Math 6H ( Period 3)

Comparing & Ordering Fractions 2.6 

When you decide tat one fraction is less than another fraction, you are comparing fractions. WHen you list fractions from least to greatest you are ordering fractions. You use the symbols > and < to replace words when you compare fractions
> means greater than
< means less than
Compare 2/5 and 3/5
Two fractions with the same denominator are easy to compare. The fraction with the lesser numerator is lesser
so 2/5 < 3/5 because 2<3 p="p">
What about the same numerator ?
 Write 2/5, 2/3, and 2/4 in order
The larger number in the denominator is the smaller sized fraction.
so 2/5 < 2/4< 2/3

What about
8/32  8/24 and 8/16
Using the same process above
8/32 < 8/24 < 8/16
and if you simplify each of the fractions ( from our lesson in 2.4
8/32 = 1/4
8/24 = 1/3
8/16 = 1/2
so we could use the same rule and still get the same results since
1/4 < 1/3< 1/2

What if you were comparing 3/4 and 5/6?
You might need to change to a common denominator
One of the common denominators is always the product of the two denominators so 4 times 6 = 24
3/4 ( 6/6) = 18/24
5/6( 4/4) = 20/24
since 18/24 < 20/24
3/4 < 5/6

OR you could use the Least Common Denominator (LCD) which is really the Least Common Multiple of the Denominators!
LCM(4, 6) = 12 so
change 3/4 into  9/12
and 5/6 = 10/12
since 9/12 , 10/12
3/4 < 5/6

One additional method
When comparing two fractions,

Algebra Honors (Periods 5 & 6)

Dividing Monomials 5-2

There are 3 basic rules used to simplify fractions made up of monomials.
Property of Quotients
if a, b, c, d are real numbers with b≠0 and d ≠0

ac/bd = a/b ⋅c/d
Our example was 15/21 = (3⋅5)/(3⋅7) = 5/7
The rule for simplifying fractions follows ( when a = b)
(bc)/(bd) = c/d
This rule lets you divide both the numerator and the denominator by the same NON ZERO number.

35/42 = 5/6
-4xy/10x = -2y/5 which can also be written (-2/5)x as well as with out the (((HUGS)))

c⁷/c⁴ = c⁴c³/c4 = c³
another way we proved this was to write out all the c's
c⋅c⋅c⋅c⋅c⋅c⋅c⋅/c⋅c⋅c⋅c = and we realized we were left with
c⋅c⋅c = c³
In addition, we noticed that
c⁷/c⁴ = = c^7-4 = c³
THen we considered
c⁴/c⁷ =
c⋅c⋅c⋅c/c⋅c⋅c⋅c⋅c⋅c⋅c = 1/c⋅c⋅c = 1/c³ = c^-3

Since we all agreed that any number divided by itself was = 1
(our example was b⁵/b⁵ ), we proved the following
1 = b⁵/b⁵ = b^5-5 = b⁰

We finally arrived at the Rule of Exponents for Division

if m > n
a^m/aⁿ = a ^m-n

If n > m
a^m/aⁿ = 1/a ^n-m
and if m = n
a^m/aⁿ = 1

A quotient of monomials is simplified when
1)each base appears only once in the fraction,
2) there are NO POWERS of POWERS and
3)when the numerator and denominator are relatively prime, that is, they have no common factor other than 1.

35x³yz⁶/ 56x⁵yz
5z⁵/8x²

Finding the missing factor when you are given the following
48x³y²z⁴ = (3xy²z)⋅ (______)
we find that

48x³y²z⁴ = (3xy²z)⋅ (16x²z³)

Algebra Honors (Periods 5 & 6)

Factoring Integers 5-1

When we write 56= 8⋅7 or 56 = 4⋅14 we have factored 56
to factor a number over a given set, you write it as a product of integers in that set ( the factor set).
When integers are factored over the set of integers, the factors are called integral factors.

We used the T- charts (students learned in 6th grade) to first find the positive integer factors
56 = 1, 2, 4, 7, 8, 14, 28, 56

A prime number is an integer greater than 1 that has no positive integral factors other than itself and 1.
The first ten prime numbers are
2, 3, 5, 7, 11, 13, 17, 19, 23, 29

To find prime factorization of a positive integer, you express it as a product of primes. We used inverted division (again taught in 6th grade)

504
Try to find the primes in order as divisors.
Divide each prime as many times as possible before going on to the next prime
we found 504 - 2⋅2⋅2⋅3⋅3⋅7
which we write as 2³⋅3²⋅7
Exponents are generally used for prime factors
The prime factorization is unique--> and the order should be from the smallest prime to the largest.
A factor of two or more integers is called a common factor of the integers.
The greatest common factor (GCF) of two or more integers is the greatest integer that is a factor of all the given integers.

