Algebra ( Period 4)

CHAPTER 2-1: STRATEGIES FOR TRANSLATING

TO ALGEBRAIC EQUATIONS

The notes are below but you should already know this!

Try the homework and come with questions.

The only part of this section that will not be familiar:

TRANSLATING GIVEN INFORMATION INTO YOUR OWN WORD PROBLEM…YOU’RE WRITING A STORY!
I’ve included example 5 from your book below (p. 77)
Maxine’s time every time she drove = t
Tia’s time every time she drove = t + 4
Given: 2t + (t + 4) = 28
Write a word problem or story

Maxine and Tia took a trip together and took turns driving.
Maxine took 2 turns driving and Tia only 1, but when Tia drove, she drove 4 more hours than Maxine drove on each of her turns.
The trip took 28 hours.
How long did each of them drive?

Algebraic expressions just are the ones that have variables
Numeric expressions have only numbers
Equations must have an = sign while expressions do not

STRATEGY #1:TRANSLATE WORD BY WORD
You did this in Chapter 1!
Always try this first.
Just be careful of less THAN and subtracted FROM because these are switched from the order that you read/say them:
A number less THAN 12 is 12 – n but if you say a number less 12, this would be n – 12
12 subtracted FROM a number is n – 12, but 12 subtract a number would be 12 – n

The only other translation to be careful of is when you multiply a SUM or DIFFERENCE by a number or variable:
12 times the SUM of a number and 5 is 12(n + 5), but the sum of 12 times a number and 5 would  be 12n + 5
12 times the DIFFERENCE of a number and 5 is 12(n – 5), but the difference of 12 times a number and 5 would be 12n – 5

If you have 2 or more unknowns, use different variables:
The difference of a number and ANOTHER number would be x - y

STRATEGY #2: DRAWING A PICTURE
(When in doubt, draw it out! ;)
I have 5 times the number of quarters as I have dimes.
I translate to: 5Q = D
I check: If I assume that I have 20 quarters, then 5(20) = 100 dimes
Does this make sense? That would mean I have a lot more dimes than quarters.
The original problem says I have a lot more quarters!
My algebra is WRONG! I need to switch the variables.
5D = Q
I check: If I assume that I have 20 quarters, then 5D = 20
D = 4
Does this make sense? YES! I have 20 quarters and only 4 dimes.
Sometimes it helps to make a quick picture.
Imagine 2 piles of coins.
The pile of quarters is 5 times as high as the pile of dimes.
You can clearly see that you would need to multiply the number of dimes
to make that pile the same height as the number of quarters!

STRATEGY #3: MAKE A T-CHART
To translate known relationships to algebra (known as dimensional analysis), it often helps to make a T-Chart.
You always put the unknown variable on the LEFT side and what you know on the right.
Fill in the chart with 3 lines of numbers and look for the relationship between the 2 columns.
Then, you use that mathematical relationship with a variable.

EXAMPLE: The number of hours in d days
Your unknown is d days so that goes on the left side:
d days number of hours
1                   24
2                   48
3                   72
Now look at the relationship between the left column and the right column.
You must MULTIPLY the left column BY 24 to get to the right column
The last line of the chart will then use your variable d
d days number of hours
1                    24
2                   48
3                   72
d                  24d

EXAMPLE: The number of days in h hours (The flip of the first example)
Your unknown is h hours so that goes on the left side:
h hours number of days
24             1
48             2
72             3
(Why did I start with 24 and not 1 hour this time?)
Now look at the relationship between the left column and the right column.
You must DIVIDE the left column BY 24 to get to the right column
The last line of the chart will then use your variable h
h hours number of days
24           1
48           2
72           3
h        h/24
WHEN IT’S AN EQUATION AND NOT AN EXPRESSION….
You use the same strategies, but you’ll have an = sign and then you can solve for the unknown variable if there is only 1 variable:
10 less than the product of 5 and a number is 25:
5n – 10 = 25
n = 7

FORMulas
Equations that represent KNOWN RELATIONSHIPS are called formulas because there is a specific format that must be used that never changes.
For example, d = rt is a FORMula…it’s also an equation, but it has a more specific name because this is a KNOWN RELATIONSHIP in the real world.
You can translate words for formulas:
Distance is the product of the rate of speed and the time traveled.

TRANSLATING ALGEBRAIC EQUATIONS TO WORDS:
Going back the other way, you’ll have choices in the words you can use to represent the same equation.
2n = 40
You can say:
2 times a number is 40
Double a number is 40
Twice a number is 40
The product of a number and 2 is 40
The product of 2 and a number is 40 (multiplication is commutative!)

Algebra Honors ( Period 5)

Chapter 2-5: Absolute Value Equations

When you plug into an expression with absolute value,

the absolute value signs function as parentheses

in Order of Operations.
So make sure you simplify INSIDE before turning that POSITIVE!

Generally, you solve these the same way you solve regular equations.
Make sure you balance equally on both sides!

Follow the steps of a 2-step equation.
1. Add the opposite (you can subtract as well)
2. Multiply by the reciprocal (you can divide as well)

THE DIFFERENCE?
YOU HAVE 2 POSSIBLE ANSWERS!

