Objective: Add fractions with sums greater than 2. Module 3 – Lesson 10 Objective: Add fractions with sums greater than 2.

Sprint 3 + 1 = 3 + 1 2 = 3 1 2 + 1 = 4 – 2 = 4 1 2 – 2 = 4 2 3 – 2 = 6 3 4 + 3 = 6 3 4 – 3 = 2 1 6 + 3 = 2 5 6 + 7 = 10 5 6 – 3= 10 5 6 – 7 = 7 4 5 – 2 3 5 = 13 7 9 – 7 5 9 = 6 3 4 – 6 = 7 + 2 5 + 2 + 2 5 = 6 2 3 – 1 2 3 = 3 4 7 – 4 7 =

Application Problem To make punch for the class party, Mrs. Lui mixed 1 1 3 cups orange juice, 3 4 cup apple juice, 2 3 cup cranberry juice, and 3 4 cup lemon-lime soda. Mixed together, how many cups of punch does the recipe make? (Bonus: Each student drinks 1 cup. How many recipes does Mrs. Lui need to serve her 20 students? 1 1 3 Cups Orange Juice 3 4 Cups Apple Juice 2 3 Cups Cranberry Juice 3 4 Cups Lemon-Lime Soda 1 1 3 + 3 4 + 2 3 + 3 4 The recipe makes 3 1 2 cups. 1 1 3 + 2 3 + 3 4 + 3 4 1 3 3 + 6 4 = 2 + 1 2 4 = 3 2 4 or 3 1 2

Application Problem - Bonus To make punch for the class party, Mrs. Lui mixed 1 1 3 cups orange juice, 3 4 cup apple juice, 2 3 cup cranberry juice, and 3 4 cup lemon-lime soda. Mixed together, how many cups of punch does the recipe make? (Bonus: Each student drinks 1 cup. How many recipes does Mrs. Lui need to serve her 20 students? 1 1 3 Cups Orange Juice 3 4 Cups Apple Juice 2 3 Cups Cranberry Juice 3 4 Cups Lemon-Lime Soda Bonus: Think about how many students 2 recipes will serve. 2 x 3 ½ cups = 7 cups. 2 recipes will serve 7 students. Take 20 divided by 7 = 2 with 6 students left. We would need to make an extra recipe to cover those students. Take 3 and 2 because 7 students is for 2 recipes. 3 x 2 = 6 recipes.

Concept Development – Problem 1 Look over the three problems and decide what is similar and different. A) 2 1 5 + 1 1 5 B) 2 1 5 + 1 1 2 C) 2 4 5 + 1 1 2 All three add whole number plus fractional units. All three have the whole numbers of 2 and 1 being added. B and C both have the same fractional units. A has the same fractional units being added where B and Ca re different. Both A and B will have a result between 3 and 4, but C will be between 4 and 5, because 4 5 is more than 1 2 .

Concept Development – Problem 1 How could we re-write the problem 2 1 5 + 1 1 2 ? Why? 2 1 5 + 1 1 2 = 2 + 1 5 + 1 + 1 2 = 2 + 1 + 1 5 + 1 2 = 3 + 1 5 + 1 2 = We can write it this way because of the commutative and associative properties and because we know a whole number and a fractional unit can be re-written as an addition problem. Can we add the problem the way it is? Why or Why not? No because the denominators are different, so we don’t have like terms. We need to find a common denominator and then calculate equivalent fractions before adding.

