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
Thursday, April 26, 2012
Math 6 Honors ( Periods 1, 2, & 3)
Commission and Profit 9-6
Some sales jobs pay an amount based on how much you sell. This amount is called a commission.
Like a discount, the commission can be expressed as a percent or as an amount of money.
amount of commission = percent of commission X total sales.
Using the examples from our textbook,
Maria sold $42,000 word of insurance in January. If her commission is 3% of the total sales, what was the amount of her commission in January?
amount of commission = percent X total sales
0.03 X 42,000 = 1260
Her commission was $1,260.
Profit is the difference between total income and total operating costs.
profit = total income – total costs
The percent of profit is the percent of total income that is profit
percent of profit = profit/total income
A shoe store had an income of $8600 and operating costs of $7310. What percent of the store's income was profit?
profit= income- total costs = 8600 -7310 = 1290
percent of profit = profit/total income = 1290/8600 = 0.15
So the percent of profit was 15%.
Practice finding 10%-- its easy--- just move the decimal over one place.
We practiced finding 20%. Just double what you got for 10%.
MATH AT WORK:
Caterer
A caterer provides food for parties, weddings, bar/bat mitzvahs, and other events. Caterers plan the menu, buy the ingredients, and cook the food. Often they provide seating and music as well. For each event, a caterer determines the cost per guest. The catering business requires a thorough knowledge of ratios, proportions, and percents.
Some sales jobs pay an amount based on how much you sell. This amount is called a commission.
Like a discount, the commission can be expressed as a percent or as an amount of money.
amount of commission = percent of commission X total sales.
Using the examples from our textbook,
Maria sold $42,000 word of insurance in January. If her commission is 3% of the total sales, what was the amount of her commission in January?
amount of commission = percent X total sales
0.03 X 42,000 = 1260
Her commission was $1,260.
Profit is the difference between total income and total operating costs.
profit = total income – total costs
The percent of profit is the percent of total income that is profit
percent of profit = profit/total income
A shoe store had an income of $8600 and operating costs of $7310. What percent of the store's income was profit?
profit= income- total costs = 8600 -7310 = 1290
percent of profit = profit/total income = 1290/8600 = 0.15
So the percent of profit was 15%.
Practice finding 10%-- its easy--- just move the decimal over one place.
We practiced finding 20%. Just double what you got for 10%.
MATH AT WORK:
Caterer
A caterer provides food for parties, weddings, bar/bat mitzvahs, and other events. Caterers plan the menu, buy the ingredients, and cook the food. Often they provide seating and music as well. For each event, a caterer determines the cost per guest. The catering business requires a thorough knowledge of ratios, proportions, and percents.
Tuesday, April 24, 2012
Math 6 Honors ( Periods 1, 2, & 3)
Discount and Markup 9-5
A discount is a decrease in the price of an item. A markup is an increase in the price of an item. Both of these changes can be expressed as an amount of money or as a percent of the original price of the item. A store may announce a discount of $3 off the original price of $30 basketball, or a discount of 10%
A warm-up suit that sold for $42.50 is on sale at a 12% discount. What is the sale price?
Method 1: Use the formula
amount of change = percent of change X original amount
= 12% X $42.50
therefore the discount is 0.12 X 42.50 or 5.10
The amount of discount is $5.10
The sale price is 42.50 – 5.10 = $37.40
Method 2: Since the discount is 12%, the sale price is 100% - 12% = 88%.
The sale price is 0.88 X 42.50 = $ 37.40
When you know the amount of discount you subtract to find the new price. When dealing with a markup you add to find the new price.
The price of a new car model was marked up 6% over the previous year’s model. If the previous year’s model sold for $7800, what is the cost of the new car? {and what kind of a car could that be?}
Method 1: Use the formula
amount of change = percent of change X original amount
= 6% X 7800
Therefore the markup is 0.06 X7800= $468
The new price is 7800 + 468 = $8268
Method 2: Since the markup is 6% the new price is 100% + 6% or 106% of the original price. so the new price is 1.06 X7800 = $8268
This year a pair of ice skates sells for $46 after a 15% mark up over last year’s price. What was last year’s price?
This year’s price is 100 + 15 or 115% of last year’s price. Let n present last year’s price
46 = (115/100)n
46 = 1.15n
46/.15 = 1.15n/1.115
40 = n
So last year’s price was $40.
