Math 7 ( Period 4)

Solving Multi Step Equations 4.2

Before you use inverse operations to solve an equation, you should check to see whether ONE or BOTH sides of the equation can be simplified by combining like terms!

2x + 3x - 4 = 11

5x -  4 = 11   ( Combine like termsà 2x + 3x = 5x)

we used the  addition property of equality ( +  prop=)

5x + 0 = 15    
Here we have the Additive Inverse or Inverse Property of Addition  (Inv +)

We don’t want to write with “+ 0”

The Identity property of Addition (ID+) allows us to write

5x = 15

Now we need to divide both sides by 5 and the Division Property of Equality ( ÷ prop =)  allows us to do that

5x/5 = 15/5  Multiplicative Inverse or the Inverse Property of Mult (Inv×)

1x = 3  we don’t want to write 1x ->Using the Identity Property of Multiplication (ID×)

x = 3 
and we BOX our answer.

Solving

-13 = 3n + 3 + n

Involves a similar process:

we need to combine like terms first

-13 = 4n + 3 ( combining 3n + 1n= 4n)

  we used the Subtraction Property of Equality ( -prop=)

-16 = 4n +0   
Here we have the Additive Inverse or Inverse Property of Addition  (Inv +)

We don’t want to write with “+ 0”

The Identity property of Addition (ID+) allows us to write

-16 = 4n

Now we need to divide BOTH sides by 4 and the Division Property of Equality (÷prop=) allows us to do that.

-16/4 = 4n/4   

Multiplicative Inverse or the Inverse Property of Mult (Inv×)

-4 =1n   we don’t want to write 1x ->Using the Identity Property of Multiplication (ID×)

-4 = n  

and we BOX our answer. 

Algebra Honors ( Periods 6 & 7)

Solving Equations by Factoring Section 5-12

 a⋅0 = 0  and if a = 0 or b = 0 then we know that ab= 0

This is called an  “ if, then”  statement

 conversely

if ab = 0 then either a= 0 or b = 0

That is the converse.

Not all converse of true statements are true. For example

"If a polygon is a square, then it is a rectangle," is a true statement but its converse

"If a polygon is a rectangle, then it is a square"—is NOT true.

The words “ if and only if” are used to combine a statement and its converse when BOTH are true.

The Zero Products Property  is one such statement

This Zero Products Property helps us solve equations.

For all real numbers a and b,

ab= 0 if and only if a = 0 and b = 0

A product of factors is zero if and only if one or more of the factors is zero.

For example: 

(x +2)(x -5) = 0

either x + 2 must equal zero or x - 5 must equal zer0

so set each expression equal to zero and solve

x + 2 = 0

x= -2

and x-5 = 0

x = 5

{-2. 5}

5m(m-3)(m-4) = 0

now you have three expressions so set each of them to zer0

5m = 0 so m = 0

m-3 = 0 so m=3

m-4 = 0 so m=4

{0,3,4}

What happens with

3x²+ x = 2

It isn't the "2 products property" --> we need to use the ZERO Products Property,  so set the expression equal to zer0

3x²+ x - 2 = 0

Now factor

(x+1)(3x -2) = 0

set each of these equal to zero

x + 1 = 0 x = -1

3x -2 = 0 so x = 2/3

{-1, 2/3}

10x³ - 15x² = 0

factor

5x²(2x -3) = 0

again set each equal to zer0

5x² = 0 so x = 0

and

2x -3 = 0 so x = 3/2

{0, 3/2}

Polynomial equations are named by the term of highest degree

(where a≠ 0)

ax + b = 0  is a linear equation

ax² + bx + c = 0 is a quadratic equations

ax³ + bx² + cx + d = 0  is a cubic equation

Many polynomial equations can be solved by factoring and then using the zero-products property. The first step is often to transform the equation into STANDARD FORMà in which one side is zer0. The other side should be a simplified polynomial arranged in order of decreasing degree of the variable.

