Algebra Honors (Periods 5 & 6)

Inequalities in One Variable
Order of Real Numbers 10-1

This sections should be review... you have been working with less than and greater than symbols since 6th grade. A number line shows order relationships among all real numbers. The value of a variable may be unknown but you may know that is either greater than or equal to another number.
For example,
x ≥ 5 is read "x is greater than or equal to 5."
x ≥ 5 is another way of writing " x >5 or x = 5"

Translating statements into symbols is a critical concept. Practice these as review
-3 is greater than -5.    -3 > -5
and
x is less than or equal to 8  x    ≤   8

To show that x is between -4 and 2 you write
-4 < x < 2
which is read 
"-4 is less than x and x is less than 2."
or you could read it as
" x is greater than -4 AND less than 2."

The same comparisons are stated in the sentence  2 > x  > -4

When all the numbers are know you can classify the statement as true or false.
Thus,
-4 < 1 < 2   is true
but
-4 < 8 < 2 is false

An inequality is formed by placing an inequality symbol (  > ,  < ,     ≤ , or    ≥) between numerical pr variable expressions-- called the SIDES of the inequality
You solve an inequality by finding the values from the domain of the variable which make the inequality a true statement.   Such values are called the solutions of the inequality. All the solutions make up the solution set of the inequality.

Algebra Honors (Periods 5 & 6)

Puzzle  Problems  9-7

This includes digit, age, and fraction problems

Example 1: Digit problem
The sum of the digits in a 2 digit number is 12. The new number obtained when the digits are reversed is 36 more than the original number. Find the original number.

Let t = the ten's digit of the original number
Let u = the unit's digit of the original number

then the quantity relationship is t + u = 12

The value expression of the original number is 10t +u so when the digits are reversed the value would be 10u+ t

SO, the value equation becomes
10u + t = 10t + u + 36 or
9u - 9t = 36 Now, divide by 9

u - t = 4
So the two equations are
u - t = 4
u + t = 12

2u = 16
u = 8
If u = 8 then t= 4
and the original number is {48}

EXAMPLE 2: Age Problem

Mini is 4 years older than Ronald
5 years ago she was twice as old as he was. Find their ages now.

Let m = Mimi's age now
Let r = ronald's age now
The two equations created from this information are the following:
m = r + 4
and
m-5 = 2(r-5)

m -5 = 2r - 10 -->I would use substitution with this one so
(r+4) - 5 = 2r - 10
r-1 = 2r - 10
9 = r
Since Ronald is 9 years old
Mimi is 13 years old

Example 3: Fraction Problem

The numerator of a fraction is 3 less than the denominator
If the numerator and the denominator are each increased by 1, the value of the resulting fraction is 3/4 What was the original fraction?

Let n = the numerator
Let d = the denominator

n/d is the fraction

n = d -3
which can be translated to n - d = -3

and
n+1=3
d+1 4

4(n+1) = 3(d+1)
4n + 4 = 3d+ 3

4n-3d = -1
n - d = -3

Using the addition/subtraction method and multiplying the 2nd equation by (-4)

4n-3d = -1
-4n - (-4)d = -3(-4)

4n-3d = -1
-4n+4d = 12
d = 1

Since d = 11
n = 8
and the fraction is 8/11
{8/11}

We then turned to Page 447 and completed several even problems # 8-16 even.
Here are a few of the problems worked out. TUrn to our textbook to read the actual wording of each problem.

#8. Let t = ten's digit
Let u = unit's digit

t = u-5

t + u = (1/3)(10t + u)
3t + 3u = 10t + u
-7t + 2u = 0

Using substitution

-7(u-5) + 2u + 0
-7u + 35 + 2u = 0
-5u = -35
u = 7

Since u = 7 t = 2 and the number is 27
{27}

10.
Let t = ten's digit
Let u = unit's digit

The original number would be 10t + u and the reverse would be 10u + t

so 10u + t = 10t + u -54
9u -9t = -54
divide that by 9
u - t = -6
so
u - t = -6
u + t = 8
2u =2
u = 1
so the number must be 71
{71}

12.
Cecelia is 24 years younger than Joe
Let c = Cecelia's age NOW
:et j = Joe's age now
c + 24 = j
c - 6 = (1/2)(j - 6)

2c - 12 = j- 6

2c - j = 6
c - j = -24

Becomes
2c - j = 6
- c + j = 24
c = 30

So Cecelia's age is 30 years old and Joe is 54 years old

Math 6A ( Periods 2 & 4)

 Simple Interest 9-7
When you borrow money you pay the lender INTEREST for the use of the money. The amount of interest you pay is usually a percent of the amount borrowed figured on a yearly basis. This percent is called the annual rate.

