Algebra 2 Week #1A Review. It’s Friday! Week #1A – Section 4 Classwork – –He is decomposing –Buoy meets gull –Bushed Homework 1. x = 5 2. x = - 2 3.

Algebra 2

Week #1A Review

It’s Friday!

Week #1A – Section 4

• Classwork – – He is decomposing– Buoy meets gull– Bushed

Homework1. x = 5 2. x = - 23. x = - 1 4. x = 15. x = 4 6. x = 07. x = - 4 8. x = 7

Extra credit⅜a - ⅞b + ⅛a + ⅜b = ⅜a + ⅛a - ⅞b + ⅜b

= ½a - ½b

STANDARD STUDIED THIS WEEK:

• (leading to) CA STANDARD: 1 – To be able to solve equations and inequalities involving absolute value.

Vocabulary to Know

• coefficient: the number in front of a variable (like the 2 in 2x)• combining like terms: adding terms with the same variable and

exponent• distributive property: a(b + c) = ab + ac• equation: an algebraic expression with a = in it• integers: whole numbers (positive, negative, and zero)• like terms: terms with same variable and exponent• PEMDAS: order of operations• true/false statements: an equation that is true/not true

How to:A. Do arithmetic with integers (or positive and negative numbers)

1. Multiplication or division

Same signs? Answer is positive.

Different signs? Answer is negative.

2. Addition or subtraction

Same signs? ADD and keep the sign.

Different signs? SUBTRACT and keep the sign of thelarger number.

EXAMPLES:

- 20 ÷ - 4 =

EXAMPLES:

- 20 ÷ - 4 = 5

- 20 – 4 = - 24

How to:B. Use the order of operations (PEMDAS).

1. parentheses

2. exponents

3. multiplication/division

4. addition/subtraction

EXAMPLE: 5 – 3(2 + 1)2 + 16 =

1. parentheses

2. exponents

EXAMPLE: 5 – 3(2 + 1)2 + 16 = 5 – 3(3)2 + 16

1. parentheses

2. exponents

EXAMPLE: 5 – 3(2 + 1)2 + 16 = 5 – 3(3)2 + 16 = 5 – 3(9) + 16

1. parentheses

2. exponents

EXAMPLE: 5 – 3(2 + 1)2 + 16 = 5 – 3(3)2 + 16 = 5 – 3(9) + 16 = 5 – 27 + 16

1. parentheses

2. exponents

EXAMPLE: 5 – 3(2 + 1)2 + 16 = 5 – 3(3)2 + 16 = 5 – 3(9) + 16 = 5 – 27 + 16 = 21 – 27

1. parentheses

2. exponents

EXAMPLE: 5 – 3(2 + 1)2 + 16 = 5 – 3(3)2 + 16 = 5 – 3(9) + 16 = 5 – 27 + 16 = 21 – 27 = - 6

How to:C. Solve one variable equations.

1. Variable on each side of the =? Move smaller variable to the same

side of the = (don’t forget to change the sign) and add it to the

variable that’s already there.

2. Number on the same side of the = as the variable? Move it to the

other side of the = (don’t forget to change the sign) and add it with any number that’s already there.

3. Divide both sides if the variable has a coefficient.

EXAMPLES

2x – 5 = 10

EXAMPLES

2x – 5 = 10

2x = 15

EXAMPLES

2x – 5 = 10

2x = 15

x = 15/2

EXAMPLES

- 6x – 4 = 3x - 67

EXAMPLES

6x – 4 = 3x – 67

3x – 4 = - 67

EXAMPLES

6x – 4 = 3x – 67

3x – 4 = - 67

3x = - 63

EXAMPLES

6x – 4 = 3x – 67

3x – 4 = - 67

3x = - 63

x = - 21

How to:D. Get an equation ready to be solved (should be done FIRST).

1. If you see ( )s, multiply.

2.Combine any like terms on each side of the = sign. (Do NOT use opposite operations.)

3.Get rid of fractions. Multiply the whole equation by the number in the denominator. OR use the least common denominator.

EXAMPLES