Warm Up Problem of the Day Lesson Presentation Lesson Quizzes.

Slides:



Advertisements
Similar presentations
Do Now Solve. 1. x – 17 = y + 11 = = x = x – 9 = 20 x = 49 y = 30 w 5 w = 90 x = 9 x = 29 Hwk: p29 odd, Test this Friday.
Advertisements

Preview Warm Up California Standards Lesson Presentation.
REVIEW Inverse Operations
Solve one step equations. You can add the same amount to both sides of an equation and the statement will remain true = = 9 x = y +
Warm Up  – Evaluate.  (0.29)
Lesson 1.1 Objective: To solve equations using addition, subtraction, multiplication, and division Are operations that undo each other such as addition.
01/05/11 Fraction and Decimal Equations Today’s Plan: -Fraction and Decimal Equations -Assignment Learning Target: -I can solve fraction and decimal equations.
3-5 Solving Equations Containing Decimals Warm Up Warm Up Lesson Presentation Lesson Presentation Problem of the Day Problem of the Day Lesson Quizzes.
AF1.1 Write and solve one-step linear equations in one variable.
1-8 Solving Equations by Adding or Subtracting Warm-up: Fluency Quiz. Sit with paper blank side up. Homework: Page 41 #4-8 even, #12-16 even, even,
Warm Up Solve. 1. x – 17 = y + 11 = = x = 108
Holt CA Course Solving Equations Containing Decimals AF1.1 Write and solve one-step linear equations in one variable. California Standards.
Today’s Plan: -Solving Equations w/ decimals -Assignment 11/29/10 Solving Equations with Decimals Learning Target: -I can solve equations containing decimals.
2-7 Solving Equations with Rational Numbers Learn to solve equations with rational numbers.
One step equations Add Subtract Multiply Divide  When we solve an equation, our goal is to isolate our variable by using our inverse operations.  What.
Algebra 1 Chapter 2 Section : Solving One-Step Equations An equation is a mathematical statement that two expressions are equal. A solution of an.
Warm Up Solve. 1. 3x = = z – 100 = w = 98.6 x = 34 y = 225 z = 121 w = 19.5 y 15.
3-4 Solving Equations Containing Decimals Do Now Solve. 1. x – 17 = 322. y + 11 = = x = x – 9 = 20 x = 49y = 30 w = 90 x = 9 x = 29.
Solve one step equations. You can add the same amount to both sides of an equation and the statement will remain true = = 9 x = y +
PRE-ALGEBRA. Lesson 2-6 Warm-Up PRE-ALGEBRA What is the “Division Property of Equality”? Rule: Division Property of Equality: If you divide both sides.
Warm Up Solve. 1. x + 5 = 9 2. x – 34 = 72 = x – 39 x = 4 x = 106
2.2 Solving Two- Step Equations. Solving Two Steps Equations 1. Use the Addition or Subtraction Property of Equality to get the term with a variable on.
* Collect the like terms 1. 2a = 2a x -2x + 9 = 6x z – – 5z = 2z - 6.
Solving One Step Equations subtract 3 Adding or subtracting the same number from each side of an equation produces an equivalent equation. Addition.
Warm Up Simplify each expression. 1.10c + c 2. 5m + 2(2m – 7) 11c 9m – 14.
Solving Two-Step and Multi-Step Equations Warm Up Lesson Presentation
Lesson 7.4 Solving Multiplication and Division Equations 2/3/10.
4-6 Solving Equations Containing Decimals Course 2 Warm Up Warm Up Problem of the Day Problem of the Day Lesson Presentation Lesson Presentation.
Solving Absolute-Value Equations
2-7 Warm Up Problem of the Day Lesson Presentation
Solve. Use an inverse operation! 1. x – 17 = y + 11 = = 18
Solving One-Step Equations
Notes 7.1 Day 1– Solving Two-Step Equations
Warm Up Problem of the Day Lesson Presentation Lesson Quizzes.
SOLVING EQUATIONS, INEQUALITIES, AND ALGEBRAIC PROPORTIONS
Warm Up Find the GCF of each set of numbers and , 45 and 30
 .
Bell Ringer.
Warm Up Problem of the Day Lesson Presentation Lesson Quizzes.
Warm up 11/1/ X ÷ 40.
Solving 1-Step Integer Equations
Objective Solve equations in one variable that contain more than one operation.
Lesson 2.1 How do you use properties of addition and multiplication?
Solving Two-Step Equations Lesson 2-2 Learning goal.
Objective Solve one-step equations in one variable by using multiplication or division.
Simplifying Algebraic Expressions
Algebra /19-20/16 EQ: How do I solve Multi-Step Equations?
Objective Solve equations in one variable that contain more than one operation.
Objective Solve inequalities that contain variable terms on both sides.
Subtract the same value on each side
Warm Up Determine if the given numbers are solutions to the given equations. 1. x = 2 for 4x = 9 2. x = 5 for 8x + 2 = x = 4 for 3(x – 2) = 10 no.
Solve the following problems.
Objective Solve one-step equations in one variable by using multiplication or division.
Objective Solve equations in one variable that contain more than one operation.
3-6 Warm Up Problem of the Day Lesson Presentation
Bell Ringer.
Solving Absolute-Value Equations
Objective Solve equations in one variable that contain more than one operation.
Warm Up Problem of the Day Lesson Presentation Lesson Quizzes.
Warm Up Problem of the Day Lesson Presentation Lesson Quizzes.
Solving Absolute-Value Equations
Learning Objective Students will be able to: Solve equations in one variable that contain absolute-value expressions.
Lesson 1.1 Objective: To solve equations using addition, subtraction, multiplication, and division Vocab: Inverse operations: Are operations that undo.
Do Now Solve. 1. x – 17 = y + 11 = = x = -108
Unit 2B/3A Solving Equations
Solving basic equations
Solving Equations by 2-1 Adding or Subtracting Warm Up
Warm Up Problem of the Day Lesson Presentation Lesson Quizzes.
Warm Up Problem of the Day Lesson Presentation Lesson Quizzes.
Warm Up Problem of the Day Lesson Presentation Lesson Quizzes.
Presentation transcript:

Warm Up Problem of the Day Lesson Presentation Lesson Quizzes

Warm Up Solve. 1. x – 17 = 32 2. y + 11 = 41 x = 49 3. = 18 3. = 18 4. 12x = 108 5. x – 9 = 20 x = 49 y = 30 w 5 w = 90 x = 9 x = 29

Problem of the Day A bowling league has 156 players evenly distributed on 39 teams. If 40 players switch to another league and the number of teams is reduced to 29, will there be the same number of players on each team? yes; 4

Learn to solve one-step equations that contain decimals.

Students in a physical education class were running 40-yard dashes as part of a fitness test. The slowest time in the class was 3.84 seconds slower than the fastest time of 7.2 seconds. You can write an equation to represent this situation. The slowest time s minus 3.84 is equal to the fastest time of 7.2 seconds. s – 3.84 = 7.2

Remember! You can solve an equation by performing the same operation on both sides of the equation to isolate the variable.

Additional Example 1: Solving Equations by Adding and Subtracting Solve. A. n – 2.75 = 8.3 Use the Addition Property of Equality. Add 2.75 to both sides. n – 2.75 = 8.30 + 2.75 + 2.75 n = 11.05 B. a + 32.66 = 42 a + 32.66 = 42.00 Use the Subtraction Property of Equality. Subtract 32.66 from both sides. –32.66 –32.66 a = 9.34

Check It Out: Example 1 Solve. A. n – 1.46 = 4.7 Use the Addition Property of Equality. Add 1.46 to both sides. n – 1.46 = 4.70 + 1.46 + 1.46 n = 6.16 B. a + 27.51 = 36 Use the Subtraction Property of Equality. Subtract 27.51 form both sides. a + 27.51 = 36.00 –27.51 –27.51 a = 8.49

Additional Example 2A: Solving Equations by Multiplying and Dividing Solve. x = 5.4 4.8 x = 5.4 4.8 Use the Multiplication Property of Equality. Multiply by 4.8 on both sides. x · 4.8 = 5.4 · 4.8 4.8 x = 25.92

