Algebra Honors (Periods 5 & 6)

 Negative Exponents 7-9
If a is a nonzero real number and n is a positive integer
a^-n = 1/aⁿ
so
10^-3 = 1/10³ = 1/1000
5^-4= 1/5⁴ = 1/625

16^-1 = 1/16

The rule of exponents for division (page 190) will help you understand why
a^-n = 1/aⁿ
recall that for m > n a^m/aⁿ= a^m-n
For example
a⁷/a³ = a^7-3= a⁴
you can also apply this rule when m < n that is when m - n becomes a negative number. For example a³/a⁷ = a^3-7 = a^-4
since
a⁷/a³ and a³/a⁷ are reciprocals then
a⁴ and a^-4 must also be reciprocals.
Thus
a^-4= 1/a⁴
a⁵/ a⁵ = a^5-5 = a⁰
But you already know that a⁵/a⁵ = 1
SO, definition of a⁰
a⁰ = 1
However, the expression 0⁰ has no meaning

All the rules for positive exponents also hold for zero and negative exponents.

Summary of Rules for Exponents
Let m and n be any integers
Let a and b be any non zero integers
Review—>But you should really know these because of our Powers Project
Products of Powers
b^mbⁿ = b^m+n
Example with negative exponents
2³⋅2^-5 = 2^3+(-5) = 2^-2 = 1/2² = 1/4
Quotient of Powers
b^m ÷ bⁿ = b^m-n
Example with negative exponents
6³÷6⁷= 6^3-7= 6^-4= 1/6⁴= 1/1296
Power of Powers
(b^m)ⁿ = b^mn
Example with negative exponents
(2³)^-2 = 2^-6 = 1/2⁶ = 1/64
Power of a Product
(ab)^m= a^mb^m
Example with negative exponents
(3x)^-2 = 3^-2 ⋅x^-2 = 1/3²⋅1/x² = 1/9x²
Power of a Quotient
(a/b)^m= a^m/b^m
Example with negative exponents
(3/5)^-2= 3^-2/5^-2= (1/3²)/ (1/5²)= 1/3² ÷ 1/5² which means
1/3² ⋅5²/1= 5²/3²= 25/9

Algebra Honors (Periods 5 & 6)

Work Problems 7-8

To solve work problems use the following formula
work rate × time = work done
or rt = w
Work rate means the fractional part of a job done in a given unit of time.
For example if it take you 3 hours to clean up your room, what part of the job can be done in 1 hour? That's easy... 1/3
To finish a job the sum of the fractional parts of the work done must be 1.
( for one whole job completed)

Josh can split a cord of wood in 4 days. His father can split a cord in 2 days. How long will it take them to split a cord of wood if they work together?
Let x = the number of days needed to do the job together.
Josh and his father will each work x days
Using those great tables from class fill in with the information you know
Since Josh can do the whole job in 4 days his work rate is 1/4 job per day.
His father's work rate is 1/2 job per day.

**posting the TABLE HERE**

Josh's part of the job = x/4
His father's part of the job = x/2
so the sum of that would equal the job completed
OR

Josh's part of the job + His father's part of the job = Whole JOB
x/4 + x/2 = 1
Clear the equation of fractions by multiplying by the LCD
4(x/4 + x/2) = 4(1)
x + 2x = 4
3x = 4
x= 4/3
It would take them 1 1/3 days to do the job together.

Robot A takes 6 minutes to weld a fender. Robot B takes only 5 1/2 minutes. If they work together for 2 minutes, how long will it take Robot B to finish welding the fender by itself?

Let x = the number of minutes needed for Robot B to finish the work.
Robot B's work rate is 1/5.5 or 1/(11/2) = 2/11

***posting the TABLE HERE***
Robot A's part is (1/6)(2)
Robot B's part is (2/11)(2 +x)
A's part of the job + B's part of the Job = Whole JOB

1/3 + (2/11)(2 + x) = 1
Multiply by the LCD, which is 33

(33)[1/3 + (2/11)(2 + x)] = 33(1)
11 + 6(2 + x) = 22
11 + 12 + 6x = 33
6x = 10
x = 5/3
It will take 1 2/3 minutes for Robot B to finish welding.
The charts or tables for work problems look similar to the charts and tables used for other problems. The following formulas show the similarities among some types of problems you have studied

Work done by A + work done by B = TOTAL work done
Acid in solution A + acid in solutions B = TOTAL acid in mixture
Interest from banks + Interest from Bonds = TOTAL Interest
Distance by bike + Distance by car = TOTAL distance traveled

Math 6A (Periods 2 & 4)

