Distributive Property 2.2 LESSON DO NOW: IF YOU WERE ASKED TO DISTRIBUTE MATERIALS IN CLASS, EXPLAIN WHAT YOU THINK YOUR JOB MIGHT REQUIRE YOU TO DO?

Distributive Property: Multiply the number outside the parentheses with all terms (numbers and variables) inside the parentheses. Simplify like terms if possible. Distributive Property 2.2 LESSON

Use the distributive property to evaluate or write an equivalent expression. 3. –2(5 + 12) = –2(5) + (–2)(12) Distributive property = –10 + (–24) = –34 Multiply. Add. Distributive Property 2.2 LESSON

Use the distributive property to evaluate or write an equivalent expression. 4. –4(–7 – 10) = –4(–7) – (–4)(–10) Distributive property = = 68 Multiply. Add. Distributive Property 2.2 LESSON

a. –5(x + 10) = –5x + (–5)(10) Distributive property = –5x + (–50) = –5x – 50 Multiply. Simplify. b. 3[1 – 20 + (–5)] = 3(1) – 3(20) + 3(–5) = 3 – 60 + (–15) = 3 + (–60) + (–15) = –72 Distributive property Multiply. Add the opposite of 60. Add. Distributive Property 2.2 LESSON

Use the distributive property to evaluate or write an equivalent expression. 6. –8(z + 25) = –8z + (–8)(25) Distributive property Multiply. Simplify. = –8z – 200 = –8z +(–200) Distributive Property 2.2 LESSON

Use the distributive property to evaluate or write an equivalent expression. 5. 2(w – 8) = 2w – (2)(8) Distributive property = 2w – (16) Multiply. Simplify. = 2w – 16 Distributive Property 2.2 LESSON

Distributive Property 2.2 LESSON CLOSING: WRITE AN EQUIVALENT VARIABLE EXPRESSION FOR 4(10+X)

Distributive Property 2.2 LESSON DO NOW: THINK OF EXAMPLES OF CONSTANT AND VARIABLE (CHANGING) NUMBERS IN EVERYDAY LIFE.

VOCABULARYDEFINITION TERMTHE PARTS OF AN EXPRESSION CONSTANTSA NUMBER WITHOUT A VARIABLE COEFFICIENTSTHE NUMBER IN FRONT OF THE VARIABLE Simplifying Variable Expression 2.3 LESSON *When a variable has no number in front, there is always and imaginary 1*

IDENTIFYING TERMS –Split the expression before each operation TERMS: 16, -8k, 9k, -8 Simplifying Variable Expression 2.3 LESSON

IDENTIFYING CONSTANTS –Split the expression before each operation –FIND THE NUMBERS WITHOUT VARIABLE CONSTANTS: 16, -8 Simplifying Variable Expression 2.3 LESSON

IDENTIFYING COEFFICIENTS –Split the expression before each operation –FIND THE NUMBERS WITH A VARIABLE *COEFFICIENT IS THE NUMBER ONLY* COEFFICIENTS: -8, 9 Simplifying Variable Expression 2.3 LESSON

Simplifying Variable Expression 2.3 LESSON CLOSING: EXPLAIN WHAT YOU THINK ARE EXAMPLES OF LIKE TERMS IN MATH.

Simplifying Variable Expression 2.3 LESSON DO NOW: SIMPLIFY 3x x - 5

14t t – 6 Split expression into terms Like terms - terms that have the same variable or no variable at all. Combine like terms by adding them Like terms: 14t + -2t t 9 Combined Like Terms: 12t + 9 Simplifying Variable Expression 2.3 LESSON

1) 6 – 2m – 7 + 3m2) 10 – 4y + 5y – 8 3) -p p – 74) x + 4 – 2x – 10 Simplifying Variable Expression 2.3 LESSON

1) Distribute – multiply by the number in front of the parentheses 2) Split – draw a line before each operation 3) Combine Like Terms – Find like terms then add Simplifying Variable Expression 2.3 LESSON

Example: 6 ( 2 – j ) + 2j – 5 12 – 6j + 2j – 5 -4j +7 Why can you not combine the expression any further?

1)12v - 3 ( 2 + 4v ) + 3 2) 2 ( 4s – 7 ) + 3s 3) 3 ( n – p ) + 2n – 4p Simplifying Variable Expression 2.3 LESSON

Simplifying Variable Expression 2.3 LESSON CLOSING: EXPLAIN THE PROCEDURE FOR SIMPLIFYING AN EXPRESSION WITH SEVERAL TERMS.