A department store advertised eclectic shavers at a sale price of $36.
If this is a 20% discount, what was the original price?
The sale price is 100 - 20 or 80% of the original price. Let n represent the original price.
36 = (80/100)n
36 = .8n
36/.8 = .8n/.8
45 = n
The original price was $45.
Check to see that your answers are logical and reasonable.
Try these: A service station (that’s gas station, now—they no longer provide service!!) give cash customers a 5% discount on the price of gasoline. If gasoline regularly sells for $3.00 a gallon, what is the discounted price?
A store marks up the price of a $5 item to $12. What is the percent of markup?
A discount is a decrease in the price of an item. A markup is an increase in the price of an item. Both of these changes can be expressed as an amount of money or as a percent of the original price of the item. A store may announce a discount of $3 off the original price of $30 basketball, or a discount of 10%
A warm-up suit that sold for $42.50 is on sale at a 12% discount. What is the sale price?
Method 1: Use the formula
amount of change = percent of change X original amount
= 12% X $42.50
therefore the discount is 0.12 X 42.50 or 5.10
The amount of discount is $5.10
The sale price is 42.50 – 5.10 = $37.40
Method 2: Since the discount is 12%, the sale price is 100% - 12% = 88%.
The sale price is 0.88 X 42.50 = $ 37.40
When you know the amount of discount you subtract to find the new price. When dealing with a markup you add to find the new price.
The price of a new car model was marked up 6% over the previous year’s model. If the previous year’s model sold for $7800, what is the cost of the new car? {and what kind of a car could that be?}
Method 1: Use the formula
amount of change = percent of change X original amount
= 6% X 7800
Therefore the markup is 0.06 X7800= $468
The new price is 7800 + 468 = $8268
Method 2: Since the markup is 6% the new price is 100% + 6% or 106% of the original price. so the new price is 1.06 X7800 = $8268
This year a pair of ice skates sells for $46 after a 15% mark up over last year’s price. What was last year’s price?
This year’s price is 100 + 15 or 115% of last year’s price. Let n present last year’s price
46 = (115/100)n
46 = 1.15n
46/.15 = 1.15n/1.115
40 = n
So last year’s price was $40.
A department store advertised eclectic shavers at a sale price of $36.
If this is a 20% discount, what was the original price?
The sale price is 100 - 20 or 80% of the original price. Let n represent the original price.
36 = (80/100)n
36 = .8n
36/.8 = .8n/.8
45 = n
The original price was $45.
Check to see that your answers are logical and reasonable.
Try these: A service station (that’s gas station, now—they no longer provide service!!) give cash customers a 5% discount on the price of gasoline. If gasoline regularly sells for $3.00 a gallon, what is the discounted price?
A store marks up the price of a $5 item to $12. What is the percent of markup?
Monday, April 23, 2012
Math 6 Honors ( Periods 1, 2, & 3)
Percent of Increase or Decrease 9-4
Let's say we have an iPod that originally sold for $260. It is on sale for $208. What is the amount of change? "How much did you save?"
Just subtract
260-208 = 52
$52.
What is the percent of change?
The percent of change = amount of change/original
52/260 - x/100
or just divide 52 by 260 = .2
which is 20%
REMEMBER: The denominator in the formula is ALWAYS the ORIGINAL AMOUNT.
Amount of change = percent of change X the original amount.
Find the new number when 75 is decreased by 26%
Amount of change - 26% (75
= .26(75)
=19.5
Now take the difference (the amount of change) and subtract THAT from 75
75- 19.5 = 55.5
Remember the circle with the various parts of this formula?
Difference or amount of change
% of change X original amount
Difficult to show here so if you missed these notes make sure to ask a fellow student to see this!! IT is a great way to remember what to do!!
State the increase or decrease. Tell what the amount of change is and the percent of change.
from:
10 to 12
increase
amount of increase: 2
% of change : 20%
4 to 3
decrease
amount of decrease:
% of change : 25%
2 to 5
increase
amount of increase: 3
% of change : 3/2 = 1.5 = 150%
12 to 6
decrease
amount of decrease: 6
% of change : 50%
6 to 12
increase
amount of increase: 6
% of change : 6/6 = 1 = 100%
Find the new number produced when the given number is increased or decrease by the given percent.
