1. Course 2 1-9 Simplifying Algebraic Expressions 1-9 Simplifying Algebraic Expressions Course 2 Warm Up Warm Up Problem of the Day Problem of the Day Lesson Presentation Lesson Presentation

2. Course 2 1-9 Simplifying Algebraic Expressions Warm Up Evaluate each expression for y = 3. 1. 3y + y 2. 7y 3. 10y – 4y 4. 9y 5. y + 5y + 6y 6. 10y 12 21 18 27 36 30

3. Course 2 1-9 Simplifying Algebraic Expressions Problem of the Day Emilia saved nickels, dimes, and quarters in a jar. She had as many quarters as dimes, but twice as many nickels as dimes. If the jar had 844 coins, how much money had she saved? $94.95

4. Course 2 1-9 Simplifying Algebraic Expressions Learn to simplify algebraic expressions.

5. Course 2 1-9 Simplifying Algebraic Expressions Vocabulary term coefficient

6. Course 2 1-9 Simplifying Algebraic Expressions In the expression 7x + 9y + 15, 7x, 9y, and 15 are called terms. A term can be a number, a variable, or a product of numbers and variables. Terms in an expression are separated by + and –. 7x + 5 – 3y 2 + y + x3x3 term In the term 7x, 7 is called the coefficient. A coefficient is a number that is multiplied by a variable in an algebraic expression. A variable by itself, like y, has a coefficient of 1. So y = 1y. Coefficient Variable term

7. Course 2 1-9 Simplifying Algebraic Expressions Like terms are terms with the same variable raised to the same power. The coefficients do not have to be the same. Constants, like 5,, and 3.2, are also like terms. 1212 Like Terms Unlike Terms 3x and 2x 5x 2 and 2x The exponents are different. 3.2 and n Only one term contains a variable 6a and 6b The variables are different w and w7w7 5 and 1.8

8. Course 2 1-9 Simplifying Algebraic Expressions Identify like terms in the list. Additional Example 1: Identifying Like Terms 3t 5w 2 7t 9v 4w 2 8v Look for like variables with like powers. 3t 5w 2 7t 9v 4w 2 8v Like terms: 3t and 7t 5w 2 and 4w 2 9v and 8v Use different shapes or colors to indicate sets of like terms. Helpful Hint

9. Course 2 1-9 Simplifying Algebraic Expressions Check It Out: Example 1 Identify like terms in the list. 2x 4y 3 8x 5z 5y 3 8z Look for like variables with like powers. Like terms: 2x and 8x 4y 3 and 5y 3 5z and 8z 2x 4y 3 8x 5z 5y 3 8z

10. Course 2 1-9 Simplifying Algebraic Expressions x Combining like terms is like grouping similar objects. += x x x xx x xx xxxx x xxxx 4x4x+5x5x =9x9x To combine like terms that have variables, add or subtract the coefficients.

11. Course 2 1-9 Simplifying Algebraic Expressions Simplify. Justify your steps using the Commutative, Associative, and Distributive Properties when necessary. Additional Example 2: Simplifying Algebraic Expressions A. 6t – 4t 6t – 4t 2t2t 6t and 4t are like terms. Subtract the coefficients. B. 45x – 37y + 87 In this expression, there are no like terms to combine.

12. Course 2 1-9 Simplifying Algebraic Expressions Additional Example 2: Simplifying Algebraic Expressions C. 3a 2 + 5b + 11b 2 – 4b + 2a 2 – 6 3a 2 + 5b + 11b 2 – 4b + 2a 2 – 6 5a 2 + b + 11b 2 – 6 Identify like terms. Add or subtract the coefficients. (3a 2 + 2a 2 ) + (5b – 4b) + 11b 2 – 6 Group like terms. Simplify. Justify your steps using the Commutative, Associative, and Distributive Properties when necessary.

13. Course 2 1-9 Simplifying Algebraic Expressions Check It Out: Example 2 Simplify. Justify your steps using the Commutative, Associative, and Distributive Properties when necessary. A. 5y + 3y 5y + 3y 8y8y 5y and 3y are like terms. Add the coefficients. B. 2(x 2 – 13x) + 6 There are no like terms to combine. 2x – 26x + 6 2 Distributive Property.

14. Course 2 1-9 Simplifying Algebraic Expressions Check It Out: Example 2 C. 4x 2 + 4y + 3x 2 – 4y + 2x 2 + 5 9x 2 + 5 Identify like terms. Add or subtract the coefficients. 4x 2 + 4y + 3x 2 – 4y + 2x 2 + 5 Group like terms. (4x 2 + 3x 2 + 2x 2 )+ (4y – 4y) + 5 Simplify. Justify your steps using the Commutative, Associative, and Distributive Properties when necessary.

