Pre Algebra (Period 2 & 4)

RATIOS AND RATES: 6-1



Ratios = fractions with meaning (it's all about the labels!)
3 ways to write a ratio:


EXAMPLE: 16 girls and 14 boys at a party


16 girls to 14 boys or

16 girls: 14 boys or
16 girls/14 boys


You can simplify this just like a fraction:
8 girls to 7 boys


In fact, anything you can do with a fraction, you can do with a ratio!



Rates = ratios with 2 DIFFERENT LABELS

Miles per gallon, miles per hour

10 out of 16 girls went to my party is not a rate.
Not 2 different labels! (but it is a ratio)



Unit rates = rates with a denominator of 1
(SO MUST HAVE 2 DIFFERENT LABELS)


I drive 150 miles in 3 hours is a rate

To change it to a UNIT RATE, simply DIVIDE the numerator by the denominator
150 miles/3 hours = 50 miles per hour

NOW IT'S A UNIT RATE


People focus on MPGs these days when they buy cars!

A Honda Civic = 40 mpg while a Hummer = 8 mpg




A special unit rate called the UNIT PRICE:
I USE UNIT RATES ALL THE TIME WHEN I TRY TO DECIDE WHETHER IT'S WORTH GOING TO COSTCO INSTEAD OF PAVILIONS
Goldfish = $7.99 at Pavilions for 33.5 oz and $10.99 at Costco for 48 oz.



If you divide $/oz you get a unit rate know as UNIT PRICE
MONEY MUST BE THE NUMERATOR!!!!






IN CLASS: Chapter 6-2: Proportions
A proportion = 2 equal ratios (2 equivalent fractions)

Solve using equivalent fractions or


Cross multiplication and then a one-step equation
(see if you can simplify the fractions before multiplying)

Example: Solve the proportion for y:


4/3 = y
/21

EQUIVALENT FRACTION APPROACH:


Multiply both top and bottom by 7, y = 28



CROSS PRODUCTS APPROACH:
You'll get 3y = (21)(4)
Now divide each side by 3.

Do this before multiplying on the right side!

Why? Because a lot of the time you'll be able to simplify and keep the numbers smaller!


3y/3 = (21)(4) /3

See how the 3 cross cancels into the 21?
so y = 28


ALWAYS SIMPLIFY THE FRACTIONS FIRST!

Algebra (Period 1)

Add and subtract rational expressions with LIKE DENOMINATORS: 10-4
Add and subtract with UNLIKE DENOMINATORS: 10-5
When adding with LIKE DENOMINATORS,
simply add the numerators,
simplify

When subtracting with LIKE DENOMINATORS,
CHANGE THE SIGNS
OF EACH TERM IN THE NUMERATOR AFTER THE SUBTRACTION SIGN,
THEN ADD (double check!!!)

When adding or subtracting with UNLIKE DENOMINATORS,
find the common denominator (the least common multiple of all denominators),
then use equivalent fractions to restate each numerator using the new common denominator.
EXAMPLE:
2 + x
x² - 16 x – 4

First, factor the denominators if possible



2x + x
(x + 4)(x - 4) x – 4

Find the LCM = (x + 4)(x - 4);

2x + (x + 4)(x)
(x + 4)(x - 4) (x + 4)(x - 4)


Restate both fractions with the LCM

2x + x² + 4x
(x + 4)(x - 4)


Now add the numerators

x² + 6x =
(x + 4)(x - 4)

Simplify and re-factor numerator, if possible
x(x + 6)
(x + 4)(x - 4)

Math 6 Honors (Period 6 and 7)

Solving Equations 11-7 (cont'd)


2- STEP EQUATIONS

What about

-3x - ^- 15 = 9
add the opposite first and you get
- 3x + 15 = 9
In order to solve this 2 step equation
we need to do the reverse of PEMDAS-- as we did with unwrapping the present so many months ago
-3x + 15 = 9
subtract 15 from both sides of the equation
-3x + 15 = 9
- 15 = - 15
Wait a minute... we have different signs... what is the rule? Ask your self.. "Who wins? and by how much?" Use a sidebar and stack them and take their difference. ( Can't stack well on this blog, sorry)
15
- 9
6 but you know that this part is -6
-3x = -6
now divide by by -3 on both sides of the equation

-3x/-3 = -6/-3
x = 2


3x + 15 = -9
3z - ^- 15 = -9
add the opposite first and you get
3x + 15 = -9
In order to solve this 2 step equation
we need to do the reverse of PEMDAS-- as we did with unwrapping the present so many months ago
3x + 15 = -9
subtract 15 from both sides of the equation
3x + 15 = -9
- 15 = - 15

This time the sides are the same-- so just add them and use their sign
3x + 15 = -9
- 15 = - 15
3x = -24
Now divide both sides by 3
3x/3 = -24/3

x = -8

Make sure to BOX your answer!!
What about this one
(1/2)(x) + 3 = 0
subtract 3 from both sides
(1/2)x = -3

Multiple by the reciprocal of 1/2 which is 2/1
(2/1)(1/2)x = -3(2/1)
x = -6
Again box your answer.

