60 r w
= 50mph ∴ r p
= 11(50) or 550mph

(see XXVI)
Ex1) a) 2x – 3y ≤ 12 b) 3x + 5y ≥ 15
X
0
6
Equation A
Y
-4
0
PPM
Equation B
X
0
5
Y
3
0
The crosshatched yellow region is the solution set
.

(Continued)
Ex2) a) 2x + y = 2 b) y = x c) x = 0
The yellow shaded area is the
Solution Set.

A) In multiplication of like bases, add the exponents and write the base just once.
Ex1) x 2 × x 3 = x 2+3 or x 5
Ex2) x 3x – 4y × x 5 – 2x + 6y x 3x -4y + 5 – 2x
A) In division of like bases, write the base just once then subtract the exponents (usually top from bottom).
B) In raising a power inside to a power outside, write the base just once and multiply the exponents.
C) In taking roots (radicals), divide the index of the radical (the number in the crotch) into the exponent of the base.

Ex 1) 3 4 * 3 6 = 3 10 (exponential form)
Ex 2) 10 4 =
10 4 – 2 or 10 2 = 100 (expanded form)
10 2
Ex 3) (m + 8) 20
(m + 8)
=
(m + 8) 20 – 1c = (m + 8) 19
Ex 4) x 4p * y 2x x p * y -3x
= x 4p – p * y 2x – (-3x) = x 3p * y 5x
Ex 5) x 3y (x -3 * y 4 + x 2y ) = x 3y – 3 + y 4 + x 5y
Ex 6) (s 3 t 3 ) 4
(st) 2
= s 12 t 12 = s 12 – 2 t 12 – 2 = s 10 t 10 = (st) 10 s 2 t 2

Ex1) 3x 2 y – xy 2 + 4xy 2 – 5x 2 y
-2x 2 y + 3xy 2
Ex 2) x 2 y(3x 2 y – 4xy 2 ) Contrast example
3x 4 y 2 – 4x 3 y 3
Operation Exponent
+ CLT
-
×
÷
CLT
+
Rp
Tr
-
×
÷
Ex 1) When exponents are the same x 4 – x 4 then x 4 – 4 = x 0 = 1
This is true, because anything divided by itself will result in one.

(Continued)
Ex 2) When the exponent on bottom is greater than the exponent on top
X 4 = x 4 – 12 = x -8 = 1
X 12 x 8
This rule is known as the
Negative exponent corollary and can be written as a -n = 1 a n
This might look a little unusual, but from the definition of exponents in algebra, the exact same problem could be written xxxx _ xxxxxxxxxxxx
Once everything has been canceled, a one will be left on top while eight x’s will be left on bottom
Or
1 x 8

XXXVIII
Example of Zero & Negative Exponent Corollaries
Ex 1) x
4
-2
-6 y 0 z
5 -2 ( ) x -8 y 12 z -10 x 4 y 0 z 10 x -8 – 4 y 12 - 0 z -10 – 10 x -12 y 12 z -20
Steps
1)Use the raising to a power rule
2)Use the quotient rule
3)Use the negative exponent rule y 12 x 12 z 20

a) LARGE numbers are characterized with positive exponents
Ex 1) 93,000,000======= 9.3 × 10 7
Standard notation Scientific Notation b) SMALL numbers are characterized with negative exponents
Ex 1) .000032 ======== 3.2 × 10 -5
Standard Notation Scientific Notation
Ex 2) 6 × 10 -8 ======== .00000006
Scientific Notation Standard Notation c) Operating in Scientific notation
1) Multiplication problems
Ex1) 300,000(.00042)
(3 × 10 5 )(4.2 × 10 -4 )

(Continued) c) Operating in Scientific notation
Ex 1) Continued
(3 × 4.2)(10 5 × 10 -4 )
12.6 × 10 5 – 4
12.6 × 10 1
126 Standard Notation
1.26 × 10 2 Scientific Notation
2) Division Problems
Ex 1) 48,000
.0000008
(4.8 × 10 4 )
(8 × 10 -7 )
If asked to put the answer into scientific notation, only this answer is correct
.6 × 10 4 – (-7)
Or
.6 × 10 11
Or
6 × 10 10 Scientific Notation