120; 20% decrease
120(.20) = 24 120 -24 = 96
30; 10% decrease
30(.10) = 3 30 -3 = 27
48: 50% increase
48(.50) = 24 48 + 24 = 72
128: decrease by 25%, then increased by 25%
What... why multiply by .25 if you can use a fraction and work smarter?
128(1/4) = 32
128 - 32 = 96
then 96 ( 1/4) = 24
96 + 24 = 120
Did you think it would be the starting number? Why wasnn't it?
Let's say we have an iPod that originally sold for $260. It is on sale for $208. What is the amount of change? "How much did you save?"
Just subtract
260-208 = 52
$52.
What is the percent of change?
The percent of change = amount of change/original
52/260 - x/100
or just divide 52 by 260 = .2
which is 20%
REMEMBER: The denominator in the formula is ALWAYS the ORIGINAL AMOUNT.
Amount of change = percent of change X the original amount.
Find the new number when 75 is decreased by 26%
Amount of change - 26% (75
= .26(75)
=19.5
Now take the difference (the amount of change) and subtract THAT from 75
75- 19.5 = 55.5
Remember the circle with the various parts of this formula?
Difference or amount of change
% of change X original amount
Difficult to show here so if you missed these notes make sure to ask a fellow student to see this!! IT is a great way to remember what to do!!
State the increase or decrease. Tell what the amount of change is and the percent of change.
from:
10 to 12
increase
amount of increase: 2
% of change : 20%
4 to 3
decrease
amount of decrease:
% of change : 25%
2 to 5
increase
amount of increase: 3
% of change : 3/2 = 1.5 = 150%
12 to 6
decrease
amount of decrease: 6
% of change : 50%
6 to 12
increase
amount of increase: 6
% of change : 6/6 = 1 = 100%
Find the new number produced when the given number is increased or decrease by the given percent.
120; 20% decrease
120(.20) = 24 120 -24 = 96
30; 10% decrease
30(.10) = 3 30 -3 = 27
48: 50% increase
48(.50) = 24 48 + 24 = 72
128: decrease by 25%, then increased by 25%
What... why multiply by .25 if you can use a fraction and work smarter?
128(1/4) = 32
128 - 32 = 96
then 96 ( 1/4) = 24
96 + 24 = 120
Did you think it would be the starting number? Why wasnn't it?
Thursday, April 19, 2012
Algebra Honors (Period 6 & 7)
Solving Problems with Two Variables 9-3
Although the textbook uses charts and tables for these word problems, I think they work easily without the charts...
John has 15 coins -- all dimes and quarters , worth $2.55 How many dimes and quarters does he have?
let d = the number of dimes and let q = the number of quarters
we know d + q = 15 and we know 10d + 25q = 255
using a system of equations and the substitution method
since d + q = 15 we know d = 15- q
10d + 25q - 255
10(15-q) + 25 q = 255
150 -10q + 25q = 255
150 + 15q = 255
15q = 105
q = 7
He has 7 quarters and 8 dimes
Ann and Betty together have $ 60. Ann has $9 more than twice Betty's amount. How much money does each have?
Let a = the amount Ann has and let b = the amount Betty has.
we know
a + b = 60
and we know
a= 2b + 9
so using a + b = 60
(2b+9) + b = 60
3b = 51
b = 17
Betty has $17 and Ann has (60-17) = $43
Joan Wu invested $8000 in stocks and bonds. the stocks pay 4% interest and the onds pay 7% interest . The annual interest from the stocks and bonds is $500.
How much is invested in bonds?
let s = the amount invested in stocks
let b = amount invested in bonds
s + b = 8000
0.04s + 0.07b = 500
clear the decimals
4s + 7b = 50000
but we know s + b = 8000 or s = 8000 -b
4(8000 -b) + 7b = 50000
32000 - 4b + 7b = 50000
3b = 18000
b = 6000
She invested $6000 in bonds.
Although the textbook uses charts and tables for these word problems, I think they work easily without the charts...
John has 15 coins -- all dimes and quarters , worth $2.55 How many dimes and quarters does he have?
let d = the number of dimes and let q = the number of quarters
we know d + q = 15 and we know 10d + 25q = 255
using a system of equations and the substitution method
since d + q = 15 we know d = 15- q
10d + 25q - 255
10(15-q) + 25 q = 255
150 -10q + 25q = 255
150 + 15q = 255
15q = 105
q = 7
He has 7 quarters and 8 dimes
Ann and Betty together have $ 60. Ann has $9 more than twice Betty's amount. How much money does each have?
