{"id":79092,"date":"2022-06-10T09:07:15","date_gmt":"2022-06-10T03:37:15","guid":{"rendered":"https:\/\/ccssanswers.com\/?p=79092"},"modified":"2022-05-10T14:14:39","modified_gmt":"2022-05-10T08:44:39","slug":"eureka-math-grade-7-module-3-lesson-6","status":"publish","type":"post","link":"https:\/\/ccssanswers.com\/eureka-math-grade-7-module-3-lesson-6\/","title":{"rendered":"Eureka Math Grade 7 Module 3 Lesson 6 Answer Key"},"content":{"rendered":"

Engage NY Eureka Math 7th Grade Module 3 Lesson 6 Answer Key<\/h2>\n

Eureka Math Grade 7 Module 3 Lesson 6 Example Answer Key<\/h3>\n

Example 1.
\nRewrite the expression in standard form by collecting like terms.
\n\\(\\frac{2}{3}\\) n-\\(\\frac{3}{4}\\) n+\\(\\frac{1}{6}\\) n+2 \\(\\frac{2}{9}\\) n
\nAnswer:
\n\\(\\frac{2}{3}\\) n-\\(\\frac{3}{4}\\) n+\\(\\frac{1}{6}\\) n+2 \\(\\frac{2}{9}\\) n
\n\\(\\frac{24}{36}\\) n-\\(\\frac{27}{36}\\) n+\\(\\frac{6}{36}\\) n+2 \\(\\frac{8}{36}\\) n
\n2 \\(\\frac{11}{36}\\) n<\/p>\n

\u2192 What are various strategies for adding, subtracting, multiplying, and dividing rational numbers?
\n\u2192 Find common denominators; change from mixed numbers and whole numbers to improper fractions, and then convert back.<\/p>\n

Example 2.
\nAt a store, a shirt was marked down in price by $10.00. A pair of pants doubled in price. Following these changes, the price of every item in the store was cut in half. Write two different expressions that represent the new cost of the items, using s for the cost of each shirt and p for the cost of a pair of pants. Explain the different information each one shows.
\nAnswer:
\nFor the cost of a shirt:
\n\\(\\frac{1}{2}\\) (s-10) The cost of each shirt is \\(\\frac{1}{2}\\) of the quantity of the original cost of the shirt, minus 10.
\n\\(\\frac{1}{2}\\) s-5 The cost of each shirt is half off the original price, minus 5, since half of 10 is 5.
\nFor the cost of a pair of pants:
\n\\(\\frac{1}{2}\\) (2p) The cost of each pair of pants is half off double the price.
\np The cost of each pair of pants is the original cost because \\(\\frac{1}{2}\\) is the multiplicative inverse of 2.<\/p>\n

\u2192 Describe a situation in which either of the two expressions in each case would be more useful.
\n\u2192 Answers may vary. For example, p would be more useful than \\(\\frac{1}{2}\\)(2p) because it is converted back to an isolated variable, in this case the original cost.<\/p>\n

Example 3.
\nWrite this expression in standard form by collecting like terms. Justify each step.
\n5 \\(\\frac{1}{3}\\)-(3 \\(\\frac{1}{3}\\))(\\(\\frac{1}{2}\\) x-\\(\\frac{1}{4}\\))
\nAnswer:
\n\\(\\frac{16}{3}\\)+(-\\(\\frac{10}{3}\\))(\\(\\frac{1}{2}\\) x)+(-\\(\\frac{10}{3}\\))(-\\(\\frac{1}{4}\\)) Write mixed numbers as improper fractions, then distribute.
\n\\(\\frac{16}{3}\\)+(-\\(\\frac{5}{3}\\) x)+\\(\\frac{5}{6}\\) Any grouping (associative) and arithmetic rules for multiplying rational numbers
\n–\\(\\frac{5}{3}\\) x+(\\(\\frac{36}{6}\\)+\\(\\frac{5}{6}\\)) Commutative property and associative property of addition, collect like terms
\n–\\(\\frac{5}{3}\\) x+\\(\\frac{37}{6}\\) Apply arithmetic rule for adding rational numbers<\/p>\n

\u2192 A student says he created an equivalent expression by first finding this difference: 5 \\(\\frac{1}{3}\\)-3 \\(\\frac{1}{3}\\) . Is he correct? Why or why not?
\n\u2192 Although they do appear to be like terms, taking the difference would be incorrect. In the expression (3 \\(\\frac{1}{3}\\))(\\(\\frac{1}{2}\\) x-\\(\\frac{1}{4}\\)), 3 \\(\\frac{1}{3}\\) must be distributed before applying any other operation in this problem.<\/p>\n

\u2192 How should 3 \\(\\frac{1}{3}\\) be written before being distributed?
\n\u2192 The mixed number can be rewritten as an improper fraction \\(\\frac{10}{3}\\). It is not necessary to convert the mixed number, but it makes the process more efficient and increases the likelihood of getting a correct answer.<\/p>\n

Example 4.
\nModel how to write the expression in standard form using rules of rational numbers.
\n\\(\\frac{x}{20}\\)+\\(\\frac{2x}{5}\\)+\\(\\frac{(x+1)}{2}\\)+\\(\\frac{(3x-1)}{10}\\)<\/p>\n