VARIABLES AND EQUATIONS 2.3 LESSON DO NOW: SOLVE: 6x + 8 = -2x + 72

Writing Verbal Sentences as Equations Verbal SentenceEquation The sum of x and 6 is 9. x + 6 = 9 The difference of 12 and y is – y = 15 The product of –4 and p is 32. –4p = 32 The quotient of n and 2 is 9. = 9 n2n2 EXAMPLE 1 Variables and Equations 2.4 LESSON

Checking Possible Solutions EXAMPLE 2 Tell whether 9 or 7 is a solution of x – 5 = 2. Substitute 9 for x. x – 5 = 2 4 ≠ 2 9 – 5 = 2 ? Substitute 7 for x. x – 5 = 2 2 = 2 7 is a solution. 7 – 5 = 2 ? ANSWER 9 is not a solution. ANSWER Variables and Equations 2.4 LESSON

VARIABLES AND EQUATIONS 2.4 LESSON CLOSING: DEFINE THE DIFFERENCE BETWEEN AND EQUATION AND AN EXPRESSION. HOW IS SOLVING EQUATIONS DIFFERENT FROM SIMPLIFYING EXPRESSIONS?

Solving Equations Using Addition and Subtraction 2.5 LESSON DO NOW: A BOX CONTAINING BOOKS WEIGHS 3 POUNDS, THE BOX WEIGHS 1 POUND ALONE. HOW MUCH DO THE BOOKS WEIGH ALONE?

Solving Equations Using Addition and Subtraction 2.5 LESSON Inverse Operations OperationInverse Operation Isolate a variable by using inverse operations which "undo" operations on the variable. An equation is like a balanced scale. To keep the balance, perform the same operation on both sides. AdditionSubtraction Addition

Solving Equations Using Addition and Subtraction 2.5 LESSON Solve the equation. Check your answer. Since 8 is subtracted from y, add 8 to both sides to undo the subtraction. y – 8 = y = 32 Check y – 8 = 24 To check your solution, substitute 32 for y in the original equation. 32 –

Solving Equations Using Addition and Subtraction 2.5 LESSON Solve the equation. Check your answer. Since 6 is subtracted from k, add 6 to both sides to undo the subtraction. –6 = k – = k Check –6 = k – 6 –6 0 – 6 –6 To check your solution, substitute 0 for k in the original equation.

Solve the equation. Check your answer. Since 9 is subtracted from m, add 9 to both sides to undo the subtraction. 16 = m – = m Check 16 = m – – 9 16 To check your solution, substitute 25 for m in the original equation. Solving Equations Using Addition and Subtraction 2.5 LESSON

Solve the equation. Check your answer. Since 17 is added to m, subtract 17 from both sides to undo the addition. m + 17 = 33 – 17 –17 m = 16 Check m + 17 = To check your solution, substitute 16 for m in the original equation. Solving Equations Using Addition and Subtraction 2.5 LESSON

Solving Equations Using Addition and Subtraction 2.5 LESSON Solve the equation. Check your answer. Since 5 is added to k, subtract 5 from both sides to undo the subtraction. –5 = k + 5 – 5 – 5 –10 = k Check –5 = k + 5 –5 – –5 To check your solution, substitute –10 for k in the original equation.

Solve the equation. Check your answer. Since 6 is added to t, subtract 6 from both sides to undo the addition. 6 + t = 14 – 6 t = 8 Check 6 + t = To check your solution, substitute 8 for t in the original equation. Solving Equations Using Addition and Subtraction 2.5 LESSON

Solve –11 + x = 33. Check your answer. Since –11 is added to x, add 11 to both sides. –11 + x = x = 44 Check –11 + x = 33 – To check your solution, substitute 44 for x in the original equation. Solving Equations Using Addition and Subtraction 2.5 LESSON

Solving Equations Using Addition and Subtraction 2.5 LESSON Solve each equation. 1. r – 4 = –8 2. m + 13 = –5 + c = This year a high school had 578 sophomores enrolled. This is 89 less than the number enrolled last year. Write and solve an equation to find the number of sophomores enrolled last year.

Solving Equations Using Multiplication and Division 2.6 LESSON Inverse Operations OperationInverse Operation MultiplicationDivision Multiplication Solving an equation that contains multiplication or division is similar to solving an equation that contains addition or subtraction. Use inverse operations to undo the operations on the variable.

Solving Equations Using Multiplication and Division 2.6 LESSON Solve the equation. Since j is divided by 3, multiply both sides by 3 to undo the division. –24 = j –8 –8 = j 3 –8 –24 3 Check –8 = j 3 To check your solution, substitute –24 for j in the original equation.

Solve the equation. Check your answer. Since p is divided by 5, multiply both sides by 5 to undo the division. p = To check your solution, substitute 50 for p in the original equation. = 10 p Check = 10 p 5 Solving Equations Using Multiplication and Division 2.6 LESSON

Solving Equations Using Multiplication and Division 2.6 LESSON Solve the equation. Check your answer. Since y is divided by 3, multiply both sides by 3 to undo the division. –39 = y To check your solution, substitute –39 for y in the original equation. –13 = y 3 –13 –39 3 y Check –13 = 3

Solve the equation. Check your answer. Since c is divided by 8, multiply both sides by 8 to undo the division. c = 56 To check your solution, substitute 56 for c in the original equation. = 7 c 8 Check = 7 c 8 Solving Equations Using Multiplication and Division 2.6 LESSON