Concept Development – Problem 1 How can we find a common denominator? By multiplying the denominators together or listing the multiples of both denominators until we have one in common. What is the common denominator in the following problem? 2 1 5 + 1 1 2 10 (5 x 2 = 10) How do we find equivalent fractions? Multiply the numerator and denominator by the same number. What will we time 1 2 and 1 5 by? 1 2 by 5 5 and 1 5 by 2 2 What is the problem with the equivalent fractions with the common denominator? 2 2 10 + 1 5 10 What is our answer? 3 7 10

Concept Development – Problem 2 Solve using the standard algorithm and/or a number line. 2 4 5 + 1 1 2 What is the first thing we must do before we can add the two mixed numbers? Find a common denominator? What is the common denominator for the problem? 10 Now re-write the problem with the common denominator. 4 5 * 2 2 = 8 10 = 2 8 10 1 2 * 5 5 = 5 10 = 1 5 10 2 8 10 + 1 5 10 = 3 13 10 Can we stop at 3 13 10 ? Why or Why not? No because it is not in simplest form. What is the answer in simplest form? 13 10 = 1 3 10 . 3 + 1 3 10 = 4 3 10

Concept Development – Problem 3 Solve: 2 2 3 + 5 2 5 (Show all your steps.) 2 2 3 + 5 2 5 2 ( 2 3 * 5 5 ) + 5 ( 2 5 * 3 3 ) 2 10 15 + 5 6 15 7 16 15 = 7 15 15 + 1 15 7 + 1 + 1 15 = 8 1 15

Concept Development – Problem 4 Solve 3 5 7 + 6 2 3 (show all your work.) 3 5 7 + 6 2 3 3 ( 5 7 * 3 3 ) + 6 ( 2 3 * 7 7 ) 3 15 21 + 6 14 21 9 29 21 9 21 21 + 8 21 = 10 + 8 21 10 8 21

Concept Development – Problem 5 Solve 3 1 2 + 4 7 8 (show all your work.) 3 1/2 + 4 7 8 Multiples of 2 – 2, 4, 6, 8, 10 Multiples of 8 – 8, 16 Common denominator 8 3 ( 1 2 * 4 4 ) + 4 7 8 3 4 8 + 4 7 8 7 11 8 7 + 1 3 8 8 3 8 3 1 2 + 4 7 8 3 ( 1 2 * 8 8 ) + 4 ( 7 8 * 2 2 ) 3 8 16 + 4 14 16 7 22 16 7 + 1 6 16 or 7 16 16 + 6 16 8 6 16 8 ( 6 16 divided by 2 2 ) 8 3 8 Divided by 2 because that is a common factor of 6 and 16.

Concept Development – Problem 6 Solve 15 5 6 + 7 9 10 23 11 15

End of Lesson Activities Student Debrief Problem Set Exit Ticket Homework

Problem Set Add 2 1 4 + 1 1 5 B) 2 3 4 + 1 2 5 C) 1 1 5 + 2 1 3 D) 4 2 3 + 1 2 5 E) 3 1 3 + 4 5 7 F) 2 6 7 + 5 2 3 G) 15 1 5 + 3 5 8 H) 15 5 8 + 5 2 5 Erin jogged 2 1 4 miles on Monday. Wednesday she jogged 3 1 3 miles, and on Friday she jogged 2 2 3 miles. How far did Erin jog altogether. Darren bought some paint. He used 2 1 4 gallons painting his living room. After that, he had 3 5 6 gallons left. How much paint did he buy? Clayton says that 2 1 2 + 3 3 5 will be more than 5 but less than 6 since 2 + 3 is 5. Is Clayton’s reasoning correct? Prove him right or wrong.

Exit Ticket Solve the problems. 3 1 2 + 1 1 3 4 5 7 + 3 3 4

Homework Add A) 2 1 2 + 1 1 5 B) 2 1 2 + 1 3 5 C) 1 1 5 + 3 1 3 D) 3 2 3 + 1 3 5 E) 2 1 3 + 4 4 7 F) 3 5 7 + 4 2 3 G) 15 1 5 + 4 3 8 H) 18 3 8 + 2 2 5 Angela practiced piano for 2 1 2 hours on Friday, 2 1 3 hours on Saturday, and 3 2 3 hours on Sunday. How much time did Angela practice during the weekend? String A is 3 5 6 meters long. String B is 2 1 4 long. What’s the total length of both strings? Matt says that 5 – 1 1 4 will be more than 4, since 5-1 is 4. Draw a picture to prove that Matt is wrong.