Let a = the amount Ann has and let b = the amount Betty has.
we know
a + b = 60
and we know
a= 2b + 9
so using a + b = 60
(2b+9) + b = 60
3b = 51
b = 17
Betty has $17 and Ann has (60-17) = $43
Joan Wu invested $8000 in stocks and bonds. the stocks pay 4% interest and the onds pay 7% interest . The annual interest from the stocks and bonds is $500.
How much is invested in bonds?
let s = the amount invested in stocks
let b = amount invested in bonds
s + b = 8000
0.04s + 0.07b = 500
clear the decimals
4s + 7b = 50000
but we know s + b = 8000 or s = 8000 -b
4(8000 -b) + 7b = 50000
32000 - 4b + 7b = 50000
3b = 18000
b = 6000
She invested $6000 in bonds.
Math 6 Honors ( Periods 1, 2, & 3)
Computing with Percents 9-3
The statement 20% of 300 is 60 can be translated into the following equations
20/100(300) = 60 or 0.20 •300 = 60
EQUATION METHOD:
Notice the following relationship between the words and the symbols
20% of 300 is 60
0.20 • 300 = 60
WRITE THE PROBLEM OUT AND THEN DIRECTLY UNDER THE "IS" WRITE AN EQUAL SIGN. DIRECTLY UNDER THE WORD 'OF" WRITE A MULTIPLICATION SIGN. iF YOU ARE GIVEN A % CHANGE IT FIRST TO A DECIMAL. THEN BRING DOWN ALL THE OTHER NUMBERS GIVEN IN YOUR PROBLEM. LET x OR n REPRESENT YOUR VARIABLE... THAT IS THE "WHAT " PART OF YOUR PROBLEM.
A similar relationship occurs whenever a statement or a question involves a number that is a percent of another number
What is 8% of 75?
Let n represent the number asked for
What number is 8% of 75?
n = 0.08 • 75
solve
What percent of 40 is 6?
let n represent the percent asked for.
What percent of 40 is 6?
n% • 40 = 6
n% • 40 = 6
n% (40)/40 = 6/40
n% = 6/40
n/100 = 6/40
(100) n/100 = (100) 6/40
n=15 so 15% of 40 is 6
140 is 35 % of what number?
let n represent the number asked for
140 is 35% of what number?
140 = 0.35 • n
140 = 0.35n
140/0.35 = 0.35n/0.35 divide carefully!! Watch those decimals!!
400 = n
so 140 is 35% of 400
Always check to see if your answer is logical.
PROPORTION METHOD
In these types of percent problems you are always know three parts of the following proportion
n/1oo = a/b
or better yet
n/100 = is/ of
The n represents the %
Read the problems carefully and you can easily determine which is the "is" and which represents the 'of"
For example:
What percent of 40 is 6?
What percent -- from the problem above indicates that we DO NOT know the n
of 40-- hmm... then 40 must be the 'of' and
similarly is 6 means that 6 represents the 'is'
n/100 = 6/40 solve as a proportion
and you get n= 15 but since it asked us to state the 5 your answer is 15%
140 is 35 % of what number?
In this problem I notice 35% right away so that is the n!!
Then I read the problem again and notice 140 is... hmmm.. THat says 140 must be the is
35/100 = 140/ x I do not know the 'of'
Solve again
x = 400
The statement 20% of 300 is 60 can be translated into the following equations
20/100(300) = 60 or 0.20 •300 = 60
EQUATION METHOD:
Notice the following relationship between the words and the symbols
20% of 300 is 60
0.20 • 300 = 60
WRITE THE PROBLEM OUT AND THEN DIRECTLY UNDER THE "IS" WRITE AN EQUAL SIGN. DIRECTLY UNDER THE WORD 'OF" WRITE A MULTIPLICATION SIGN. iF YOU ARE GIVEN A % CHANGE IT FIRST TO A DECIMAL. THEN BRING DOWN ALL THE OTHER NUMBERS GIVEN IN YOUR PROBLEM. LET x OR n REPRESENT YOUR VARIABLE... THAT IS THE "WHAT " PART OF YOUR PROBLEM.
A similar relationship occurs whenever a statement or a question involves a number that is a percent of another number
What is 8% of 75?