15. Course 2 1-9 Simplifying Algebraic Expressions Write an expression for the perimeter of the triangle. Then simplify the expression. Additional Example 3: Geometry Application x 2x + 33x + 2 2x + 3 + 3x + 2 + x (x + 3x + 2x) + (2 + 3) 6x + 5 Write an expression using the side lengths. Identify and group like terms. Add the coefficients.

16. Course 2 1-9 Simplifying Algebraic Expressions Check It Out: Example 3 x 2x + 1 x + 2x + 1 + 2x + 1 5x + 2 Write an expression using the side lengths. Identify and group like terms. Add the coefficients. Write an expression for the perimeter of the triangle. Then simplify the expression. (x + 2x + 2x) + (1 + 1)

17. Course 2 1-9 Simplifying Algebraic Expressions Lesson Quiz: Part I Identify like terms in the list. 1. 3n 2 5n 2n 3 8n 2. a 5 2a 2 a 3 3a 4a 2 Simplify. Justify your steps using the Commutative, Associative, and Distributive Properties when necessary. 3. 4a + 3b + 2a 4. x 2 + 2y + 8x 2 2a 2, 4a 2 5n, 8n 6a + 3b 9x 2 + 2y

18. Course 2 1-9 Simplifying Algebraic Expressions Lesson Quiz: Part II 5. Write an expression for the perimeter of the given figure. 6x + 8y 2x + 3y x + y

19. Course 2 1-9 Simplifying Algebraic Expressions 1-10 Equations and Their Solutions Course 2 Warm Up Warm Up Problem of the Day Problem of the Day Lesson Presentation Lesson Presentation

20. Course 2 1-9 Simplifying Algebraic Expressions Warm Up Evaluate each expression for x = 12. 1. x + 2 2. 3. x – 8 4. 10x – 4 5. 2x + 12 6. 5x + 7 14 3 4 Course 2 1-10 Equations and Their Solutions x4x4 116 36 67

21. Course 2 1-9 Simplifying Algebraic Expressions Problem of the Day Alicia buys buttons at a cost of 8 for $20. She resells them for $5 each. How many buttons does Alicia need to sell for a profit of $120? 48 buttons Course 2 1-10 Equations and Their Solutions

22. Course 2 1-9 Simplifying Algebraic Expressions Learn to determine whether a number is a solution of an equation. Course 2 1-10 Equations and Their Solutions

23. Course 2 1-9 Simplifying Algebraic Expressions Vocabulary equation solution Insert Lesson Title Here Course 2 1-10 Equations and Their Solutions

24. Course 2 1-9 Simplifying Algebraic Expressions Ella has 22 CDs. This is 9 more than her friend Kay has. This situation can be written as an equation. An equation is a mathematical statement that two expressions are equal in value. An equation is like a balanced scale. Right expressionLeft expression Number of CDs Ella has 22 is equal to = 9 more than Kay has j + 9 Course 2 1-10 Equations and Their Solutions

25. Course 2 1-9 Simplifying Algebraic Expressions Just as the weights on both sides of a balanced scale are exactly the same, the expressions on both sides of an equation represent exactly the same value. When an equation contains a variable, a value of the variable that makes the statement true is called a solution of the equation. 22 = j + 9 j = 13 is a solution because 22 = 13 + 9. 22 = j + 9 j = 15 is not a solution because 22  15 + 9. The symbol ≠ means “is not equal to.” Reading Math Course 2 1-10 Equations and Their Solutions

26. Course 2 1-9 Simplifying Algebraic Expressions Determine whether the given value of the variable is a solution of t + 9 = 17. Additional Example 1A: Determining Whether a Number is a Solution of an Equation 26 26 + 9 = 17 ? 35 = 17 ? 26 is not a solution of t + 9 = 17.  Substitute 26 for t. t + 9 = 17 Course 2 1-10 Equations and Their Solutions

27. Course 2 1-9 Simplifying Algebraic Expressions Additional Example 1B: Determining Whether a Number is a Solution of an Equation Determine whether the given value of the variable is a solution of t + 9 = 17. 8 8 + 9 = 17 ? 17 = 17 ? 8 is a solution of t + 9 = 17. Substitute 8 for t. t + 9 = 17 Course 2 1-10 Equations and Their Solutions