Percents and Fractions 9-1

The word “percent” is derived from the Latin “per centum” meaning “per hundred” or “out of one hundred” so 28% means 28 out of 100

A percent is a ratio that compares a number to 100. Therefore you can write a percent as a fraction with a denominator of 100, so 28% is also 28/100

Our book’s example is as follows;
During basketball season, Alice made 17 out of 25 free throws, while Nina made 7 out of 10. To see who did better, we compare the fractions representing each girl’s successful free throws. 17/25 or 7/10
We have calculated this type of problem before.. this time when we compare fractions use the common denominator 100, even if 100 is not the LCD of the fractions.
17/25 = 68/100 and
7/10 = 70/100

Since Alice makes 68 free throws per 100 and Nina makes 70 per hundred, Nina is the better free throw shooter.
the ratio of a number to 100 is called a percent. We write percents by using the symbol %
so
17/25 =68% and
7/10= 70%

Rule
To express the fraction a/b
as a percent, solve the equation

a/b = n/100

for the variable n and write n%


Express 17/40 as a percent
n/100 = 17/40 multiply both sides by 100 100 ( n/100) = 17(100)/40
n = 17(100)/40 n = 85/2 n= 42½

Therefore, 17/40 = 42 1/2 %


Rule
To express n% as a fraction, write the fraction

n/100 in lowest terms


Express 7 ½ % as a fraction in lowest terms

7 ½ % = 7.5% = 7.5/100 How do we get rid of the decimal?
multiply the numerator and the denominator by 10
7.5(10)/100(10) simplify
Similarly, you could change a mixed numebr into an improper fraction
5 3/8% becomes 43/8 % and to change that to a fraction simple divide by 100
That looks messy but if you remember that to divide by 100 you are actually multiplying by 1/100
(43/8) (1/100) = 43/800 and you are finished with your calculations!! EASY!!

Since a percent is the ratio of a number to 100, we can have percents that are greater than or equal to 100%

1 = 100/100 = 100%

165/100 = 165 %

Write 250% as a mixed number in simple form

250% = 250/100

250/100 = 2 50/100 = 2 1/2

The town of Wonderful spends 42% of its budget on education. What percent is used for other purposes?
the whole budget is represented by 100%. Therefore, the part used for other purposes is

100 - 42 or 58%

Percents and Decimals 9-2

By looking at the following examples, you will be able to see a general relationship between decimals and percents

57% = 57/100
0.79 = 79/100 = 79%
113% = 113/100 = 1 13/100
0.06 = 6/100 = 6%

Rules

To express a percent as a decimal, move the decimal point two places to the left and remove the percent sign

57% = 0.57
113% = 1.13

To express a decimal as a percent, move the decimal point two places to the right and add a percent sign

0.79 = 79%
0.06 = 6%


In 9-1 you learned one method of changing a fraction into a percent. Here is an alternative method

Rule
To express a fraction as a percent, first express the fraction as a decimal
and then as a percent

Express 7/8 as a percent

Divide 7 by 8
7/8 = 0.875 = 87.5%

Express 1/3 as a percent
divide 1 by 3
0.33333….. it’s a repeating decimal
express the decimal as a percent 0.333… = 33 1/3%
so, to the nearest tenth of a percent = 33.3% but it is much more accurate to keep the 1/3 and write 33 1/3%

Math 6A ( Periods 2 & 4)

 Graphing Inequalities  

You will need to look at the graphs in your textbook. .. page 397

Whenever we graph relations that are inequalities we must be aware of all the facts that can influence your work. You need to ask yourself, "What kind of numbers is the solution supposed to be?"
When you graphed inequalities such as
-3 < x < 2 where x was an integer we used a point on the number line for each integer that could be a solution to that inequality. To show every number in x < 2 we would use a number line and place an Open Dot at 2 indicating that 2 was NOT part of the solution and then draw a darkened ray away from 2 indicating 1, 0, -1, -2... were all part of the solution.
To show that this line has infinite solutions in that direction, you MUST place an arrow at the end of that darkened ray.

If the inequality was a " less than or equal to" " ≤" you would use a Closed Dot at 2 to indicate that 2 was part of the solution.


We graphed y ≥ x - 3
Notes will be added ...

WE will continue this lesson on Monday, April 15th....

We can now graph inequalities such as y ≥ x + 2
first you find the BOUNDARY LINE which is just y = x + 2 and you can use the 3 column table as we have done before or use a T chart as shown in class.

Remember you only need 2 points to determine a line---> but 3 points will help you make sure you have 3 correct points on the line!!