Let n represent the number asked for
What number is 8% of 75?
n = 0.08 • 75
solve
What percent of 40 is 6?
let n represent the percent asked for.
What percent of 40 is 6?
n% • 40 = 6
n% • 40 = 6
n% (40)/40 = 6/40
n% = 6/40
n/100 = 6/40
(100) n/100 = (100) 6/40
n=15 so 15% of 40 is 6
140 is 35 % of what number?
let n represent the number asked for
140 is 35% of what number?
140 = 0.35 • n
140 = 0.35n
140/0.35 = 0.35n/0.35 divide carefully!! Watch those decimals!!
400 = n
so 140 is 35% of 400
Always check to see if your answer is logical.
PROPORTION METHOD
In these types of percent problems you are always know three parts of the following proportion
n/1oo = a/b
or better yet
n/100 = is/ of
The n represents the %
Read the problems carefully and you can easily determine which is the "is" and which represents the 'of"
For example:
What percent of 40 is 6?
What percent -- from the problem above indicates that we DO NOT know the n
of 40-- hmm... then 40 must be the 'of' and
similarly is 6 means that 6 represents the 'is'
n/100 = 6/40 solve as a proportion
and you get n= 15 but since it asked us to state the 5 your answer is 15%
140 is 35 % of what number?
In this problem I notice 35% right away so that is the n!!
Then I read the problem again and notice 140 is... hmmm.. THat says 140 must be the is
35/100 = 140/ x I do not know the 'of'
Solve again
x = 400
Wednesday, April 18, 2012
Algebra Honors (Period 6 & 7)
Solving Systems of Linear Equations
The Graphing Method 9-1
Two or more equations in the same variables form a system of equations. The solution of a system of two equations in two variables is a pair of values x and y that satisfies each equation in the system. The point corresponding to the ordered pair (x, y) must lie on the graph of both equations.
Solve the system by graphing
2x - y = 8
x + y = 1
Solution:
Graph both 2x - 7 = 8 and x + y = 1 in the same coordinate plane.
We did this in class by transforming both equations to slope-intercept form (y = mx +b)
and then graphed them. We noticed that the only point on BOTH lines is the intersection point ( 3, -2)
The only solution of both equations is (3, -2).
You can check that ( 3, -2) is a solution fof the system by substituting x = 3 and y = -2 in BOTH eqquations.
Solve the system by graphing
x - 2y = -6
x -2y = 2
When you graph the equations in the same coordinate plane, you see that the lines have the same slope but different y-intercepts. The graphs are parallel lines. SInce the lines do not intersect, there is no point that represents a solution of both equations.
Therefore, the system has NO SOLUTION.
Solve the system by graphing
2x + 3y = 6
4x + 6y = 12
When you graph the equations in the same coordinate plane, you see that the graphs coincide. The equations are equivalent. Every point on the line represents a solution of BOTH equations.
Therefore, the system has infinitely many solutions.
The Graphing Method in review:
To solve a system of linear equations in two variables, draw the graph of each linear equation in the same coordinate plane...
--> if the lines interset there is only one solutions, namely the intersection point.
--> if the lines are parallel, there is no solution
--> if the lines coincide, there are infinitely many solutions.
The Substitution Method 9-2
There are several ways to solve a system of equations, In the substitution method we use either equation to solve for one variable in terms of the other.
Solve
x + y = 15
4x + 3y = 38
Solve the first equation for y
x + y = 15
becomes
y = -x + 15
Substitute this expression for y in the other equation, and solve for x
4x + 3y = 38
4x + 3(-x+15) = 38
4x -3x + 45 = 38
x + 45 = 38
x = -7
Substitute the value of x in the equation in your first step and solve for y
y = -x + 15
y = -(-7) + 15
y = +7 +15
y = 22
CHeck x = -7 and y = 22 on BOTH equations
x + y = 15
(Here let ?=? represent having a ? above the equals sign)
-7 + 22 ?=? 15
15 = 15
and
4x + 3y = 38
4(-7) + 3(22) ?=? 38
-28 + 66 ?=? 38
38 = 38
It checks for both equations so the solution is (-7, 22)
Solve
2x - 3y = 4
x + 4y = -9
Using the 2nd equation is easier to manipulate so solve for x since x has a coefficient of 1
x = -4y - 9
substitute this expression for x in the other equation and solve for y
2x - 3y = 4
2(-4y-9) - 3y = 4
-8y -18 -3y = 4
-11y = 22
y = -2
Substitute the value of y in the equation in step 1 and solve for x
x = -4y -9
x = -4(-2) -9
x = 8 -9 = -1
Check both equations... and you discover that the solution is ( -1, -2)
The substitution method is most convenient to use when the coefficient of one of the variables is 1 or -1.
