{"id":88196,"date":"2021-04-10T08:27:40","date_gmt":"2021-04-10T02:57:40","guid":{"rendered":"https:\/\/ccssmathanswers.com\/?p=88196"},"modified":"2022-04-01T10:25:43","modified_gmt":"2022-04-01T04:55:43","slug":"eureka-math-grade-6-module-4-lesson-27","status":"publish","type":"post","link":"https:\/\/bigideasmathanswers.com\/eureka-math-grade-6-module-4-lesson-27\/","title":{"rendered":"Eureka Math Grade 6 Module 4 Lesson 27 Answer Key"},"content":{"rendered":"

Engage NY Eureka Math Grade 6 Module 4 Lesson 27 Answer Key<\/h2>\n

Eureka Math Grade 6 Module 4 Lesson 27 Example Answer Key<\/h3>\n

Example 1:<\/p>\n

Solve 3z = 9 using tape diagrams and algebraically. Then, check your answer. First, draw two tape diagrams, one to represent each side of the equation.
\nAnswer:<\/p>\n

\"Eureka<\/p>\n

If 9 had to be split into three groups, how big would each group be?
\nAnswer:
\n3<\/p>\n

Demonstrate the value of z using tape diagrams.
\nAnswer:
\n\"Eureka<\/p>\n

How can we demonstrate this algebraically?
\nAnswer:
\nWe know we have to split 9 into three equal groups, so we have to divide by 3 to show this algebraically.
\n3z \u00f7 3 = 9 \u00f7 3<\/p>\n

How does this get us the value of z?
\nAnswer:
\nThe left side of the equation will equal z because we know the identity property, where a . b \u00f7 b = a, so we con use this identity here.<\/p>\n

The right side of the equation will be 3 because 9 \u00f7 3 = 3.
\nTherefore, the value of z is 3.<\/p>\n

How can we check our answer?
\nAnswer:
\nWe can substitute the value of z into the original equation to see if the number sentence is true.
\n3(3) = 9; 9 = 9. This number sentence is true, so our answer is correct.<\/p>\n

Example 2:<\/p>\n

Solve \\(\\frac{y}{4}\\) = 2 using tape diagrams and algebraically. Then, check your answer. First, draw two tape diagrams, one to represent each side of the equation.
\nAnswer:
\n\"Eureka<\/p>\n

If the first tape diagram shows the size of y \u00f7 4, how can we draw a tape diagram to represent y?
\nAnswer:
\nThe tape diagram to represent y should be four sections of the size y \u00f7 4.<\/p>\n

Draw this tape diagram.
\nAnswer:
\n\"Eureka<\/p>\n

What value does each y \u00f7 4 section represent? How do you know?
\nAnswer:
\nEach y \u00f7 4 section represents a value of 2. We know this from our original tape diagram.<\/p>\n

How can you use a tape diagram to show the value of y?
\nAnswer:
\nDraw four equal sections of 2, which will give y the value of 8.
\n\"Eureka<\/p>\n

How can we demonstrate this algebraically?
\nAnswer:
\n\\(\\frac{y}{4}\\) . 4 = 2 . 4. Because we multiplied the number of sections in the original equation by 4, we know the identity
\n\\(\\frac{a}{b}\\) . b = a can be used here.<\/p>\n

How does this help us find the value of y?
\nAnswer:
\nThe left side of the equation will equal y, and the right side will equal 8. Therefore, the value of y is 8.<\/p>\n

How can we check our answer?
\nAnswer:
\nSubstitute 8 into the equation for y, and then check to see if the number sentence is true.
\n\\(\\frac{8}{4}\\) = 2. This is a true number sentence, so 8 is the correct answer.<\/p>\n

Eureka Math Grade 6 Module 4 Lesson 27 Exercise Answer Key<\/h3>\n

Exercises:<\/p>\n

Exercise 1.
\nUse tape diagrams to solve the following problem: 3m = 21.
\nAnswer:
\n\"Eureka
\ncheck: 3(7) = 21; 21 = 21. This number sentence is true, so 7 is the correct solution.<\/p>\n

Exercise 2.
\nSolve the following problem algebraically: 15 = \\(\\frac{n}{5}\\)
\nAnswer:
\n15 = \\(\\frac{n}{5}\\)
\n15 . 5 = \\(\\frac{n}{5}\\) . 5
\n75 = n
\nCheck: 15 = \\(\\frac{75}{5}\\); 15 = 15. This number sentence is true, so 75 is the correct solution.<\/p>\n

Exercise 3.
\nCalculate the solution of the equation using the method of your choice: 4p = 36.
\nAnswer:
\nTape Diagrams:
\n\"Eureka<\/p>\n

Algebraically:
\n4p = 36
\n4p \u00f7 3 = 36 \u00f7 4
\np = 9
\nCheck:
\n4(9) = 36; 36 = 36. This number sentence is true, so 9 is the correct solution.<\/p>\n

Exercise 4.
\nExamine the tape diagram below, and write an equation it represents. Then, calculate the solution to the equation using the method of your choice.<\/p>\n

\"Eureka<\/p>\n

Answer:<\/p>\n

\"Eureka<\/p>\n

Algebraically:
\n7q = 70
\n7q \u00f7 7 = 70 \u00f7 7
\nq = 10<\/p>\n

70 = 7q
\n70 \u00f7 7 = 7q \u00f7 7
\nq = 10<\/p>\n

Check:
\n7(10) = 70, 70 = 7(10); 70 = 70. This number sentence is true, so 10 is the correct answer.<\/p>\n