3u - 1 = -7
+ 1 = + 1
3u = -6
divide both sides by 3 ( or multiple by the reciprocal of 3 which is 1/3)
3u/3 = -6/3
u = -2

What about x = -6 + 3x
OH dear... we have variables on BOTH sides of the equations... we need to get the variables on one side all the constants on the other.
We need to isolate the variable!!

x = -6 + 3x
What if we add six to both sides
x = -6 + 3x
+6 = + 6
x + 6 = 3x
now we need to subtract x from both sides
x + 6 = 3x
- x - x
6 = 2x
so now divide both sides by 2
6/2 = 2x/2
3 = x

How about this one
3 - r = -5 + r
- 3 = - 3
-r = -8 + r
if subtract r from both sides, I will get rid of the +r on the right side

-r = -8 + r
- r = -r
-2r = -8
Now divide by -2 on both sides

-2r/-2 = -8/-2
r = 4

Algebra (Period 1)

Simplify, Multiply, and Divide RATIONAL EXPRESSIONS 10-1, 10-2, and 10-3

Rational Expressions = Expressions in fraction format (division) with a variable in the denominator

You have already been simplifying, multiplying and dividing these throughout this year!

SIMPLIFY: 10-1

You will need to FACTOR (Chapter 6) both the numerator and denominator and "cross out" common factors in both (their quotient is 1!)
EXAMPLE: Simplify
y² + 3y + 2 =
y² - 1


(y + 2)(y + 1) =
(y - 1)(y + 1)

y + 2
y - 1

MULTIPLY: 10-2

FACTOR if possible, cross cancel if possible, multiply numerators, then denominators, simplify
EXAMPLE:
(y + 4)³[y² + 4y + 4] =
[(y + 2) ³(y² + 8y + 16)]

(y + 4) ³][ (y + 2) ² =
(y + 2) ³(y + 4) ²

y + 4
y + 2



DIVIDE: 10-3


Same as the previous example, only this time you will need to

“FLIP the SECOND” fraction,

FACTOR then
MULTIPLY!!!!!
EXAMPLE:
x + 1 ÷ x + 1 =
x² - 1 x² - 2x + 1


( x + 1) ( x2 - 2x + 1) =
(x + 1)( x - 1) (x + 1)

( x + 1)( x - 1)(x - 1) =
(x + 1)( x - 1)(x + 1)


x - 1
x + 1

Math 6 Honors (Period 6 and 7)

Quotients of Integers 11-6

We all remember 2⋅ 5 = 10
and we know corresponding information
10÷ 5 = 2

The quotient of two positive OR two negative integers is POSITIVE!!

The quotient of a positive AND a negative integer is NEGATIVE!!

The same rules of multiplication apply to division. The same life story!! :-)

Although we talked about the sum of two integers always being an integer and
the difference of two integers always being an integers...

the QUOTIENT of two integers is NOT ALWAYS an integer!!

10/4 = 2 1/2 --> which is NOT an integer!!

We also reviewed:

10/0 --> is undefined!!
whereas,
0/10 = 0

98/-14 = -7

Simplify the following:
6 × 8 + -3 × 5
7 × 5 + -3 × 8

Make sure you perform operations using PEMDAS
(also called Aunt Sally's rules or even Order of Operation O³)

6 × 8 + -3 × 5
7 × 5 + -3 × 8

= 3

Solving Equations 11-7

Now that we have learned about negative integers, we can solve an equation such as
x + 7 = 2
We need to subtract 7 from both sides of the equation
x + 7 = 2
- 7 = - 7

to do this use a side bar and use the rules for adding integers
Notice the signs are different so
ask yourself... Who wins? and By How Much?
stack the winner on top and take the difference
so
x + 7 = 2
- 7 = - 7
x = -5

y -^- 6 = 4
add the opposite and you get
y + 6 = 4
now you need to subtract 6 from both sides of the equation
y + 6 = 4
- 6 = - 6
Again the signs are different -- ask your self those all important questions
"Who Wins? and "By How Much?"
Use a side bar, stack the winner on top and take the difference. Make sure to use the winner's sign in your answer!!
y = 2

2- STEP EQUATIONS

What about
3z - ^- 15 = 9
add the opposite first and you get
3x + 15 = 9
In order to solve this 2 step equation
we need to do the reverse of PEMDAS-- as we did with unwrapping the present so many months ago
3x + 15 = 9
subtract 15 from both sides of the equation
3x + 15 = -9
- 15 = - 15

This time the sides are the same-- so just add them and use their sign
3x + 15 = -9
- 15 = - 15
3x = -24
Now divide both sides by 3
3x = -24
3 3

x = -8

Make sure to BOX your answer!!
What about this one
(1/2)(x) + 3 = 0
subtract 3 from both sides
(1/2)x = -3

Multiple by the reciprocal of 1/2 which is 2/1
(2/1)(1/2)x = -3(2/1)
x = -6
Again box your answer.

3u - 1 = -7
+ 1 = + 1
3u = -6
divide both sides by 3 ( or multiple by the reciprocal of 3 which is 1/3)
3u/3 = -6/3
u = -2

What about x = -6 + 3x
OH dear... we have variables on BOTH sides of the equations... we need to get the variables on one side all the constants on the other.
We need to isolate the variable!!

x = -6 + 3x
What if we add six to both sides
x = -6 + 3x
+6 = + 6
x + 6 = 3x
now we need to subtract x from both sides
x + 6 = 3x
- x - x
6 = 2x
so now divide both sides by 2
6/2 = 2x/2
3 = x

How about this one
3 - r = -5 + r
- 3 = - 3
-r = -8 + r
if subtract r from both sides, I will get rid of the +r on the right side

-r = -8 + r
- r = -r
-2r = -8
Now divide by -2 on both sides

-2r/-2 = -8/-2
r = 4