EXAMPLE: 2 lxl + 1 = 15
2 lxl  + 1 - 1 = 15 - 1
2 lxl  = 14
1/2 ( 2 lxl  ) = 1/2 (14)
 lxl  = 7
x = {-7, 7}

REMEMBER, IF YOU AFTER YOU GET THE ABSOLUTE VALUE ALONE ON ONE SIDE, YOU FIND THAT THE CONSTANT ON THE OTHER SIDE IS NEGATIVE,  
THE ANSWER IS THE NULL SET!

EXAMPLE: 2 lxl + 16 = 15
2 lxl + 16 - 16 = 15 - 16
2 lxl = -1
1/2 ( 2 lxl ) = 1/2 (-1)
 lxl = -1/2
NOT POSSIBLE! So the answer is the null set ∅

We’ll go over how to create an absolute value equation from a graph and how they’re used in real world problems.

The constant inside the absolute value sign is where the middle of the solution is.

If it’s positive, the center is actually a negative number.

If it’s negative, the center is a positive number.

The solution on the other side of the = sign is the DISTANCE from the center on each side of the center.

EXAMPLE:

lx +1l = 12

The center is -1 and the distance from this center is 12 so one solution is -1 + 12 = 11 and the other solution -1 - 12 = -13.

If there is no constant in the absolute value signs,

the center is 0 and the two solutions will be ± the DISTANCE on the other side of =

EXAMPLE:

lxl = 12

The center is 0 and the distance from this center is 12 so one solution is 0 + 12 = 12 and the other solution 0 - 12 = -12 or we can write x = ±12

THE HARDER SKILL IS LOOKING AT A GRAPH AND DETERMINING THE ABSOLUTE VALUE EQUATION THAT IT CAME FROM.

If you really understand the meaning of the parts of the equation, this skill is not as difficult!

HOW DO YOU KNOW IT’S AN ABSOLUTE VALUE EQUATION?

There will generally be TWO solutions graphed.

HOW DO YOU CREATE THE ABSOLUTE VALUE EQUATION?

1. Set up the absolute value signs.
2. Place x inside the absolute value signs.
3. Determine where the center is between the two solutions, FLIP THE SIGN, and place it inside with x.
4. Count the TOTAL DISTANCE between the two solutions, divide it by 2 and place that on the right side of the =.

        This number should always be positive since absolute value       

        is always positive since it’s a distance concept.

Math 8

CHAPTER 2-3: WRITING TWO-STEP EQUATIONS

Algebraic expressions just are the ones that have variables

Numeric expressions have only numbers

Equations must have an = sign while expressions do not

STRATEGY #1:TRANSLATE WORD BY WORD

Always try this first.

Just be careful of less THAN and subtracted FROM because these are switched from the order that you read/say them:

A number less THAN 12 is 12 – n but if you say a number less 12, this would be n – 12

12 subtracted FROM a number is n – 12, but 12 subtract a number would be 12 – n

The only other translation to be careful of is when you multiply a SUM or DIFFERENCE by a number or variable:

12 times the SUM of a number and 5 is 12(n + 5), but the sum of 12 times a number and 5 would  be 12n + 5

12 times the DIFFERENCE of a number and 5 is 12(n – 5), but the difference of 12 times a number and 5 would be 12n – 5

If you have 2 or more unknowns, use different variables:

The difference of a number and ANOTHER number would be x - y

STRATEGY #2: DRAWING A PICTURE

(When in doubt, draw it out! ;)

I have 5 times the number of quarters as I have dimes.

I translate to: 5Q = D

I check: If I assume that I have 20 quarters, then 5(20) = 100 dimes

Does this make sense? That would mean I have a lot more dimes than quarters.

The original problem says I have a lot more quarters!

My algebra is WRONG! I need to switch the variables.

5D = Q

I check: If I assume that I have 20 quarters, then 5D = 20

D = 4

Does this make sense? YES! I have 20 quarters and only 4 dimes.

Sometimes it helps to make a quick picture.

Imagine 2 piles of coins.

The pile of quarters is 5 times as high as the pile of dimes.

You can clearly see that you would need to multiply the number of dimes

to make that pile the same height as the number of quarters!

STRATEGY #3: MAKE A T-CHART (really and x-y table)

Let’s say you know that every bagel you buy costs the same amount of money, $.65. You buy bagels and spend $4.55. Write the algebraic equation and then solve for the number of bagels that you purchased.

Number of Bagels	Price
1	.65
2	1.30
3	1.95
b	.65b

By looking at the pattern from the left column to the right column, you find that the number of bagels TIMES the unit rate of the price/bagel gives you the total purchase price. Now you use a variable like b to come up with the algebraic expression for the purchase of any number of bagels, .65b.

Finally, set up the algebraic equation for the amount of money you spent (given in the problem):

.65b = 4.55

You can also do this for two-step equations. Let’s say you’re joining a gym and there is an initiation fee of $50 and then a monthly fee of $20 a month. Focus on the amount that is happening repeatedly because that is going to be your coefficient of the variable…since that is the amount that will “vary”, as opposed to the $50 one-time fee that will never change.

You can do this with or without a table!

Number of Months	Price
0	50
1	70
2	90
3	110
m	20m + 50

$20m + $50 = your cost

Say the problem asks how many months you’ve been going if you’ve paid $170:

20m + 50 = 170

Solve for the number of months:

20m = 120

m = 6 months