Additional Example 2B: Solving Equations by Multiplying and Dividing Solve. 9 = 3.6d 9 = 3.6d Use the Division Property of Equality. Divide by 3.6 on both sides. 9 3.6d = 3.6 3.6 9 = d Think: 9 ÷ 3.6 = 90 ÷ 36 3.6 2.5 = d

· · 3.5 Check It Out: Example 2A Solve. x = 2.4 3.5 x = 2.4 3.5 Use the Multiplication Property of Equality. Multiply by 3.5 on both sides. x · 3.5 = 2.4 · 3.5 3.5 x = 8.4

Check It Out: Example 2B Solve. 9 = 2.5d 9 = 2.5d Use the Division Property of Equality. Divide by 2.5 on both sides. 9 2.5d = 2.5 2.5 9 = d Think: 9 ÷ 2.5 = 90 ÷ 25 2.5 3.6 = d

Understand the Problem Additional Example 3: Problem Solving Application A board-game box is 2.5 inches tall. A toy store has shelving measuring 15 inches vertically in which to store the boxes. How many boxes can be stacked in the space? 1 Understand the Problem Rewrite the question as a statement. Find the number of boxes that can be placed on the shelf. List the important information: • Each board-game box is 2.5 inches tall. • The store has shelving space measuring 15 inches.

Additional Example 3 Continued 2 Make a Plan The total height of the boxes is equal to the height of one box times the number of boxes. Since you know how tall the shelf is you can write an equation with b being the number of boxes. 2.5b = 15

Additional Example 3 Continued Solve 3 2.5b = 15 2.5b = 15 Use the Division Property of Equality. 2.5 2.5 b = 6 Six boxes can be stacked in the space.

Additional Example 3 Continued 4 Look Back You can round 2.5 to 3 and estimate how many boxes will fit on the shelf. 15 ÷ 3 = 5 So 6 boxes is a reasonable answer.

Understand the Problem Check It Out: Example 3 A canned good is 4.5 inches tall. A grocery store has shelving measuring 18 inches vertically in which to store the cans. How many cans can be stacked in the space? 1 Understand the Problem Rewrite the question as a statement. Find the number of cans that can be placed on the shelf. List the important information: • Each can is 4.5 inches tall. • The store has shelving space measuring 18 inches.

Check It Out: Example 3 Continued 2 Make a Plan The total height of the cans is equal to the height of one can times the number of cans. Since you know how tall the shelf is you can write an equation with c being the number of cans. 4.5c = 18

Check It Out: Example 3 Continued Solve 3 4.5c = 18 4.5c = 18 Use the Division Property of Equality. 4.5 4.5 c = 4 Four cans can be stacked in the space.

Check It Out: Example 3 Continued 4 Look Back You can round 4.5 to 5 and 18 to 20 estimate how many cans will fit on the shelf. 20 ÷ 5 = 4 So 4 cans is a reasonable answer.

Lesson Quizzes Standard Lesson Quiz Lesson Quiz for Student Response Systems

Lesson Quiz Solve. 1. x – 14.23 = 19.5 2. 12.6c = –103.32 3. 4. m + 12.97 =–14.35 5. The French Club is selling coupon books for $8.25 each. How many books must be sold to bring in $5,940? x = 33.73 c = –8.2 x = 6.1 x = 56.73 9.3 m = –27.32 720 books

Lesson Quiz for Student Response Systems 1. Solve. m + 13.55 = –15.45 A. m = –28 B. m = –29 C. m = –30 D. m = –31

Lesson Quiz for Student Response Systems 2. Solve. x – 16 = 18.7 A. x = –35.13 B. x = –2.27 C. x = 2.27 D. x = 35.13

Lesson Quiz for Student Response Systems 3. Solve. 14.8a = –102.12 A. a = 7.3 B. a = 6.9 C. a = –6.9 D. a = –7.3

Lesson Quiz for Student Response Systems 4. Solve. — = 7.2 A. x = 56.13 B. x = 61.92 C. x = 63.02 D. x = 64 x 8.6

Lesson Quiz for Student Response Systems 5. A fashion show ticket costs $7.75 per person. How many fashion show tickets must be sold to generate $4,805? A. 600 tickets B. 620 tickets C. 640 tickets D. 660 tickets