The Substitution Method in review:
To solve a system of linear equations in two variables:
--> Solve one equation for one of the variables
--> Substitute this expression in the other equation and solve fore the other variable.
--> Substitute this value n the equation in step 1 and solve
--> Check the alues in BOTH equations.
Solve by the substitution method
2x -8y = 6
x - 4y = 8
x = 4y + 8
2x-8y = 6
2(4y+8) - 8y = 6
8y + 16 -8y = 6
16= 6 WAIT that's FALSE
The false statement indicates that there is NO ordered pair (x, y) that satisfies BOTH equations. If you had graphed the equations you would see that these lines are actually parallel.
Solve by substitution method
y/2 = 2 -x
6x + 3y = 12
The first equation is easy to change to y = 4 - 2x by multiplying both sides by 2 to solve for y
6x + 3y = 12
6x + 3(4-2x) = 12
6x + 12 - 6x = 12
12 = 12 WAIT THat's TRUE... always
Every ordered pair (x, y) that satisfies one of the equations aso satisfies the other. IF you graph these two equations you will see that the lines coincide
Therefore, the system has infinitely many solutions.
The Graphing Method 9-1
Two or more equations in the same variables form a system of equations. The solution of a system of two equations in two variables is a pair of values x and y that satisfies each equation in the system. The point corresponding to the ordered pair (x, y) must lie on the graph of both equations.
Solve the system by graphing
2x - y = 8
x + y = 1
Solution:
Graph both 2x - 7 = 8 and x + y = 1 in the same coordinate plane.
We did this in class by transforming both equations to slope-intercept form (y = mx +b)
and then graphed them. We noticed that the only point on BOTH lines is the intersection point ( 3, -2)
The only solution of both equations is (3, -2).
You can check that ( 3, -2) is a solution fof the system by substituting x = 3 and y = -2 in BOTH eqquations.
Solve the system by graphing
x - 2y = -6
x -2y = 2
When you graph the equations in the same coordinate plane, you see that the lines have the same slope but different y-intercepts. The graphs are parallel lines. SInce the lines do not intersect, there is no point that represents a solution of both equations.
Therefore, the system has NO SOLUTION.
Solve the system by graphing
2x + 3y = 6
4x + 6y = 12
When you graph the equations in the same coordinate plane, you see that the graphs coincide. The equations are equivalent. Every point on the line represents a solution of BOTH equations.
Therefore, the system has infinitely many solutions.
The Graphing Method in review:
To solve a system of linear equations in two variables, draw the graph of each linear equation in the same coordinate plane...
--> if the lines interset there is only one solutions, namely the intersection point.
--> if the lines are parallel, there is no solution
--> if the lines coincide, there are infinitely many solutions.
The Substitution Method 9-2
There are several ways to solve a system of equations, In the substitution method we use either equation to solve for one variable in terms of the other.
Solve
x + y = 15
4x + 3y = 38
Solve the first equation for y
x + y = 15
becomes
y = -x + 15
Substitute this expression for y in the other equation, and solve for x
4x + 3y = 38
4x + 3(-x+15) = 38
4x -3x + 45 = 38
x + 45 = 38
x = -7
Substitute the value of x in the equation in your first step and solve for y
y = -x + 15
y = -(-7) + 15
y = +7 +15
y = 22
CHeck x = -7 and y = 22 on BOTH equations
x + y = 15
(Here let ?=? represent having a ? above the equals sign)
-7 + 22 ?=? 15
15 = 15
and
4x + 3y = 38
4(-7) + 3(22) ?=? 38
-28 + 66 ?=? 38
38 = 38
It checks for both equations so the solution is (-7, 22)
Solve
2x - 3y = 4
x + 4y = -9
Using the 2nd equation is easier to manipulate so solve for x since x has a coefficient of 1
x = -4y - 9
substitute this expression for x in the other equation and solve for y
2x - 3y = 4
2(-4y-9) - 3y = 4
-8y -18 -3y = 4
-11y = 22
y = -2
Substitute the value of y in the equation in step 1 and solve for x
x = -4y -9
x = -4(-2) -9
x = 8 -9 = -1
Check both equations... and you discover that the solution is ( -1, -2)
The substitution method is most convenient to use when the coefficient of one of the variables is 1 or -1.