28. Course 2 1-9 Simplifying Algebraic Expressions Check It Out: Example 1 Insert Lesson Title Here Determine whether each number is a solution of x – 5 = 12. A. 22 22 – 5 = 12 ? 17 = 12 ? 22 is not a solution of x – 5 = 12.  Substitute 22 for x. B. 8 8 – 5 = 12 ? 3 = 12 ? 8 is not a solution of x – 5 = 12. Substitute 8 for x. x – 5 = 12  Course 2 1-10 Equations and Their Solutions

29. Course 2 1-9 Simplifying Algebraic Expressions Mrs. Jenkins had $32 when she returned home from the supermarket. If she spent $17 at the supermarket, did she have $52 or $49 before she went shopping? Additional Example 2: Writing an Equation to Determine Whether a Number is a Solution $52 m – 17 = 32 52 - 17 = 32 ? 35 = 32 ?  Substitute 52 for m. Course 2 1-10 Equations and Their Solutions You can write an equation to find the amount of money Mrs. Jenkins had before she went shopping. If m represents the amount of money she had before she went shopping, then m - 17 = 32.

30. Course 2 1-9 Simplifying Algebraic Expressions Additional Example 2 Continued $49 m – 17 = 32 49 - 17 = 32 ? 32 = 32 ? Substitute 49 for m. Course 2 1-10 Equations and Their Solutions You can write an equation to find the amount of money Mrs. Jenkins had before she went shopping. If m represents the amount of money she had before she went shopping, then m - 17 = 32. Mrs. Jenkins had $49 before she went shopping. Mrs. Jenkins had $32 when she returned home from the supermarket. If she spent $17 at the supermarket, did she have $52 or $49 before she went shopping?

31. Course 2 1-9 Simplifying Algebraic Expressions Mr. Rorke had $12 when he returned home from buying a hat. If he spent $47 at the hat store, did he have $61 or $59 before he bought the hat? Check it Out: Additional Example 2 $61 m – 47 = 12 61 - 47 = 12 ? 14 = 12 ? Substitute 61 for h. Course 2 1-10 Equations and Their Solutions You can write an equation to find the amount of money Mr. Rorke had before he purchased a hat. If m represents the amount of money he had before he purchased a hat, then m – 47 = 12. 

32. Course 2 1-9 Simplifying Algebraic Expressions Mr. Rorke had $12 when he returned home from buying a hat. If he spent $47 at the hat store, did he have $59 or $61 before he bought the hat? Check it Out: Additional Example 2 Continued $59 m – 47 = 12 59 - 47 = 12 ? 12 = 12 ? Substitute 59 for h. Course 2 1-10 Equations and Their Solutions You can write an equation to find the amount of money Mr. Rorke had before he purchased a hat. If m represents the amount of money he had before he purchased a hat, then m – 47 = 12. Mr. Rorke had $59 before he purchased a hat.

33. Course 2 1-9 Simplifying Algebraic Expressions Which problem situation best matches the equation 5 + 2x = 13? Additional Example 3: Deriving a Real-World Situation from an Equation Situation A: Admission to the county fair costs $5 and rides cost $2 each. Mike spent a total of $13. How many rides did he go on? Course 2 1-10 Equations and Their Solutions $2 per ride 2x Mike spent $13 in all, so 5 + 2x = 13. Situation A matches the equation. $5 for admission 5 +

34. Course 2 1-9 Simplifying Algebraic Expressions Which problem situation best matches the equation 5 + 2x = 13? Additional Example 3 Continued Situation B: Course 2 1-10 Equations and Their Solutions Admission to the county fair costs $2 and rides cost $5 each. Mike spent a total of $13. How many rides did he go on? $5 per ride 5x Since 5x is not a term in the given equation, Situation B does not match the equation. The variable x represents the number of rides that Mike bought.

35. Course 2 1-9 Simplifying Algebraic Expressions Which problem situation best matches the equation 13 + 4x = 25? Check It Out: Additional Example 3 Situation A: Course 2 1-10 Equations and Their Solutions Admission to the baseball game costs $4 and souvenir hats cost $13 each. Trina spent a total of $25. How many souvenir hats did she buy? $13 per souvenir hat 13x Since 13x is not a term in the given equation, Situation A does not match the equation. The variable x represents the number of souvenir hats Trina bought.

36. Course 2 1-9 Simplifying Algebraic Expressions Which problem situation best matches the equation 13 + 4x = 25? Check It Out: Additional Example 3 Continued Situation B: Course 2 1-10 Equations and Their Solutions $4 per souvenir hat 4x Trina spent $25 in all, so 13 + 4x = 25. Situation B matches the equation. $13 for admission 13 + Admission to the baseball game costs $13 and souvenir hats cost $4 each. Trina spent a total of $25. How many souvenir hats did she buy?