Exercise 5.
\nWrite a multiplication equation that has a solution of 12. Use tape diagrams to prove that your equation has a solution of 12.
\nAnswer:
\nAnswers will vary.<\/p>\n

Exercise 6.
\nWrite a division equation that has a solution of 12. Prove that your equation has a solution of 12 using algebraic methods.
\nAnswer:
\nAnswers will vary.<\/p>\n

Eureka Math Grade 6 Module 4 Lesson 27 Problem Set Answer Key<\/h3>\n

Question 1.
\nUse tape diagrams to calculate the solution of 30 = 5w. Then, check your answer.
\nAnswer:
\n\"Eureka<\/p>\n

Check: 30 = 5(6); 30 = 30. This number sentence is true, so 6is the correct solution.<\/p>\n

Question 2.
\nSolve 12 = \\(\\frac{x}{4}\\) algebraically. Then, check your answer.
\nAnswer:
\n12 = \\(\\frac{x}{4}\\)
\n12 . 4 = \\(\\frac{x}{4}\\) . 4
\n48 = x
\nCheck:
\n12 = \\(\\frac{48}{4}\\); 12 = 12. This number sentence is true, so 48 is the correct solution.<\/p>\n

Question 3.
\nUse tape diagrams to calculate the solution of \\(\\frac{y}{5}\\) = 15. Then, check your answer.
\nAnswer:
\n\"Eureka<\/p>\n

Check:
\n\\(\\frac{75}{5}\\) = 15; 15 = 15. This number sentence is true, so 75 is the correct solution.<\/p>\n

Question 4.
\nSolve 18z = 72 algebraically. Then, check your answer.
\nAnswer:
\n18z = 72
\n18z \u00f7 18 = 72 \u00f7 18
\nz = 4
\nCheck:
\n18(4) = 72; 72 = 72. This number sentence is true, so 4 is the correct solution.<\/p>\n

Question 5.
\nWrite a division equation that has a solution of 8. Prove that your solution is correct by using tape diagrams.
\nAnswer:
\nAnswers will vary.<\/p>\n

Question 6.
\nWrite a multiplication equation that has a solution of 8. Solve the equation algebraically to prove that your solution is correct.
\nAnswer:
\nAnswers will vary.<\/p>\n

Question 7.
\nWhen solving equations algebraically, Meghan and Meredith each got a different solution. Who is correct? Why did the other person not get the correct answer?
\n\"Eureka
\nAnswer:
\nMeghan is correct. Meredith divided by 2 to solve the equation, which is not correct because she would end up with \\(\\frac{y}{4}\\) = 2. To solve a division equation, Meredith must multiply by 2 to end up with y because the identity states.
\ny \u00f7 2 . 2 = y.<\/p>\n

Eureka Math Grade 6 Module 4 Lesson 27 Exit Ticket Answer Key<\/h3>\n

Calculate the solution to each equation below using the indicated method. Remember to check your answers.<\/p>\n

Question 1.
\nUse tape diagrams to find the solution of \\(\\frac{r}{10}\\) = 4.
\nAnswer:
\n\"Eureka<\/p>\n

Check:
\n\\(\\frac{40}{10}\\) = 4; 4 = 4. This number sentence is true, so 40 is the correct solution.<\/p>\n

Question 2.
\nFind the solution of 64 = 16u algebraically.
\nAnswer:
\n64 = 16u
\n64 \u00f7 16 = 16u \u00f7 16
\n4 = u
\nCheck:
\n64 = 16(4); 64 = 64. This number sentence is true, so 4 is the correct solution.<\/p>\n

Question 3.
\nUse the method of your choice to find the solution of 12 = 3v.
\nAnswer:
\nTape Diagrams:
\n\"Eureka<\/p>\n

Algebraically:
\n12 = 3v
\n12 \u00f7 3 = 3v \u00f7 3
\n4 = v<\/p>\n

Check:
\n12 = 3(4); 12 = 12. This number sentence is true, so 4 is the correct solution.<\/p>\n","protected":false},"excerpt":{"rendered":"

Engage NY Eureka Math Grade 6 Module 4 Lesson 27 Answer Key Eureka Math Grade 6 Module 4 Lesson 27 Example Answer Key Example 1: Solve 3z = 9 using tape diagrams and algebraically. Then, check your answer. First, draw two tape diagrams, one to represent each side of the equation. Answer: If 9 had … Read more<\/a><\/p>\n","protected":false},"author":3,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":""},"categories":[6],"tags":[],"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/bigideasmathanswers.com\/wp-json\/wp\/v2\/posts\/88196"}],"collection":[{"href":"https:\/\/bigideasmathanswers.com\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/bigideasmathanswers.com\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/bigideasmathanswers.com\/wp-json\/wp\/v2\/users\/3"}],"replies":[{"embeddable":true,"href":"https:\/\/bigideasmathanswers.com\/wp-json\/wp\/v2\/comments?post=88196"}],"version-history":[{"count":1,"href":"https:\/\/bigideasmathanswers.com\/wp-json\/wp\/v2\/posts\/88196\/revisions"}],"predecessor-version":[{"id":91323,"href":"https:\/\/bigideasmathanswers.com\/wp-json\/wp\/v2\/posts\/88196\/revisions\/91323"}],"wp:attachment":[{"href":"https:\/\/bigideasmathanswers.com\/wp-json\/wp\/v2\/media?parent=88196"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/bigideasmathanswers.com\/wp-json\/wp\/v2\/categories?post=88196"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/bigideasmathanswers.com\/wp-json\/wp\/v2\/tags?post=88196"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}