The Substitution Method in review:
To solve a system of linear equations in two variables:
--> Solve one equation for one of the variables
--> Substitute this expression in the other equation and solve fore the other variable.
--> Substitute this value n the equation in step 1 and solve
--> Check the alues in BOTH equations.
Solve by the substitution method
2x -8y = 6
x - 4y = 8
x = 4y + 8
2x-8y = 6
2(4y+8) - 8y = 6
8y + 16 -8y = 6
16= 6 WAIT that's FALSE
The false statement indicates that there is NO ordered pair (x, y) that satisfies BOTH equations. If you had graphed the equations you would see that these lines are actually parallel.
Solve by substitution method
y/2 = 2 -x
6x + 3y = 12
The first equation is easy to change to y = 4 - 2x by multiplying both sides by 2 to solve for y
6x + 3y = 12
6x + 3(4-2x) = 12
6x + 12 - 6x = 12
12 = 12 WAIT THat's TRUE... always
Every ordered pair (x, y) that satisfies one of the equations aso satisfies the other. IF you graph these two equations you will see that the lines coincide
Therefore, the system has infinitely many solutions.
Math 6 Honors ( Periods 1, 2, & 3)
Percent 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%
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%
Wednesday, April 4, 2012
Algebra Honors (Period 6 & 7)
Scientific Notation 7-10
Scientific notation makes it easier to work with either very large numbers or very small numbers. To write a positive number in scientific notation, you express it as the product of a number greater than or equal to 1 BUT less than 10 AND an integral power of 10.
When a positive number greater than or equal to 10 is written in scientific notation, the power of 10 is positive. When the number is less than 1, the power of 10 will be negative.
Scientific notation consists of two parts
1≤ x< 10 × POWER10
for example
58,120,000,000
Move the decimal point left 10 spaces to between the 5 and the 8 to get a number BETWEEN 1 and 10
5.8
so 5.8 × 1010
0.00000072
Move the decimal point 7 places to the right to get a number between 1 and 10
7.2
7.2 × 10-7
Numbers written in scientific notation can be multiplied and divided easily by using the rules of exponents
Simplify. Keep your answers in scientific notation
3.2 × 107/2.0 × 104
= 3.2/2.0 ×107/104
=1.6 × 107-4
=1.6 × 103
(2.5 × 103)(6.0 × 102)
=(2.5 ×6.0) × (103 × 102) Add exponents
= (15)(103+2)
=15 × 105
But wait 15 is not between 1 and 10
15 itself is 1.5 × 101
so it becomes
1.5 ×101× 105
1.5 × 101+5
=1.5 × 106
The distance from the sun to Mercury is approximately 6 × 108 km.
The distance from the sun to Pluto is approximately 5.9 × 109 km
Find the ratio of the first distance to the second distance
Distance from sun to Mercury
Distance from sun to Pluto
6 × 108
5.9 × 109
= 6/5.9 × (108)/109
= 6/5.9 ×108-9
6/5.9 ×10-1
=6/5.9 ×1/10
= 6/59
You learned expanded notation in 6th grade... but a review of writing numbers in expanded notation using powers of 10 follows:
8572 = 8000 + 500 + 70 + 2
=8(103) + 5(102) + 7(101) + 2(100)
0.3946 = 0.3 + 0.09 + 0.004 + 0.0006
= 3(10-1) + 9(10-2) +4(10-3) + 6(10-4)
25.03 = 2(101) +5(100) + 0(10-1) + 3(10-2)
The metric system is also based on powers of TEN.
To change from one metric unit to another, you simply multiply by a power of 10.
Remember:
King Henry Died By Drinking Chocolate Milk
1 km = 103m = 1000m
1mL =10-3 L = 1/1000 L
Scientific notation makes it easier to work with either very large numbers or very small numbers. To write a positive number in scientific notation, you express it as the product of a number greater than or equal to 1 BUT less than 10 AND an integral power of 10.
When a positive number greater than or equal to 10 is written in scientific notation, the power of 10 is positive. When the number is less than 1, the power of 10 will be negative.