I am going to try to set up a T chart using "I" to separate the x and y
X I Y
-2 I 0
-1 I 1
0 I 2
1 I 3
2 I 4

(Note: it doesn't line up well here.. but hopefully you get the idea)

Plot those points on the graph and you have what appears to be a straight line. Since we are graphing y ≥ x + 2 we ARE including the line so we draw a solid line.

But.. what points are included?
Well, we know that (-2,0) works but we also see if we plug into our inequality that (-2,1) and (-2,2) work as well.

We need to shade the part above the line to indicate all those points are part of the solution as well.

Three set method for graphing an inequality

(1) Determine the boundary line. Draw it--
use a solid line if the boundary line is part of the graph (≤ or ≥)
use a dashed line if the boundary line is NOT part of the graph (< or >)

(2) Shaded either the part above the boundary line or the part below the boundary line.
If the inequality reads y > or y ≥ shade ABOVE the line.
If the inequality reads y < or y ≤ shade BELOW the line

(3) Always CHECK- choose a point you think works within the shaded region and see if it does work.. or use (0,0) and determine if it is part of the solution or not!!

Algebra Honors ( Periods 5 & 6)

 Percents 7.5
You have been doing this since at least 6th grade so this portion should really be review.
The word percent means hundredths or divided by 100

Some examples:
29 percent = 29% = 29/100 = 0.29
2.6 percent = 2.6 % = 2.6/100 = 26/1000 = 0.026
637 percent = 637% = 637/100 = 6 37/100 = 6.37
0.02 percent = 0.02% = 0.02/100 = 2/10000 = 0.0002
1/4 percent = 1/2% = 0.25 % = .25/100 = 25/10000 = 0.0025
33 1/3 percent = 33 1/3 % = 100/3% = 100/3 ÷ 100 = 1/3

Most calculators have a % key which will enable you to check your work!!
Remember in percent problems, the word, OF means Multiply and the word IS means EQUALS.

You can set up your problems either with an equation or with a proportion
Example 1:
15% of 180 is what number?
15/100 × 180 = x or
.15 × 180 = x

or set up a proportion
%/100 = 'is'/'of'
15/100 = x/180

In both cases x = 27

Example 2:
23 is 25% of what number?
23 = .25×n

or as a proportion
25/100 = 23/x

Again in both cases
x = 92

Example 3:

What Percent of 64 is 48?
x% × 64 = 48
or
x/100 × 64 = 48
64x/100 = 48

Set up as a proportion
x/100 = 48/64

x = 75
Remember to look at the question to make sure you have answered it. In this case you need to make sure you answer with 75%

When you solve with decimal coefficients, you can multiply both sides of the equation by a power of 10 (10, 100, 100 and so on) to get an equivalent equation with integral coefficients.

Example 4:
1.2x = 36 + 0.4x
Multiply both sides by 10 because the coefficients are tenths
12x = 360 + 4x
8x = 360
x = 45
{45}

Example 5:

94 = 0.15x + 0.08(1000 - x)
Multiply both sides by 100 because the coefficients are hundredths

9400 = 15x + 8(1000 - x)
9400 = 15x + 8000 - 8x
1400 = 7x
200 = x
{200}

Word Problem Example:
During a sale, a sporting goods store gave a 40% discount on sleeping bags. How much did Ross pay for a sleeping bag with an original price of $75?

Two different methods to find the same solution:
Method 1
Find 40% of $75? 0.40 × 75 = 30
Subtract the amount of discount from the original price
75-30 = $45
Ross paid $45 for the sleeping bag

Method 2
If the sleeping bag was discounted 40%, it then cost 100% -40% or 60% of its original price
Find 60% of 75 0.60 × 75 = 45
Ross paid $45 for the sleeping bag.

Math 6A ( Periods 2 & 4)

Graphs of Equations 11-9

An equation in two variables y = x + 2      produces an infinite number of ordered pairs
If we give x the value of 1, a corresponding value of y is determined
y = (1) + 2 = 3
The ordered pair is (1, 3)

If we let x = 4
y = (4) + 2 = 6
and we get the ordered pair (4, 6)

What happens if x = 0
y = (0) + 2 = 2 ( 0, 2)
or x = -1
y = (-1) + 2 = 1 ( -1, 1)

I like to remember ordered pairs---> ( ordered, pairs)

We graphed the line on a mini graph stickie.
The line is graphed using a ruler and connecting all the points we plotted. Put arrows at each end (since a line continues with out end) and write the line's equation right above the line.