37. Course 2 1-9 Simplifying Algebraic Expressions Lesson Quiz Determine whether the given value of the variable is a solution of 5 + x = 47. 1. x = 42 2. x = 52 Determine whether the given value of the variable is a solution of 57 – y = 18. 3. y = 75 4. y = 39 5. Kwan has 14 marbles. This is 7 more than Drue has. Does Drue have 21 or 7 marbles? no yes Insert Lesson Title Here no yes 7 Course 2 1-10 Equations and Their Solutions

38. Course 2 1-9 Simplifying Algebraic Expressions 2-6 Rates, Ratios, and Proportions Holt Algebra 1 Lesson Quiz Lesson Quiz Lesson Presentation Lesson Presentation Warm Up Warm Up

39. Course 2 1-9 Simplifying Algebraic Expressions Warm Up Solve each equation. Check your answer. 1. 6x = 36 2. 3. 5m = 18 4. 5. 8y =18.4 Multiply. 6.7. 6 48 3.6 –63 2.3 710

40. Course 2 1-9 Simplifying Algebraic Expressions Write and use ratios, rates, and unit rates. Write and solve proportions. Objectives

41. Course 2 1-9 Simplifying Algebraic Expressions ratio proportion rate cross products scale scale drawing unit rate scale model conversion factor Vocabulary

42. Course 2 1-9 Simplifying Algebraic Expressions A ratio is a comparison of two quantities by division. The ratio of a to b can be written a:b or, where b ≠ 0. Ratios that name the same comparison are said to be equivalent. A statement that two ratios are equivalent, such as, is called a proportion.

43. Course 2 1-9 Simplifying Algebraic Expressions Example 2: Finding Unit Rates Raulf Laue of Germany flipped a pancake 416 times in 120 seconds to set the world record. Find the unit rate. Round your answer to the nearest hundredth. The unit rate is about 3.47 flips/s. Write a proportion to find an equivalent ratio with a second quantity of 1. Divide on the left side to find x.

44. Course 2 1-9 Simplifying Algebraic Expressions Check It Out! Example 2 Cory earns $52.50 in 7 hours. Find the unit rate. The unit rate is $7.50. Write a proportion to find an equivalent ratio with a second quantity of 1. Divide on the left side to find x.

45. Course 2 1-9 Simplifying Algebraic Expressions A rate such as in which the two quantities are equal but use different units, is called a conversion factor. To convert a rate from one set of units to another, multiply by a conversion factor.

46. Course 2 1-9 Simplifying Algebraic Expressions Example 3A: Converting Rates Serena ran a race at a rate of 10 kilometers per hour. What was her speed in kilometers per minute? Round your answer to the nearest hundredth. The rate is about 0.17 kilometer per minute. To convert the second quantity in a rate, multiply by a conversion factor with that unit in the first quantity.

47. Course 2 1-9 Simplifying Algebraic Expressions Helpful Hint In example 3A, “ 1 hr ” appears to divide out, leaving “ kilometers per minute, ” which are the units asked for. Use this strategy of “ dividing out ” units when converting rates.

48. Course 2 1-9 Simplifying Algebraic Expressions Example 3B: Converting Rates A cheetah can run at a rate of 60 miles per hour in short bursts. What is this speed in feet per minute? Step 1 Convert the speed to feet per hour. The speed is 316,800 feet per hour. To convert the first quantity in a rate, multiply by a conversion factor with that unit in the second quantity. Step 2 Convert the speed to feet per minute. The speed is 5280 feet per minute. To convert the first quantity in a rate, multiply by a conversion factor with that unit in the first quantity.

49. Course 2 1-9 Simplifying Algebraic Expressions Example 3B: Converting Rates The speed is 5280 feet per minute. Check that the answer is reasonable. There are 60 min in 1 h, so 5280 ft/min is 60(5280) = 316,800 ft/h. There are 5280 ft in 1 mi, so 316,800 ft/h is This is the given rate in the problem.

50. Course 2 1-9 Simplifying Algebraic Expressions To convert the first quantity in a rate, multiply by a conversion factor with that unit in the second quantity. Step 1 Convert the speed to feet per hour. Check It Out! Example 3 A cyclist travels 56 miles in 4 hours. What is the cyclist ’ s speed in feet per second? Round your answer to the nearest tenth, and show that your answer is reasonable. Change to miles in 1 hour. The speed is 73,920 feet per hour.

51. Course 2 1-9 Simplifying Algebraic Expressions Check It Out! Example 3 Step 2 Convert the speed to feet per minute. To convert the second quantity in a rate, multiply by a conversion factor with that unit in the first quantity. The speed is 1232 feet per minute. Step 3 Convert the speed to feet per second. The speed is approximately 20.5 feet per second. To convert the second quantity in a rate, multiply by a conversion factor with that unit in the first quantity.