2x² + 5x = 12

becomes

2x² +5x - 12 = 0

(x + 4)(2x-3) = 0

so x = -4 and x = 3/2

{-4, 3/2}

18y³ + 8y + 24y² = 0

Rearrange first

18y³+ 24y ² + 8y = 0

Then factor the GCF

2y(9y² +12y +4) = 0

WAIT--> its a PERFECT trinomial SQ

2y(3y +2)² = 0

2y = 0 so y = 0

and 3y + 2 = 0 so y = -2/3

-2/3 is a double or multiple root but you only list it once in solution set.

That is,

{-2/3, 0}

Find an equation in standard form with integral coefficients that has the following solutions set: {2/3, -4}

This is just working backwards…

Since we know that the solution set must be {2/3, -4}, we know that

( x – 2/3)(x + 4) = 0  because we get the solution set from those two sets of hugs by using the Zero Products Property

( x – 2/3)(x + 4) = 0   is NOT in standard form so we need to use the BOX METHOD   However… since we have a fraction in one of the two sets of parenthesis.. we may want to clear that BEFORE we start using FOIL, or the BOX method. If we multiply BOTH SIDES by 3, we haven’t changed the value so  3( x – 2/3)(x + 4) = 0   Now distribute the 3( x - 2/3) becomes (3x -2)  so now we have (3x – 2)(x + 4) = 0   Now FOIL, or BOX, or FIREWORKS…. and we get 3x²+ 10x – 8 = 0

(x-1)(x+3) = 12

We can’t immediately solve this—there isn’t a “12 products property.” That is, we need to set this to “zer0.”

(x-1)(x+3)  - 12 = 0

Now FOIL, or BOX, or FIREWORKS…. à  x² -2x – 3 -12 = 0

which is x² -2x – 15 = 0

FACTOR

(x -3)(x + 5) = 0

so

x = 3 and x = -5

{-5, 3}

y = x² + x - 12

solve for the roots means you set this quadratic equal to ZERO

so

x² + x - 12 = 0

(x+4)(x -3) = 0

Math 7 (Period 4)

Solving Two Step Equations 4.1

Solving equations may involve using more than one inverse operation to isolate the variable (putting the variable alone on one side of the equation)

Since we are undoing the equation, we literally do PEMDAS in reverse!

3x + 8 = 23

   

 we used the  subtraction property of equality ( -  prop=)

3x + 0 = 15    

Here we have the Additive Inverse or Inverse Property of Addition  (Inv +)

We don’t want to write with “+ 0”

The Identity Property of Addition (ID+) allows us to write

3x = 15

Now we need to divide both sides by 3 and the Division Property of Equality ( ÷ prop =)  allows us to do that

3x/3 = 15/3   Multiplicative Inverse or the Inverse Property of Mult (Inv×)

1x = 5  we don’t want to write 1x ->Using the Identity Property of Multiplication (ID×)

x = 5 

and we BOX our answer.

Solving

x/4 – 12 = 1

Involves a similar process:

We need to add 12 to both sides

  we used the Addition Property of Equality ( +prop=)

x/4 + 0 = 13  

Here we have the Additive Inverse or Inverse Property of Addition  (Inv +)

We don’t want to write with “+ 0”

The Identity property of Addition (ID+) allows us to write

x/4 = 13

Now we need to multiply BOTH sides by 4/1 and the Multiplication Property of Equality (×prop=) allows us to do that.

Notice where the 4/1 is located and how it has  HUGS ( ) surrounding it! That’s required in my class! It shows that you are understanding the process of multiplying by the reciprocal of ¼.

  

Multiplicative Inverse or the Inverse Property of Mult (Inv×)

1x = 52  we don’t want to write 1x ->Using the Identity Property of Multiplication (ID×)

x = 52 

and we BOX our answer. 