For each value of x there is EXACTLY 1 value of y.
set of ordered pairs in which no two ordered pairs have the same x is called a FUNCTION


y = x + 2
in the future you will see it written as
f(x) = x + 2
so if x = 2
f(2) = (2) + 2 = 4
if x = 5
f(5) = (5) +4 = 9



We used a three column chart to compute our ordered pairs.
Please refer to the work sheet glued into your spiral notebook for the examples we completed from the class exercises found on Page 393 -- if you were absent, please come in one morning and I will review that chart with you.


The following equations create curves that are called PARABOLAS!! Notice the difference in these equations from our previous equations
y = x² +1
when we create your three column table using integers from -2 to 2
we notice
y = (-2)² +1 = 4 + 1 = 5 ordered pair (-2, 5)
y = (-1)² +1 = 1 + 1 = 2 ordered pair (-1, 2)
y = (0)² +1 = 0 + 1 = 1 ordered pair (0, 1)
y = (1)² +1 = 1 + 1 = 2 ordered pair (1, 2)
y = (2)² +1 = 4 + 1 = 5 ordered pair (-2, 5)

When you graph this... you get a "U" shaped graph.

Remember linear equations LINEar equations are lines!
and look like y = x + 2

PARABOLAS have the form y = x² or y = -x²

Let's try
y = 2 - x²
With our 3 column table
for values of x from -2 to 2
we find
y = 2 -(-2)² = 2 -(4) = -2 and the ordered pair is (-2,-2)
y = 2 -(-1)² = 2 - (1) = 1 and the ordered pair is ( -1, 1)
y = 2 -(0)² = 2 - 0 = 2 and the ordered pair is (0, 2)
y = 2 -(1)² = 2 -1 = 1 and the ordered pair is (1, 1)
y = 2 -(2)² = 2 - (4) = -2 and the ordered pair is (2, -2)

When you graph these ordered points you find you have an upside down U
hmmm... y = -x² results in a sad face parabola
and y = x² results in a happy face parabola!!

Algebra Honors (Periods 5 & 6)

Fractional Equations 7.4

The total resistance R of an electrical circuit with two resistors R₁ and R₂, that are connected in parallel is given by the formula
1/ R₁ +1/ R₂ = 1/R
What do you notice about the difference between this equation and those with fractional coefficients?

This formula is an example of a fractional equation.
An equation with a variable in the denominator of one or more terms is called a fractional equation. To solve a fractional equation, you can multiply BOTH sides of the equation by the LCD or you could use the method of solving a proportion when the equation consists of one fraction equal to another fraction.
3/x -1/4 = 1/12
The LCD of the fractions is 12x
Multiply BOTH sides of the equation by the LCD, 12x
Notice that x ≠ 0 because in this case 3/0 is undefined
12x(3/x – ¼) = (1/12)(12x)
36 -3x = x
36 = 4x
x = 9
{9}

(2-x)/(3-x) = 4/9
There are two different ways to solve this
First by finding the LCD, which is 9(3-x) Notice that x ≠3 Why?
9(3-x)[(2-x)/(3-x) = (4/9)[9(3-x)]
18-9x =12-4x
6 = 5x
6/5 = x
{6/5}
OR solve as proportion
(2-x)/(3-x) = 4/9

(2-x)(9) = (4)(3-x)
18- 9x = 12- 4x
we are at the same spot as with the first method and we arrive at the same solution
6 = 5x
6/5 = x
{6/5}

The following gets a little more complicated to do and to display here...
Solve
(2/b² - b) – 2/(b-1) = 1
Find the LCD by first factoring the denominators first
b² - b = b(b-1) so the LCD of the two fractions in this equation is in fact
b² - b BUT use it in factored form b(b-1) Notice: b ≠ 1 why?
(2/b² - b) – 2/(b-1) = 1
[b(b-1][ (2/b² - b) – 2/(b-1) ]= 1[b(b-1)]
which separates to
[b(b-1) (2/b² - b)] – [b(b-1)2/(b-1)] = 1
[b(b-1) (2/b(b - 1)] – [b(b-1)2/(b-1)] = 1
2 -2b=b(b-1)
or
2 – 2b = b² - b
solve for b now
0 = b² - b + 2b -2
0 = b² + b – 2
0 = (b-1)(b+2)
b = 1 and b = -2
Remember in this case b ≠ 1 because of the ORIGINAL EQUATION
the solution set is only
{-2)

Multiplying both sides of an equation by a variable expression sometimes results in an equation that has an extra root. You must check each root of the transformed equation to see if it satisfies the original equation.