52. Course 2 1-9 Simplifying Algebraic Expressions Check It Out! Example 3 Check that the answer is reasonable. The answer is about 20 feet per second. There are 60 seconds in a minute so 60(20) = 1200 feet in a minute. There are 60 minutes in an hour so 60(1200) = 72,000 feet in an hour. Since there are 5,280 feet in a mile 72,000 ÷ 5,280 = about 14 miles in an hour. The cyclist rode for 4 hours so 4(14) = about 56 miles which is the original distance traveled.

53. Course 2 1-9 Simplifying Algebraic Expressions In the proportion, the products a d and b c are called cross products. You can solve a proportion for a missing value by using the Cross Products property. Cross Products Property WORDSNUMBERS ALGEBRA In a proportion, cross products are equal. 2 6 = 3 4 If and b ≠ 0 and d ≠ 0 then ad = bc.

54. Course 2 1-9 Simplifying Algebraic Expressions Example 4: Solving Proportions Solve each proportion. 3(m) = 5(9) 3m = 45 m = 15 Use cross products. Divide both sides by 3. Use cross products. 6(7) = 2(y – 3) 42 = 2y – 6 +6 48 = 2y 24 = y A.B. Add 6 to both sides. Divide both sides by 2.

55. Course 2 1-9 Simplifying Algebraic Expressions Check It Out! Example 4 Solve each proportion. y = −20 –12 4g = 23 g = 5.75 A.B. Use cross products. Divide both sides by 2. Use cross products. Subtract 12 from both sides. Divide both sides by 4. 2y = –40 2(y) = –5(8) 4(g +3) = 5(7) 4g +12 = 35

56. Course 2 1-9 Simplifying Algebraic Expressions A scale is a ratio between two sets of measurements, such as 1 in:5 mi. A scale drawing or scale model uses a scale to represent an object as smaller or larger than the actual object. A map is an example of a scale drawing.

57. Course 2 1-9 Simplifying Algebraic Expressions Example 5A: Scale Drawings and Scale Models A contractor has a blueprint for a house drawn to the scale 1 in: 3 ft. A wall on the blueprint is 6.5 inches long. How long is the actual wall? blueprint 1 in. actual 3 ft. x 1 = 3(6.5) x = 19.5 The actual length of the wall is 19.5 feet. Write the scale as a fraction. Let x be the actual length. Use the cross products to solve.

58. Course 2 1-9 Simplifying Algebraic Expressions Example 5B: Scale Drawings and Scale Models A contractor has a blueprint for a house drawn to the scale 1 in: 3 ft. One wall of the house will be 12 feet long when it is built. How long is the wall on the blueprint? blueprint 1 in. actual 3 ft. The wall on the blueprint is 4 inches long. Write the scale as a fraction. Let x be the actual length. Use the cross products to solve. 12 = 3x 4 = x Since x is multiplied by 3, divide both sides by 3 to undo the multiplication.

59. Course 2 1-9 Simplifying Algebraic Expressions Check It Out! Example 5 A scale model of a human heart is 16 ft. long. The scale is 32:1. How many inches long is the actual heart it represents? model 32 in. actual 1 in. The actual heart is 6 inches long. Write the scale as a fraction. Use the cross products to solve. 32x = 192 Since x is multiplied by 32, divide both sides by 32 to undo the multiplication. Let x be the actual length. Convert 16 ft to inches. x = 6

60. Course 2 1-9 Simplifying Algebraic Expressions Lesson Quiz: Part 1 1. In a school, the ratio of boys to girls is 4:3. There are 216 boys. How many girls are there? 162 Find each unit rate. Round to the nearest hundredth if necessary. 2. Nuts cost $10.75 for 3 pounds.$3.58/lb 3. Sue washes 25 cars in 5 hours. 5 cars/h 4. A car travels 180 miles in 4 hours. What is the car ’ s speed in feet per minute? 3960 ft/min

61. Course 2 1-9 Simplifying Algebraic Expressions Lesson Quiz: Part 2 Solve each proportion. 5. 6. 7. A scale model of a car is 9 inches long. The scale is 1:18. How many inches long is the car it represents? 6 16 162 in.

62. Course 2 1-9 Simplifying Algebraic Expressions SMUE DIRECTIONS For Smue question 18,19,20,and21, 18,write the response and tell how you got your answer 19,write the answer in expand if you has a answer in numberic form 20,21 compare and contrast the number

63. Course 2 1-9 Simplifying Algebraic Expressions Question 18