Math 6A (Periods 1 & 2)

Greatest Common Factor 5-5

If we list the factors of 30 and 42, we notice
Factors of 30: 1, 2, 3, 5, 6, 10, 15, 30
Factors of 42: 1, 2, 3, 6, 7, 14, 21, 42

We notice that 1, 2, 3, and 6 are all COMMON factors of these two numbers. The number 6 is the greatest of these and therefore is called the
GREATEST COMMON FACTOR of the two numbers. We write
GCF(30,42) = 6

Although listing the factors of two numbers and then comparing their common factors is one way to determine the greatest common factor, using prime factorization is another easy way to find the GCF

Find GCF(54, 72)
54 = 2 ⋅ 3 ⋅ 3 ⋅ 3
72 = 2 ⋅ 2 ⋅ 2 ⋅ 3 ⋅ 3
Find the greatest power of 2 that occurs IN BOTH prime factorization. The greatest power of 2 that occurs in both is just 2 ¹
Find the greatest power of 3 that occurs IN BOTH prime factorizations. The greatest power of 3 that occurs in both is 3²
Therefore
GCF(54, 72) = 2 ⋅ 3² = 18

In class we circled the common factors and realized that
GCF(54, 72) = 2 ⋅ 3 ⋅ 3 = 18


Fin the GCF( 45, 60)
45 = 3 ⋅ 3⋅ 5
60 = 2⋅ 2⋅ 3⋅ 5
Since 2 is NOT a factor of 45-- there is NO greatest power of 2 that occurs in both prime factorizations.
The greatest power of 3 is just 3¹
and the greatest power of 5 is just 5¹
Therefore,
GCF(45,60) = 3⋅ 5 = 15

The number 1 is a common factor of any two whole numbers!! If 1 is the GCF , then the two numbers are said to be RELATIVELY PRIME. Two numbers can be relatively prime even if one or both of them are composite.

Show that 15 and 16 are relatively prime
List the factors of each number
FACTORS of 15: 1, 3, 5, 15
FACTORS of 16: 1, 2, 4, 8, 16

Since the GCF(15,16) = 1. The two numbers are relatively prime!!

Math 6A ( Periods 1 & 2)

Prime Numbers & Composite Numbers 5-4

A prime number is one that has only two factors: 1 and the number itself, such as 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31...
A counting number that has more than two factors is called a composite number, such as 4, 6, 8, 9, 10...

Since one has exactly ONE factor, it is NEITHER PRIME NOR COMPOSITE!!
Zero is also NEITHER PRIME NOR COMPOSITE!!
Sieve of Eratosthenes - We did it!! :)

Every counting number greater than 1 has at least one prime factor -- which may be the number itself.
You can factor a number into PRIME FACTORS by using a factor tree or the inverted division, as shown in class.

Using the inverted division, you also start with the smallest prime number that is a factor... and work down
give the prime factors of 42
2⎣42
3⎣21
7

Fundamental Theorem of Arithmetic

Every composite number can be written as a product of prime factors in exactly one way

When we write 42 as 2⋅3⋅7 this product of prime factors is called the prime factorization of 42.

Two is the only even prime number because all the other even numbers have two as a factor.

Explain how you know that each of the following numbers must be composite...
111; 111,111; 111,111,111; and so on....
Using your divisibility rules you notice that the sums of the digits are multiples of 3.

List all the possible digits that can be the last digit of a prime number that is greater than 10.
1, 3, 7, 9.

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Algebra Honors ( Periods 6 &7)

Using Several Methods of Factoring 5-11

A polynomial is factored completely when it is expressed as the product of a monomial and one or more prime polynomials.

Guidelines for Factoring Completely

1.     Factor out the GCF first

2.    Look for a difference of two squares

3.    Look for a perfect square trinomial

4.    If a trinomial is not a square, look for a pair of binomial factors.

5.    If a polynomial has four or more terms, look for a way to group the terms in pars or in a group of three terms that is a PERFECT SQUARE TRINOMIAL

6.    Make sure that each binomial or trinomial factor is prime.

7.    Check your work by multiplying the factors