(Continued)
3) Raising to a power problems
Ex 1) (.000004) -3
(4 × 10 -6 ) -3
4 -3 × 10 18
1 × 10 18
4 3
1 or 64
× 10 18
.015625
× 10 18 Or 1.5625 × 10 16 Scientific Notation
4) A mixture of Operations
(.000008)(320,000)
.00024
1.06 ×
Or
10666.6
10 4
Once the problem has been completed, click the mouse for the answer

a) Monomials – One term
Ex 1) -3x 2 y b) Binomials – Two terms
Ex 2) 3x 2 y – 4xy 3 c) Trinomials – Three terms
Ex 3) 2x 2 – 4x + 5 d) Four or more terms are called polynomials.
e) Addition and subtraction of polynomials
Ex 1) (2x 2 – 3x) – (4x 3 – 3x 2 + 8)
2x 2 – 3x – 4x 3 + 3x 2 – 8
-4x 3 + 5x 2 – 3x – 8
It is important to remember that the highest power should always be written first

(Continued) f) Evaluating polynomials (see XII)
- x 2 – ( 10x 2 – 3x) – 4( x 2 – 5x)
3 5 3
2x 2
3
– 8x + – 8x 2 + 20x
5 3
34x 2 112x
3 5
-170x 2 + 336x
15 Evaluate when x = -2
- 170(-2) 2 + 336(-2)
15
==========
15
-1352
15

1) x -3 x 4x + 2
2) x -4 x -3 + x
3) (x -3x – 4 ) -5
4) a b a b + 1
5) a -3 b -3 + 4x
6) a 2x – 4 a 3x + 1
7) a 3b – 2 a 2 + 4b
8) (a 3b – 4 ) -2x
9) x -2 – x -1
10) i -x – 2i -1

a) Monomial multiplication
Ex 1) -3x 2 (2x – 4)
12x 2 – 6x 3 b) Binomial multiplication
Ex 2) -2x 3 y 2 (5x 2 – y
-10x 5
4 y
)
6
Ex 1) (2x – 3y)(2x 2 – 3x + 4)
4x 3 – 6x 2 + 8x – 6x 2 y + 9xy – 12y c) Trinomial Multiplication will yield twevle terms, unless it is possible to CLT d) Two very critical special cases
1) Product of conjugates ( difference of squares) expressed as same two binomials with different signs

(Continued)
Ex 2) (2x 2 y 3 – 3xy 2 )(2x 2 y 3 + 3xy 2 ) Ex 1) (2x – 3y)(2x + 3y)
4x 2 – 9y 2 4x 4 y 6 – 9x 2 y 4
2)Squaring of a binomial
Ex 1) (2x 2 y – 3xy 2 ) 2
Step 1) 4x 4 y 2
Step 2) 2x 2 y × 3xy 2
-12x 3 y 3
Step 3)
Step 4)
9x 2 y 4
∴
4x 4 y 2 – 12x 3 y 3 + 9x 2 y 4
The rules for squaring a binomial
1)Square the first term
2)Twice the cross product (TCP) a) Do the sign first b) Do the numbers 2 nd c) Do x’s next d) Do y’s e) Twice the answer
3)Square the 2 nd term
4)Combine the three answers for the final answer.

(Continued)
E) Cubes and higher powers of a binomial
Ex 1) (x – 2y) 3 or (x – 2y) (x – 2y) (x – 2y)
Treat as a square 1 st
∴
(x 2 – 4xy + 4y 2 )(x – 2y) x 3 – 4x 2 y + 4xy 2 – 2x 2 y +8xy 2 – 8y 3
Or
X 3 – 6x 2 y + 12xy 2 – 8y 3

(Continued) f) A mixture of types
Ex 1) -3(x 2 – .2y) 2 – 2x(x – 2 y 2 )(x + y 2 )
3
2
3
-3(x 4 - .4x
2 y + .04)2x(x 2 4 – y 4 )
9
-3x 4 + 1.2x
2 y – .16y
2 – 2x 3 8
+ xy 4
9

a) Division of a polynomial by a monomial
Ex 1) 2x 2 – 3xy + 4x 2 y 2
24x 3 y 2
2x 2 _ 3xy
+
4x 2 y 2
24x 3 y 2 24x 3 y 2 24x 3 y 2
1 _ 1
+
1
12xy 2 8x 2 y 6x
Ex 2) 3x -4 – 2x 5
6x 7
3x -4 _ 2x 5
6x 7 6x 7 x -11 _ x -2 or 1 _ 1
2 3 2x 11 3x 2