Algebra Honors ( Periods 5 & 6)

Equations with Fractional Coefficients 7.3

Solving an equation with fractional coefficients can be easily accomplished by using the LCD of all the fractions in the equation. Clearing the equation of all fractions BEFORE attempting to solve the equation is probably the best way

Easy examples:
x/3 + x/7 = 10

The LCD of the fractions is 21
so multiply BOTH SIDES by 21

21(x/3 + x/7) = 10 (21)
7x + 3x = 210
10x = 210
x = 21
{21}

3a/5 – a/2 = 1/20

The LCD of the fractions is 20
20(3a/5 – a/2) = 1/20(20)
4(3a) -10a = 1
12a- 10a = 1
2a = 1
a = ½
{1/2}

x/3- (x+2)/5 = 2

The LCD of the fractions is 15
15[x/3- (x+2)/5] = 2(15)
5x –(3)(x+2) = 30
5x -3x-6 = 30
2x = 36
x = 18
{18}

2n + n/3 = n/4 + 5

The LCD is 12
12(2n + n/3) = (n/4 +5)(12)
24n + 4n = 3n + 60
28n = 3n + 60
25n = 60
n = 60/25 = 12/5
{12/5}


More complicated:
(1/4)(n + 2) – (1/6)(n – 2) = 3/2

The LCD is 12
12[(1/4)(n + 2) – (1/6)(n – 2)] = (3/2)(12)
3(n+2) – 2(n-2) = 18
3n + 6 -2n + 4 = 18
n = 8
{8}

Solving some word problems:
one eighth of a number is ten less than one third of the number. Find the number.
Let x = the number.
n/8 = n/3 – 10
LCD is 24
(24)(n/8)= (n/3-10)24
3n = 8n – 240
-5n = -240
5n = 240
n = 48
{48}

Math 6A ( Periods 2 & 4)

Graphs of Ordered Pairs 11-8

A PAIR of numbers whose ORDER is important is called an
ordered pair!!
(ordered, pair)

(2,3) is not the same as (3,2)

The two perpendicular lines are called axes.
The x-axis deals with the 1st number of the ordered pair and the y-axis deals with the 2nd number of the ordered pair.
The AXES meet at a point called the Origin (0,0)
The plane is called the coordinate plane

There are 4 quadrants, Use Roman Numerals to name them!!
Quadrant I ---> both the x and y coordinates are positive
(x,y) (+,+)

Quadrant II --> the x coordinate is negative but the y is positive
(-x,y) (-,+)

Quadrant III -->. both the x and y coordinates are negative
(-x,-y) (-,-)

Quadrant IV --> the x coordinate is positive but the y coordinate is negative
(x,-y) (+,-)

Math 6 High ( Period 3)

Discount & Sales Tax 7.7

Let’s go shopping!
A reduction in the price of an item is called a discount. Discounts are often calculated using percent of the original price.
To find the price of an item that is discounted:

1) find the amount of the discount

2) subtract the discount from the original price.

You buy a pair of hiking shoes. The original price of $65 is discounted by 35%. What is the sales price?

Solution:
Find the amount of the discount

Discount = 35% of $65

Discount = 65 ( .35)
=22.75

$22.75 is what you will SAVE!! It isn’t what you will pay to purchase them so

Remember

Sales price = Original price – discount
65- 22.75 = 42.25

$42.25

Which television’s sale price is less expensive? Let’s assume they are the same quality
Television A: Discount of 25% off the original price of $575
Television B: Discount of 30% off the original price of $585.

Television A:  Sales price = 575 – (0.25 · 575)
                        575 -143.75
                        431.75

Television B: Sales price = 585 –(0.30 · 585)
                        585- 175.5
                        409.50

Television B’s sales price of $409.50 is less expensive.

Finding Sales Tax

A sales tax is an amount added to the price of a purchased item. Sales tax is often calculated using a percent of the price.
To find the total price of an item including sales tax:

1) Find the amount of the sales tax

2) Add the sales tax to the price.

Example:
You buy an ipod shuffle. The price is $40.00 and the sales tax is 8.25% What is the total price you will pay?

Solution:
Find the amount of the sales tax.

Sales tax       = 8.25% of 40
                        =0.0825(40)
                        = 3.3

Which means the sales tax is $3.30

Add the sales tax to the price

Total price     = Purchase Price + Sales Tax
                        = $40 + $3.30 = $43.30

You will pay $43.30.

                       
Finding a Total Price

Shirts in a store are on sale for 30% off. The sales tax is 7.25%.  What is the total price of a shirt that has an original price of $29.95?