When interest is computed year by year we call it
SIMPLE INTEREST
The formula is I= Prt

Let I = simple interest charges
P = principal ( amount borrowed)
r= annual rate
t = time in years

I = Prt

simple interest is calculated just on the principal.
Let's work through a few examples

How much simple interest would you owe if you borrowed $640 for 3 years at atan annual rate of 15%?

I = Prt
I = (640)(.15)(3)
I = (1920)(.15)
I = 288
$288 in interest

Sarah borrowed $3650 for 4 years at 16% How much must she repay.
Remember she will oe the amount she borrowed as well as the the interest.
I = Prt
I = (3650)(.16)(4)

I =14600(.16)
I = 2336.
Principal + Interest = 3650 + 2336
Sarah owes $5986.


$150 borrowed at 12% annual rate for 1 year

I = Prt

I = (150)(.12)(1)
I = 18
so you would owe $18 in interest after 1 year.
The total due would be $150 + 18 = $168

What if instead you borrowed the same amount but for 2 years... nothing was due until the end of two years
I = Prt
I = 150(.12)(2) = 36
You would owe $36 in interest .. so the total due was 150 + 36 = $ 186.

What if you borrowed the same amount for 3 years...
I = 150(.12)(3) = 54 or $54 in interest.
You would owe 150 + 54 = $ 204 after three years...

However, let's say you could only borrow that amount for 6 months...
I = Prt
I = (150)(.12)(.5)
Why 0.5? that is 1/2 a year.
Now you can always multiply by 1/2 as well.. in fact, sometimes that is easier
I = 150(.12)(1/2) = 9 or $ 9.00

After 6 months you would owe $159.


Dylan paid $375 in interest on a loan of $1500 principal at 12.5% interest.
What was the length of time?

Look at what it is asking and see which of the variables you have...
I= Prt
We have the interest paid, the principal and the annual rate so
375= (1500)(.125)(t)

375 = 187.5t
solve this one step equation by dividing both sides by 187.5

375 = 187.5t
187.5 187.5

t = 2

so 2 years

divide carefully...

Alexis paid $ 585 simple interest on a $6500 loan for 6 months.

what was the annual rate?

What do we know?
I = 585
P = 6500
t= 6 months ( which is 0.5 or 1/2)

I = Prt
585 = 6500 (r)(.5)
585 = 3250r
divide both sides by 3250

585 = 3250r
3250 3250

r = 0.18
which means 18%



annual--> once a year
6 months --> 1/2 or 0.5
4 months--> 1/3
3 months --> 1/4 or 0.25
8 month --> 2/3

(We did not get to this yet... but I thought I would post it... read this..it is interesting to see the difference. We will go over this after STAR testing)
Compound Interest 9-8

Compound interest is ALWAYS more than simple interest.

interest is compounded on the interest!!

$100 savings earning $10 interest/ annual.. [this only happens NOW if your dad is the one paying you... :)]

I = Prt
at the end of the first year
I = 100(.10)(1) = 10 or $10
add that to the 100
$110.
Now for the 2nd year,
$110 is your principal
so
I = Prt
I = 110(.10)(1) = 11 or $11
so at the end of 2 years you have $110 + 11 or $121

Now for the 3rd year
I = Prt
I = 121(10)(1) = $12.10
So at the end of three years you have $121 + 12.10 = $133.10

What if you had $500 at 8% compounded quarterly for one year.

quarterly means 1/4 or .25

I = Prt
I = 500(.08) (1/4)
calculate the 08(1/4) because that will be the constant you will multiply your principal by each time
(.08)(1/4) = .02
so I = 500(.02) = 10
after the first quarter it is 510
I = Prt for the 2nd quarter
I = 510 (.02) = 10.20
so after the 2nd quarter $510 + 10.20 = $520.20
I = Prt for the third quarter
I = 520.20 (0.02) = about $10.40 ( round to the nearest penny)
so after the third quarter
$520.20 + 10.40 = $530.60
I = Prt
I = 530.60(.02) = about $10.61
So at the end of 4 quarters -- or one year
530.60 + 10.61 = $541.21

compounding terms:
annually--> once a year
semiannually --> twice a year
quarterly--> four times a year
monthly--> 12 times a year
daily--> 365 times a year