(Continued) b) Division of a polynomial by a binomial
Ex 1) x 3 – 8 x – 2 x 2 + 2x + 4 x – 2)x 2 + 0x 2 + 0x 1 – 8
_ x 3 – 2x 2
2x 2 – 0x 1
2x 2 + 4x 1
4x – 8
-4x + 8
0
Rules
0) Ensure that the problem is arranged in descending (or ascending) order, with all powers mentioned
1)Divide the first term into the first term.
2)Multiply that result by both numbers of the divisor.
3)Draw a line, change signs and add
4)Bring down next term and repeat steps
1,2,3

a) Intro – In Arithmetic
Trivial factors are 1 and the number itself. In this case the trivial factors would be 1 & 15.
b) Common Monomial Factoring (Distributive Law Backwards)
Ex 1) 6x 2 y
2x
3
2
– 8x y 2
4 y 2 + 10x
(3y – 4x 2
3 y 3
+ 5xy)
Find a common factor and write just once.
In this case 2x 2 y 2 is the common factor.
c) Grouping (Multiple [usually three times] use of common monomial type)
Ex 1) 12ab – 4ac + 3bd – cd
4a(3b – c) + d(3b – c)
BLOB BLOB
(3b – c)(4a + d)

XLIV
Five Types of Factoring
(Continued)
Ex 2) 4abc + 4ac 2 – 2bc – 2c 2
2c(2ab + 2ac – b – c)
2c(2a[b + c] – [b + c])
2c([b + c][2a – 1])
In order to check the answer use the FOIL
(first, outside, inside, last) method to get back to four terms d) Trinomial factoring
Ex 1) 8x 2 – 2x – 15
Example Continued on next slide

d) Trinomial Factoring
Ex 1) 8x 2 – 2x – 15
XLIV
Five Types of Factoring
(Continued)
Step 2)
Step 3)
Step 4)
Step 5)
Step 6)
(2x ?)(4x ? )
8x 2 – 2x – 15
(2x 3)(4x 5)
12x
10x
(2x 3)(4x 5)
- 12x
+ 10x
(2x – 3)(4x + 5)
Rules
1)Pick two numbers that will multiply together to get 1 st term, and usually (probability favors this) use the two closest to each other.
2)Pick two numbers that multiply together to get the last term.
3)Multiply the inners.
4)Multiply the outers.
5)Pick signs to give the middle term.
6)Upper sign to the left, lower sign to the right.
7)Check to be sure last terms multiply to give the last term.

XLIV
Five Types of Factoring
(Continued)
Ex 2) 16r – 40r 2 + 25r 3 r(16 – 40r +25r 2 ) r(4 – 5r)(4 – 5r)
- 20r
- 20r r(4 – 5r) 2
Ex 3) 6s 5 – 26s 4 – 20s 3
2s 3 (3s 2 – 13s – 10)
2s 3 (3s ? 2s)(s ? 5)
2s
15s
2s 3 (3s + 2)(s – 5)

Ex 4) 6x 2 + x – 12
(2x )(3x )
(2x 3)(3x 4)
+ 9x
- 8x
(2x + 3)(3x – 4)
XLIV
Five Types of Factoring
(Continued)
Mac’s Rule
Never put like factors into the same parenthesis.
Ex 1) (2x + 3) is okay
(3x – 3) is not okay e)Difference of Squares (Product of conjugates)
Ex 1) 4x 2 – 81
(2x + 9)(2x – 9)
As defined earlier in these notes, the word difference can be translated mathematically into subtraction.

XLIV
Five Types of Factoring
(Continued) e) Difference of Squares
Ex 2) (16x 12 – 121y 6 )
(4x 2 + 11y 3 )(4x 2 – 11y 3 )
Any even power is a square
Ex 3) 12x 6 – 3y 2 z 4
3(4x 6 – y 2 z 4 )
3(2x 3 +yz 2 )(2x 3 – yz 2 )
Don’t be fooled by difference of squares that are masked with a simple monomial as shown in example 3

XLIV
Five Types of Factoring
(Continued)
( Cube
Root of
As problem
Rules f) Factoring of Cubes
A B different
)(A
Sign from original
AB + B
) problem 1 st term
Same
Sign
Cube
Root of
2 nd term
Ex 1) 8x 3 + 1
(2x + 1)(4x 2 -2x + 1)
Ex 2) (x 3 y 6 – 27z 9 )
(xy 2 – 3z 3 )(x 2 y 4 + 3xy 2 z 3 + 9z 6 )

1) Common Monomial – any number of terms
2) Difference of Squares – two terms
3) Trinomial – three terms
4) Grouping – four terms
5) Cubes – two terms