Solution: First find the discount and  the sale price.
Then find the sales tax and add that to the sale price to find the total price.
Careful—lots of steps here!

Finding the discount and the sales price.

Discount         = 30% of 29.95
                        = 0.3(29.95)
                        = 8.99
Sale Price     = Original Price – Discount
                        = 29.95 – 8.99 = 20.96

Sale Price is $20.96

That is the amount you calculate your sales tax upon so

Finding the Sales tax and the Total Price

Sales tax       = 7.25% of 20.96

                        = 0.0725(20.96)=1.52

Total Price     =  Sale Price + Sales Tax
                        = 20.96 + 1.52 = 22.48

The total price including sales tax is $22.48 

Math 6 High ( Period 3)

Simple Interest 7.6

When you put money in a savings account at a bank, the bank pays you interest. When you borrow money from a bank, you must pay the bank interest to the bank in addition to paying bank the money that you borrowed.

A principal is an amount borrowed, loaned or saved. An annual interest rate is the percent of the principal you earn or pay as interest for the year.
SIMPLE INTEREST is the product of the principal, the annual interest rate , and the time in years.

Interest = Principal ·rate ·time

I=Prt

Examples:

You borrow $1500 from a friend for the down payment on a car. Your friend charges you an annual interest rate of 8% ( Nice friend!) Find the simple interest you will pay in 1 year.

Solution:
Remember   I = Prt

What do you know?

I = (1500) (0.08)(1) 
Most of the time change the interest rate to a decimal. Occasionally you will want to use fractions

Carefully multiply... and discover that

I = 120

Therefore: The simple interest you will pay in 1 year is $120.

You deposit $300 in a savings account. The annual interest rate is 3% (not much). Find the simple interest you will earn in 1 month.

What do you know?   You know the principal  ( or P in the formula) is $300. You know the annual interest rate ( or r in the formula)  is 3%  But.. wait… be careful…the time must be in years  this is only for one month. The time (or t in the formula is 1/12)

Solution:

Remember   I = Prt

I = (300)(0.03)(1/12)

I would carefully simplify before I multiplied

I = 0.75

Again, think what does that mean?
The simple interest you would earn on $300 in 1 month at that rate is $0.75 

When you know the values of any three of the variables in the formula   I = Prt  you can use substitution to find the value of the fourth variable.

Your savings account earns $68 in simple interest in 1 year. The annual interest rate is 8% what is the principal?

Ask yourself, what do I know? What are the three variables  in the formula  I = Prt   that I do know?

$68 is the interest. The rate is 8% and it’s only for 1 year.

so

Remember   I = Prt

68 =P(0.08)(1)

68=0.08p

Solve this one-step equation

Divide both sides by 0.08 carefully

68/0.08 = 0.08p/0.08

850 = p

Therefore the principal is $850

You want to open an account with $100.
At Bank A you will earn $0.65 in 3 months.
At Bank B you will earn $2.25 in one year. At which bank will you open your account?
Although banks do not pay simple interest (they use compound interest)  we will assume that Bank A and B are paying only simple interest.

Figure out each bank’s simple interest rate.

Remember   I = Prt  

Bank A:   3 months is 3/12 or ¼  of a year so 

0.65 =  (100)(r) (¼)
0.65= 25r

Solve this one step equation
Divide both sides by 25

0.65/25= 25r/25

0.026= r
We need to change the decimal into a percent
That means Bank A’s rate is 2.6%

Bank B: Yes you can use the formula but look at what you know

2.25 =(100)(r)(1)
That means  r= 0.225
or that the rate for Bank B is 2.25%

You would earn more money if you deposit your savings into Bank A

Math 6A (Periods 2 & 4)

Equations: All 4 Op's Sections 8.1- 8.5 Summary 

You may be able to solve some of the equations in the homework without pencil and paper. Nevertheless, it is important to show all the steps in your work and to make sure you can tell which transformation you are using in each step.

If the replacement set for an equation is the set of whole numbers, it is not practical to use substitution to solve the equation. Instead we transform or change the given equation into a simpler, equivalent equation. When we transform the given equation, our goal is to arrive at an equivalent equation of the form
variable = number

Transformation by addition: add the same number to both sides
Transformation by subtraction: subtract the same number from both sides

solve x – 2 = 8
our goal is to find an equivalent equation of the form
x = a number

The left side of the given equation is x – 2. Recall that addition and subtraction are inverse operations. If we add 2 to both sides the left sides simplifies to x
x-2 = 8
x – 2 + 2 = 8 + 2 (We usually show the +2 right below each side of the equation)
x = 10