Scientific notation consists of two parts
1≤ x< 10 × POWER10
for example
58,120,000,000
Move the decimal point left 10 spaces to between the 5 and the 8 to get a number BETWEEN 1 and 10
5.8
so 5.8 × 1010
0.00000072
Move the decimal point 7 places to the right to get a number between 1 and 10
7.2
7.2 × 10-7
Numbers written in scientific notation can be multiplied and divided easily by using the rules of exponents
Simplify. Keep your answers in scientific notation
3.2 × 107/2.0 × 104
= 3.2/2.0 ×107/104
=1.6 × 107-4
=1.6 × 103
(2.5 × 103)(6.0 × 102)
=(2.5 ×6.0) × (103 × 102) Add exponents
= (15)(103+2)
=15 × 105
But wait 15 is not between 1 and 10
15 itself is 1.5 × 101
so it becomes
1.5 ×101× 105
1.5 × 101+5
=1.5 × 106
The distance from the sun to Mercury is approximately 6 × 108 km.
The distance from the sun to Pluto is approximately 5.9 × 109 km
Find the ratio of the first distance to the second distance
Distance from sun to Mercury
Distance from sun to Pluto
6 × 108
5.9 × 109
= 6/5.9 × (108)/109
= 6/5.9 ×108-9
6/5.9 ×10-1
=6/5.9 ×1/10
= 6/59
You learned expanded notation in 6th grade... but a review of writing numbers in expanded notation using powers of 10 follows:
8572 = 8000 + 500 + 70 + 2
=8(103) + 5(102) + 7(101) + 2(100)
0.3946 = 0.3 + 0.09 + 0.004 + 0.0006
= 3(10-1) + 9(10-2) +4(10-3) + 6(10-4)
25.03 = 2(101) +5(100) + 0(10-1) + 3(10-2)
The metric system is also based on powers of TEN.
To change from one metric unit to another, you simply multiply by a power of 10.
Remember:
King Henry Died By Drinking Chocolate Milk
1 km = 103m = 1000m
1mL =10-3 L = 1/1000 L
Algebra Honors (Period 6 & 7)
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 am/an= am-n
For example
a7/a3 = a7-3= a4
you can also apply this rule when m < n that is when m - n becomes a negative number. For example a3/a7 = a3-7 = 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 = a5-5 = 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
bmbn = bm+n
Example with negative exponents
23⋅2-5 = 23+(-5) = 2-2 = 1/22 = 1/4
Quotient of Powers
bm ÷ bn = bm-n
Example with negative exponents
63÷67= 63-7= 6-4= 1/64= 1/1296
Power of Powers
(bm)n = bmn
Example with negative exponents
(23)-2 = 2-6 = 1/26 = 1/64
Power of a Product
(ab)m= ambm
Example with negative exponents
(3x)-2 = 3-2 ⋅x-2 = 1/32⋅1/x2 = 1/9x2
Power of a Quotient
(a/b)m= am/bm
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
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 am/an= am-n
For example
a7/a3 = a7-3= a4
you can also apply this rule when m < n that is when m - n becomes a negative number. For example a3/a7 = a3-7 = 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 = a5-5 = 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
bmbn = bm+n
Example with negative exponents
23⋅2-5 = 23+(-5) = 2-2 = 1/22 = 1/4
Quotient of Powers
bm ÷ bn = bm-n
Example with negative exponents
63÷67= 63-7= 6-4= 1/64= 1/1296
Power of Powers
(bm)n = bmn
Example with negative exponents
(23)-2 = 2-6 = 1/26 = 1/64
Power of a Product
(ab)m= ambm
Example with negative exponents
(3x)-2 = 3-2 ⋅x-2 = 1/32⋅1/x2 = 1/9x2
Power of a Quotient
(a/b)m= am/bm
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
Monday, April 2, 2012
Math 6 Honors ( Periods 1, 2, & 3)
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
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 6 Honors ( Periods 1, 2, & 3)
Equations: Decimals and Fractions 8-4
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
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
Math 6 Honors ( Periods 1, 2, & 3)
Equations: All Four OP's 8-2 & 8-3
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 = 2Ï€r for r
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.
Remember when we stated the properties as well-- back in our 1st quarter!!
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 = 2Ï€r for r
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.
Remember when we stated the properties as well-- back in our 1st quarter!!
Algebra Honors (Period 6 & 7)
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
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
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