Find the GCF(882, 945)
First find the prime factorization of each integer Then form product of the smaller powers of each common prime factor.
The GCF is only the primes (and the powers) that they SHARE!!
882 = 2⋅3²⋅7²
945 = 3³⋅5⋅7
The common factors are 3 and 7
The smaller powers of 3 and 7 are 3² and 7
You combine these as a PRODUCT and get

the GCF(882, 945) = 3²⋅7 = 63

We also talked about listing ALL pairs of factors--> thus including negative integers
For example:
List all the pairs of factors of 20
(1)(20) but also (-1)(-20)
(2)(10) and (-2)(-10)
(4)(5) and (-4)(-5)

Listing all the factors of -20, we discovered
(1)(-20) but also (-1)(20)
(2)(-10) and (-2)(10)
(4)(-5) and (-4)(5)

October 10th: Powers of Ten Day & Binary Day

Check out this great Video on the Powers of Ten
POWERS OF TEN

Algebra Honors (Periods 5 & 6)

Problems Without Solutions 4-10

Not all word problems have solutions. We listed three of the reasons for this:
1) Not Enough Information ( NEI)
2) Unrealistic Results
3) Facts are contradictory

We used the following examples:
Aurenne drove at her normal speed for the first 2 hours of the trip--- but the road repairs slowed her down 10 mph slower than her normal speed. She made the trip in 3 hours. Find her normal speed.
Wait... just looking at this you realize you just dont have enough information.
We even made a chart with the information we had.. and it just was not enough

There is NO SOLUTION Not enough information or NEI

Liam has a beautiful lawn that is 8 m longer than it is wide... and it is surrounded by a wonderful flower bed which he and his brother maintain for his mother The flower bed is 5m wide all around. Find the dimensions of the lawn if the area of the flower bed is 140m²

When you try to solve this problem by letting the dimensions of the lawn be w and w + 8 you find the following equation
(w + 10)(w + 18) - (w)(w +8) = 140
however that leads us to
w = -2
Since the width of the lawn cannot be negative,
There is No SOLUTION and the given facts are unrealistic.


Drehan says he has equal number of dimes and quarters but that he has 3 times as many nickels as he has dimes. He also tells us that the value of his nickels and dimes is 50 cents more than the value of his quarters. How many of each kind of coin does he have?

Let d = the number of dimes
well if he has the same number of quarters as he has dimes
then d also can equal the number of quarters
and with the other information
3d = the number of nickels.
Now looking at the information he gave us
10d + 5(3d) = 25d + 50
But that simplifies to
25d = 25d + 50
which is impossible.
There is NO SOLUTION
The given facts are contradictory

Math 6H (Period 3)

 Least Common Multiple  2. 5 
A multiple of a number is the product of the number and any NON ZERO whole number. For example the multiples of 5 are 5, 10, 15, 20, 25,  and so on...
A common multiple for two or more numbers is a multiple shared byt he numbers. For example
20 is a common multiple of 2 and 5 because 20 = 2 times 10 and 20 = 5 times 4.

Multiples of 2: 2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24
Multiples of 3: 3, 6, 9, 12, 15, 18, 21, 24, 27, 30
Common multiples of 2 and 3  ( that are < 25) are 6, 12, 18, and 24
and the smalles is the Least Common Multiple and that is 6
LCM(2, 3) = 6
What about the LCM (4, 6) = 12

You can use Prime Factorization to find the LCM
24 and 60
You can use Factor Trees or Inverted Division


LCM is EVERY FACTOR TO ITS GREATEST POWER.

You could list all the multiples but why not use that BOX we used when finding the GCF... it works for  finding the LCM of two numbers.



LCM ( 7, 11) =77
If the two numbers are relatively prime, then the LCM is their product.
Remember: the GCF of two relatively prime numbers is 1
LCM ( 8, 9) = 72   because we know the GCF(8,9) = 1

Math 6A (Periods 2 & 4)

Problem Solving: Using Mathematical Expressions 2-6


Seventeen less than a number is fifty six

I suggest lining up and placing the "equal sign" right under the word is
then complete the right side = 56
after that take your time translating the left
Start with a "let statement."
A "let statement" tells your reader what variable you are going to use to represent the number in your equation.
So in this case Let b = the number
it becomes
b - 17 = 56
Now solve as we have been practicing for a couple of weeks.
b - 17 = 56
+17 = +17 using the +prop=
b + 0 = 73
b = 73 by the ID(+)

How could we check?
A FORMAL CHECK involves three steps:
1) Re write the equation ( from the original source)
2) substitute your solution or... "plug it in, plug it in...."
3) DO the MATH!! actually do the math to check!!

so to check the above
b - 17 = 56
substitute 73 and put a "?" above the equal sign...

73 - 17 ?=? 56
Now really do the math!! Use a side bar to DO the MATH!!
That is, what is 73- 17? it is 56
so 56 = 56


Practice some of the class exercises on Page 50, Just practice setting up the equations from the verbal sentences.


We know about > greater and as well as < less than
so now we look at
≥ which means " greater than or equal to" and
≤ which means " less than or equal to"

This time the boundary point ( or endpoint) is included in the solution set.
The good news is that we still solve these inequalities the same way in which we solved equations-- using the properties of equality.
w/4 ≥ 3
we multiply both sides by 4/1
(4/1)(w/4) ≥ 3(4/1) by the X prop =
1w ≥ 12
w ≥ 12 by the ID(x)

Which means that any number greater than 12 is part of the solution AND 12 is also part of that solution

Take the following:
b - 3 ≤ 150 ( we need to add 3 to both sides of the equation)
+3 = +3 using the + prop =
b + 0 ≤ 153
or b ≤ 153 using the ID (+)