Solve x + 6 = 17
Subtract 6 from both sides of the equation to get an equivalent equation of the form
“ x = a number”

x + 6 = 17
x + 6 – 6 = 17 – 6 (Again, we usually show the -6 right below each side of the equation)
x = 11
the solution is 11

In equations involving a number of steps, it is a good idea to check your answer. This can be done easily by substituting the answer in the original equation.
What about the following
34 – x = 27
add x to both sides
34 – x + x = 27 + x
34 = 27 + x
subtract 27 from both sides
34 – 27 = 27 – 27 + x
7 = x


If an equation involves multiplication or division, the following transformations are used to solve the equation:

Transformation by multiplication: Multiply both sides of the equation by the same nonzero number.
Transformations by division: Divide both sides of the equation by the same nonzero number.

Remember: Do undo on one side what you would do undo the other!!

Our goal is to get the variable alone and to find an equivalent equation of the form
“n = a number”
Our goal is to arrive at the “world’s easiest equation”

Solve 3n = 24
Use the fact that multiplication and division are inverse operations
3n/3 = 24/3
n = 8
Or you could have use the reciprocal of 3--> which is 1/3 and multiplied both sides by 1/3
(1/3)(3n) = 24(1/3)
n = 8 and still arrived at the SAME solution

Solve 5x = 53

5x/5 = 53/5
x = 10 3/5

Solve n/4 = 7
(4)(n/4) = 7(4)
n = 28
How could you know for sure your answer is correct?
Substitute your solution into the ORIGINAL equation
Try:
1. 3r = 57

2. 714 = 7t

3. Solve A = bh for h

4. Solve P = 4s for s

5. Solve C = 2πr for r



Remember when we stated the properties as well-- back in our 1st quarter!!  

Equations with Decimals or Fractions 
You can use transformations to solve equations which involve decimals or fractions
Solve 0.42 x = 1.05
Divide both sides by 0.42
.42x/.42 = 1.05/.42
now, do side bar and actually divide carefully and you will arrive at
x = 2.5

Solve: n/.15 = 92
multiply both sides by .15 to undo the division
(n/.15)(.15) = 92 (.15)
Again, do a sidebar for your calculations and you will arrive at
n = 13.80

How would we solve the following: (2/3)x = 6?

Let’s look at 2x = 6. What do we do?
We divide both sides by 2—or multiply both sides by the reciprocal of 2—which is ½
Remember the product of a number and its reciprocal is 1

Reminder: the ultimate objective is applying transformations to an equation is to obtain an equivalent equation in the form x = c
(where c is a constant.)

Also remember that the understood (invivisble) coefficient of x in the equation x = c is 1.

[Can you picture the poster in the front of the room?]

So to solve (2/3)x = 6 you would divide both sides by 2/3 but that is the same as multiplying by the reciprocal of 2/3, which is 3/2.

If an equation has the form

(a/b)(x) = c,
where both a and c are nonzero,
multiply both sides by b/a, the reciprocal of a/b
Solve (1/3)y = 18

(3/1)(1/3)y = 18(3/1)

y =18(3)
y = 54

Solve the equation: (6/7)n = 8
(7/6)(6/7)n = 8(7/6)

n = 8(7/6)

simplify first , then multiply
n = 28/3
n = 9 1/3
Let’s check

(6/7)n = 8
well, we said that n = 9 1/3 so substitute back, but change to 28/3 first
(6/7)(28/3) ?=? 8
[read ?=? as ‘does that equal?’]

Now really do a side bar with the left side of the equation to see what
(6/7)(28/3) really equals. Simplify before you multiply
2(4) = 8 so
8 = 8


Try:
1. (1/7)a = 13

2. b/8 = 16

3. 3.6d = 0.9

4. (3/8)f = 129


Combined Operations 8-5 

In order to solve an equation of the form
ax + b = c or ax –b = c or b – ax = c
where a, b, c are given numbers and x is the variable, we must use more than one transformation

Solve the equation 3n - 5 = 10 + 6
Simplify the numerical expression

3n - 5 = 10 + 6
3n – 5 = 16

add 5 to both sides

3n – 5 + 5 = 16 + 5
or
3n – 5 = 16
+ 5 = +5

3n = 21

divide both sides by 3 (or multiply each side by the reciprocal of 3)
3n/3 = 21/3

n = 7

General procedures for solving equations

Simplify each side of the equation
If there are still indicated additions or subtractions, use the inverse operation to undo them
If there are indicated multiplications or division involving the variable, use the inverse operations to undo them

The books says you must always perform the same operation on both sides of the equation. I say, “do to one side what you have done to the other side.”

