Seating ChartTaylor T Farhan A Cameron B

Justin T Farhan A Barrett S Esther R Cassandra R Renee P

Jarrod P Nelson P Tanner P Riley N April M Nate M

Abigail L Justine H Danyell G Sean D Austin D Cameron B

Savannah B Shauna B Tyler A Abdikani A

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(Front)

Period 2 Blue Algebra II

Seating Chart

Ty M Erika T Elijah A Zahida S Norah S Kaitlyn R

Zacchaeus E Griffin M Sydney M Muna M Lauren H Angel G

Stone C Hannah C Brooklyn B Krysta A

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Period 3 Blue Alg II Honors

Seating Chart

Halie S Kaylyn S Shamsi A Megan U Abigail S

Eric H Abdallah O Chris M Victoria M Casey M Ben H

Jamie B Sean F Joe F Cody D Kelly C Rebekah C

Erica B Ashley B Sean B Kathryn A

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Period 2 White Algebra II

Seating Chart

Cierra St. Greg A Patrick W Emily T Elizabeth T Drew St.

Megan L Lukas S Logan P Amal M Christian L Logan L

Isaiah L Ahmed H Jhovert D Emily C

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Period 3 White Algebra II

Welcome to Algebra IIWhile I take attendance and hand out some papers please try the following problems. Just try your best, if you are unsure do what you think you should do. Please show all steps you think are important.

1)Simplify 2x – 5 + 3x + 10

2)Simplify 3x – 2 – (2x – 7)

3)Solve 2x – 5 = 11

4)Solve 5x + 7 = 3x - 5

I. Please put your calculator number in the space provided on your class guide.

II. Continue to work on the 4 problems. If you have completed the 4 problems below look over the information in the class guide

1)Simplify 2x – 5 + 3x + 10

2)Simplify 3x – 2 – (2x – 7)

3)Solve 2x – 5 = 11

4)Solve 5x + 7 = 3x - 5

Lets see how you did.

5x + 5

2. Simplify 3x – 2 – (2x – 7)

3x – 2 – 2x + 7

x + 5

3. Solve 2x – 5 = 11

2x – 5 = 11 +5 +5

2x = 16

2x = 16 ÷2 ÷2

x = 8

3. Solve 2x – 5 = 11

2x = 16

x = 8

4. Solve 5x + 7 = 3x - 5

5x + 7 = 3x – 5 -3x -3x

2x + 7 = - 5 -7 -7

2x = - 12 ÷2 ÷2

x = - 6

4. Solve 5x + 7 = 3x - 5

2x + 7 = - 5

2x = - 12

x = - 6

Start of School Business

• Look at class guide a little.

• Make sure inside book cover is filled in.

• Make sure you wrote down your calculator number in your class guide.

• Look at student information sheet a little.

• Any questions so far?

Notebooks and taking notes.

• You need a ring binder.• You need to put down each days date.• You need to put down a topic title for all notes.• Directions need to be written down.• At least the first example needs to be written

down.• Lets practice.

Solving and Check Equations.

• 2x – 5y = -15 if y= 7

2x - 5(7) = -15

2x = 20x = 10

check2(10) - 5(7) =

-1520 - 35 = -

15-15 = -15

Solving and Check Equations.

• 2(x – 4) = -(x + 3)

2x - 8 = -x - 3

3x = 5

x = 5/3

check2(5/3 - 4) = -

(5/3+3)10/3 - 8 = -5/3 - 310 - 24 = -5 - 9-14 = -14

Simplify 10 + 3x – 15 – 5x

-2x – 5

Notice the answer is in descending order

Home Work

• Fill out the student information sheet, have a parent sign it and return it to me next class.

• Review unit 1 work sheet 1 and 2.

• Unit 1 work sheet 1.

Two important Math facts

• What do the following fractions equal?

€

01

=

€

10

=

€

0

€

Undefined€

zero dividedbyanynumber isalways zero.

€

Anynumber divided by zero is always undefined.

The Most Important Geometry Idea…

• Tells us what about this diagram?

€

a2+b2 =c2

€

pythagorean theorem

John has $5 more than Mike. Together they have a total of $27. How much

money does John have?

• For this problem you may solve it in any way you want. Please write down whatever steps you do so that I can follow your thinking.

John has $5 more than Mike. Together they have a total of $27. How much

money does John have?

• No idea? Guess an amount for Mike, calculate John’s amount, and add.

• If you guessed right…great!!!• If not, adjust your guess until it does work.

John has $5 more than Mike. Together they have a total of $27. How much

money does John have?

• Now lets try an algebra approach.• Instead of guessing an amount for Mike, let

Mike’s amount be x.• So John’s amount would be ….x+5• Now add them together to get 27 and solve.

WHY ALGEBRA?

• It gives us an easier, direct route to solve many problems.

Mr. Mancine has twice as many dimes as nickels. Together his coins are worth a total of $2.50. How many

dimes does Mr. Mancine have?

• For this problem you may solve it in any way you want. Please write down whatever steps you do so that I can follow your thinking.

Mr. Mancine has twice as many dimes as nickels. Together his coins are worth a total of $2.50. How many

dimes does Mr. Mancine have?

• Tell us how you solved this problem.

• What would the algebra method look like?

WHY ALGEBRA?

1. It gives us an easier, direct route to solve many problems.

2. It makes working with decimals or fractions easier.