Solving Equations Using Multiplication and Division 2.6 LESSON Solve the equation. Check your answer. Since y is multiplied by 9, divide both sides by 9 to undo the multiplication. y = To check your solution, substitute 12 for y in the original equation. 9y = 108 9(12) 108 Check 9y = 108

Solve the equation. Check your answer. Since c is multiplied by 4, divide both sides by 4 to undo the multiplication. 4 = c 16 To check your solution, substitute 4 for c in the original equation. 16 = 4c 16 4(4) Check 16 = 4c Solving Equations Using Multiplication and Division 2.6 LESSON

Solving Equations Using Multiplication and Division 2.6 LESSON Solve the equation. Check your answer. Since k is multiplied by 15, divide both sides by 15 to undo the multiplication. k = 5 75 To check your solution, substitute 5 for k in the original equation. 15k = 75 15(5) 75 Check 15k = 75

Solving Equations Using Multiplication and Division 2.6 LESSON CLOSING: EXPLAIN THE ANALOGY, AN EQUATIONS IS LIKE A SCALE.

Solving Equations Involving Decimals 2.7 LESSON DO NOW: EXPLAIN THE RULES FOR ADDING, SUBTRACTING, MULTIPLYING, AND DIVIDING DECIMALS.

Solving Equations Involving Decimals 2.7 LESSON Solve the equation. Check your answer. Since 3.2 is subtracted from n, add 3.2 to both sides to undo the subtraction. n – 3.2 = n = 8.8 Check n – 3.2 = 5.6 To check your solution, substitute 8.8 for n in the original equation. 8.8 –

Solving Equations Involving Decimals 2.7 LESSON Solve the equation. Check your answer. Since 1.8 is added to t, subtract 1.8 from both sides to undo the addition. 4.2 = t –1.8 – 1.8 Check 4.2 = t To check your solution, substitute 2.4 for t in the original equation.

Solving Equations Involving Decimals 2.7 LESSON Solve –2.3 + m = 7. Check your answer. Since –2.3 is added to m, add 2.3 to both sides. –2.3 + m = m = 9.3 Check –2.3 + m = 7 – To check your solution, substitute 9.3 for m in the original equation.

Solving Equations Involving Decimals 2.7 LESSON Solve the equation. Since n is divided by 6, multiply both sides by 6 to undo the division. n = To check your solution, substitute 16.8 for n in the original equation. = 2.8 n Check = 2.8 n 6

Solving Equations Involving Decimals 2.7 LESSON Solve the equation. Check your answer. Since v is multiplied by –6, divide both sides by –6 to undo the multiplication. 0.8 = v –4.8 To check your solution, substitute 0.8 for v in the original equation. –4.8 = –6v –4.8 –6(0.8) Check –4.8 = –6v

Solve the equation. Check your answer. Since y is multiplied by 0.5, divide both sides by 0.5 to undo the multiplication. y = –20 –10 To check your solution, substitute –20 for y in the original equation. 0.5y = –10 0.5(–20) –10 Check 0.5y = –10 Solving Equations Involving Decimals 2.7 LESSON

Solving Equations Involving decimals 2.7 LESSON CLOSING: = n

Properties of Addition and Multiplication 2.1 LESSON DO NOW: BRAINSTORM WORDS THAT BEGIN WITH COMMUT OR ASSOC

Properties of Addition and Multiplication 2.1 LESSON Commutative Property means changing the order in which you add or multiply numbers does not change the sum or product. Associative Property means changing the grouping of numbers when adding or multiplying does not change their sum or product. Grouping symbols are typically parentheses (),but can include brackets [] or Braces {}.

Properties of Addition and Multiplication 2.1 LESSON Commutative Property of addition - (Order) Commutative Property of multiplication - (order) For any numbers a and b, a + b = b + a. For any numbers a and b, a  b = b  a. 6  8 = 8  6 48 = 48

Properties of Addition and Multiplication 2.1 LESSON Associative Property of addition - (grouping symbols) Associative Property of multiplication - (grouping symbols) For any numbers a, b, and c, (a + b) + c = a + (b + c). For any numbers a, b, and c, (a + b) + c = a + (b + c). For any numbers a, b, and c, (ab) c = a (bc). For any numbers a, b, and c, (ab) c = a (bc). (2 + 4) + 5 = 2 + (4 + 5) (2  3)  5 = 2  (3  5) (6) + 5 = 2 + (9) 11 = 11 (6)  5 = 2  (15) 30 = 30

Properties of Addition and Multiplication 2.1 LESSON Any number multiplied by 1 will give you the original number. 23,487 X 1 = 23,487 Any number added by 0 will give you the original number. 23, = 23,487

Properties of Addition and Multiplication 2.1 LESSON CLOSING: EXPLAIN WHY YOU THINK THE IDENTITY PROPERTIES WERE GIVEN THAT NAME. HINT LOOK AT THE PROBLEM AND THE ANSWER.