XLVI
Uses of Algebraic Factoring a) To Solve factorable equations of degrees (powers) greater than one.
Ex 1) 2x 2 – 8 = 0
1 st degree equations are called linear equations.
2 nd degree equations are called quadratic equations.
3 rd degree equations are called cubic equations
2(x 2 – 4) = 0
4 th degree equations are known as quartic equations.
5 th degree equations are called quintic equations.
2(x + 2)(x – 2) = 0
SS { x| x = + 2} or
SS {x| x = 2, x = -2}
An Equation will yield as many answers as the highest degree of the equation. In this case, the highest power is two, and that is how many answers will be in the solution set

XLVI
Uses of Algebraic Factoring
(Continued)
Ex 2) 2x 3 – 3x 2 – 2x + 3 = 0 x 2 (2x -3) – (2x – 3)
(x 2 – 1)(2x – 3) = 0
(x – 1)(x + 1)(2x – 3) = 0
2x – 3 = 0 x – 1 = 0 x + 1 = 0
2x = 3
2 2 x = 1 x = -1 x = 3 SS {x| x = + 1, x = }
2
2
Notice that this is a cubic equation, which means it will yield three answers.

XLVI
Uses of Algebraic Factoring
(Continued)
Ex 3) x 4 – 13x 2 + 36 = 0
(x 2 – 4)(x 2 – 9) = 0
- 4x 2
- 9x 2
(x – 2)(x + 2)(x – 3)(x + 3) = 0 x – 2 = 0 x = 2 x + 2 = 0 x = -2 x – 3 = 0 x + 3 = 0 x = 3 x = -3
SS { x| x = + 2, x = + 3}

XLVI
Uses of Algebraic Factoring
(Continued) b) To Reduce Algebraic Fractions
Ex 1) x 2 – 2x + 1
3x 2 – 3
Rules
1)Prime factorize tops and bottoms.
1)Cancel Like Factors.
(x – 1)(x – 1)
3(x – 1)(x + 1)
(x – 1)
3(x + 1)

Ex 2)
XLVI
Uses of Algebraic Factoring
(Continued)
4 – b b 2 – 5b + 4
4 – b
(b – 4)(b – 1)
-1 _
(b – 1) b – 4 ≠ 4 – b
Opposites will always yield -1
Just as
6
-6
Will yield -1
Or
1 _
1 – b

XLVI
Uses of Algebraic Factoring
(Continued) c) Multiplying fractions
Ex 1)
Rules
1)Prime factorize tops and bottoms.
11m .
14 _
21 55m 3
2)Cancel any top with a like bottom.
3)Move left to right when solving.
11m
.
7 × 2 _
7 × 3 11 × 5 × m × m × m
2_
15m 2

XLVI
Uses of Algebraic Factoring
(Continued)
Ex 2) p 2 – p – 6 2p 2 – 5p – 3
3p – 9 p 2 – 3p
(p + 2)(p – 3) (2p + 1)(p – 3)
3(p – 3) p(p – 3)
(p + 2)(2p + 1)
3p

XLVI
Uses of Algebraic Factoring
(Continued) d) Division of Fractions
Ex 1) x 2 – x – 6 _
÷ x 2 – 25 _
2x 2 + 9x + 10 2x 2 + 15x + 25
÷ x 2 – x – 6 _
2x 2 + 9x + 10
2x 2 + 15x + 25 _ x 2 – 25
(x + 2)(x – 3)
(2x + 5)(x + 2)
÷
(2x + 5)(x + 5)
(x + 5)(x – 5)
Rules
1)Invert back or bottom fraction.
2)Prime factorize tops and bottoms.
3)Cancel any top with a like bottom.
4)Move left to right when solving.
x – 3 x – 5

XLVI
Uses of Algebraic Factoring
(Continued)
Ex 2) 6x 3 – 48y 6
8(x – 2y 2 )
6(x 3 – 8y 6 )
8(x – 2y 2 )
6(x – 2y 2 )(x 2 +2xy 2 + 4y 4 )
8(x – 2y 2 )
3(x 2 + 2xy 2 + 4y 4 )
4

XLVI
Uses of Algebraic Factoring
(Continued) e) Addition and Subtraction of Fractions
Rules to get LCD
Ex 1)
3x 2
2x _
– 27
+
5x 2 x – 1 _
+ 11x – 12
1)Prime factorize
Denominator only
2)Get Jealous
2x _
+ x – 1 _
3(x + 3)(x – 3) (5x – 4)(x + 3)
GET JEALOUS
(5x – 4)(2x) + 3(x – 3)(x – 1)
3(5x – 4)(x + 3)(x – 3)
13x 2 – 20x + 9
3(5x – 4)(x + 3)(x – 3)