Solve the equation
(3/2)n + 7 = 22
subtract 7 from both sides
(3/2)n + 7 - 7 = 22 - 7
(3/2)n =15

multiply both sides by 2/3, the reciprocal of 3/2

(2/3)(3/2)n = 15(2/3)
n = 10

Solve the equation
40 – (5/3)n = 15
add (5/3)n to both sides

40 – (5/3)n + (5/3)n = 15 + (5/3)n

40 = 15 + (5/3)n
subtract 15 from both sides

40 – 15 = 15-15 + (5/3)n

25 = (5/3)n multiply both sides by 3/5

(3/5)(25) = (5/3)n (3/5)

15 = n

Math 6A (periods 2 &4)

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



t - -10 = 19
becomes -- with add the opposite---
t+ + 10 = 19
which is just
t + 10 = 19
so subtract 10 from both sides
t + 10 = 19
- 10 = - 10
t = 9

w - - 26 = -44
"Add the Opposite"
w + + 26 = -44
- 26 = - 26
x = -70

Know your integer rules and it becomes easy!!
Side bars are great, if you need them with difference signs!!


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

What about -5u = 125?
Whats happening to u?
It is being multiplied by -5... so you must divide by -5

-5u = 125
-5 -5
u = -25

or written easier to read -5u/-5 = 125/-5
u = -25

(1/-9)c = 33
Need to multiply both sides by the reciprocal of (1/-9) which is (-9/1)

(-9/1)(1/-9)c = 33(-9/1)
c = -297



2- STEP EQUATIONS

What about

3u - 1 = -7
You need to do the reverse of PEMDAS... remember unwrapping the present? We did the exact opposite of what we had done to wrap the present!!
so
3u - 1 = -7
+ 1 = + 1
3u = -6
Now divide by 3 on both sides
3u/3 = -6/3
u = -2


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.



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

Math 6 High ( Period 3)

Solving Percent Problems 7.4

You can use a proportion to find what percent one number is of another number.
The statement:

“ a is p percent of b” is expressed by the proportion shown below where

a is the  part of the base      b is the base… and    p is the percent

a/b = p/100

Part of the base / base = percent/100

What percent of 40 is 15?

First think which is the base? 40

Which number is the part?  15

Now write the proportion

a/b = p/100

15/40 =  p/100

Use cross products to multiply… simplify first if you can…

15(100) = 40p

Divide both sides by 40

15(100)/40 = 40p/40

37.5 = p

Note: there are a number of ways to simplify first! You should still arrive at p = 37.5 which means 37.5%

Finding a Percent

Your team won 19 of its 25 softball games. What percent did it win?

Games won/games played = percent/100

It really is still      a/b = p/100

19/25 = p/100

19(100) = 25p

Divide both sides by 25

19(100)/25 = 25p/25

76= p

Again there are a variety of ways to simply BEFORE you multiply! You could even utilize the equivalent fraction method to determine the p percent.

Therefore your team won 76% of its games!

You can use the proportion to find a base or a part of the base.

Finding a part of the base

You buy a pair of pants on sale. The price is 80% of the regular price of $24.50, WOW! That’s a bargain! What is this incredible sales price?

Sale price / regular price = Percent/100

a/24.50 = 80/100

Multiply and divide carefully

100a = 24.50(80)

divide both sides by 100

100a/100 = (24.50)(80)/100

a = 19.6

The sale price is $19.60

Now, you could use the writing an equation method

80% of 24.50 = sale price  or

(.80) (24.50) = p

Multiply carefully

19.60 = p

$19.60 is the sale price.

Finding a Base

8 is 32% of what number?

Method 1: Write and solve a proportion

a/b = p/100

8/b = 32/100
Using cross products
(8)(100) = 32b
Divide both sides by 32

(8)(100)/32 = 32b/32

25 = b

Method 2: Write and solve an equation

8 is 32% of what number?
8 = .32(x)

8/.32 = .32x/.32
Divide carefully

25= b

8 is 32% of 25

Solving Percent Problems

a/b = p/100   means a is p percent of b

Out of 80 students taking Algebra, 95% passed the final exam. About 80% of those who passed the exam got a grade better than a C. How many students got an A or a B?

First find out how many passed the final exam

Using the proportion method for this:

a/80 = 95/100

100a = 80(95)

100a/100 = (80)(95)/100

a = 76

76 students passed the final exam

About 80% of 76 got a grade better than a C

Using the writing an equation method

(.80)(76)  =60.8

Since you can’t have a part of a student your answer would be about 61 students got an A or a B in Algebra