\u2192 What are other equivalent expressions of \\(\\frac{x}{20}\\)? How do you know?
\n\u2192 Other expressions include 1\\(\\frac{x}{20}\\) and \\(\\frac{1}{2}\\)0 x because of the arithmetic rules of rational numbers.
\n\u2192 What about \\(\\frac{1}{20x}\\)? How do you know?
\n\u2192 It is not equivalent because if x=2, the value of the expression is \\(\\frac{1}{40}\\), which does not equal \\(\\frac{1}{10}\\).
\n\u2192 How can the distributive property be used in this problem?
\n\u2192 For example, it can be used to factor out \\(\\frac{1}{20}\\) from each term of the expression.
\n\u2192 Or, for example, it can be used to distribute \\(\\frac{1}{10}\\): \\(\\frac{3x-1}{10}\\)=\\(\\frac{1}{10}\\) (3x-1) = \\(\\frac{3x}{10}\\)–\\(\\frac{1}{10}\\).
\nBelow are two solutions. Explore both with the class.<\/p>\n

\\(\\frac{x}{20}\\)+\\(\\frac{4(2x)}{20}\\)+\\(\\frac{10(x+1)}{20}\\)+\\(\\frac{2(3x-1)}{20}\\)
\n\\(\\frac{x+8 x+10 x+10+6 x-2}{20}\\)
\n\\(\\frac{25 x+8}{20}\\)
\n\\(\\frac{5}{4}\\)x + \\(\\frac{2}{5}\\)<\/p>\n

\\(\\frac{1}{20}\\)x+\\(\\frac{2}{5}\\) x+\\(\\frac{1}{2}\\) x+\\(\\frac{1}{2}\\)+\\(\\frac{3}{10}\\) x-\\(\\frac{1}{10}\\)
\n(\\(\\frac{1}{20}\\)+\\(\\frac{2}{5}\\)+\\(\\frac{1}{2}\\)+\\(\\frac{3}{10}\\))x+(\\(\\frac{1}{2}\\)–\\(\\frac{1}{10}\\))
\n(\\(\\frac{1}{20}\\)+\\(\\frac{8}{20}\\)+\\(\\frac{10}{20}\\)+\\(\\frac{6}{20}\\))x+(\\(\\frac{5}{10}\\)–\\(\\frac{1}{10}\\))
\n\\(\\frac{5}{4}\\) x+\\(\\frac{2}{5}\\)<\/p>\n

Ask students to evaluate the original expression and the answers when x=20 to see if they get the same number.<\/p>\n

Evaluate the original expression and the answers when x=20. Do you get the same number?
\n\\(\\frac{x}{20}\\)+\\(\\frac{2x}{5}\\)+\\(\\frac{x+1}{2}\\)+\\(\\frac{3x-1}{10}\\)
\n\\(\\frac{20}{20}\\)+\\(\\frac{2(20)}{5}\\)+\\(\\frac{20+1}{2}\\)+\\(\\frac{3(20)-1}{10}\\)
\n1+8+\\(\\frac{21}{2}\\)+\\(\\frac{59}{10}\\)
\n9+\\(\\frac{105}{10}\\)+\\(\\frac{59}{10}\\)
\n9+\\(\\frac{164}{10}\\)
\n9+\\(\\frac{164}{10}\\)
\n25 \\(\\frac{2}{5}\\)
\n\\(\\frac{5}{4}\\) x+\\(\\frac{2}{5}\\)
\n\\(\\frac{5}{4}\\) (20)+\\(\\frac{2}{5}\\)
\n25+\\(\\frac{2}{5}\\)
\n25 \\(\\frac{2}{5}\\)<\/p>\n

Example 5.
\nRewrite the following expression in standard form.
\n\\(\\frac{2(3 x-4)}{6}-\\frac{5 x+2}{8}\\)
\nAnswer:
\n\u2192 How can we start to rewrite this problem?
\n\u2192 There are various ways to start rewriting this expression, including using the distributive property, renaming \\(\\frac{2}{6}\\), rewriting the subtraction as an addition, distributing the negative in the second term, rewriting each term as a fraction (e.g., \\(\\frac{2}{6}\\)(3x-4)-(\\(\\frac{5x}{8}\\)+\\(\\frac{2}{8}\\))), or finding the lowest common denominator.
\n\"Engage<\/p>\n

Eureka Math Grade 7 Module 3 Lesson 6 Opening Exercise Answer Key<\/h3>\n

Solve each problem, leaving your answers in standard form. Show your steps.
\na. Terry weighs 40 kg. Janice weighs 2 \\(\\frac{3}{4}\\) kg less than Terry. What is their combined weight?
\nAnswer:
\n40+(40-2 \\(\\frac{3}{4}\\))=80-2\\(\\frac{3}{4}\\)=78-\\(\\frac{3}{4}\\)=77 \\(\\frac{1}{4}\\). Their combined weight is 77 \\(\\frac{1}{4}\\) kg.<\/p>\n

b. 2\\(\\frac{2}{3}\\)-1 \\(\\frac{1}{2}\\)–\\(\\frac{4}{5}\\)
\nAnswer:
\n\\(\\frac{8}{3}\\)–\\(\\frac{3}{2}\\)–\\(\\frac{4}{5}\\)
\n\\(\\frac{80}{30}\\)–\\(\\frac{45}{30}\\)–\\(\\frac{24}{30}\\)
\n\\(\\frac{11}{30}\\)<\/p>\n

c. \\(\\frac{1}{5}\\)+(-4)
\nAnswer:
\n-3 \\(\\frac{4}{5}\\)<\/p>\n

d. 4(\\(\\frac{3}{5}\\))
\nAnswer:
\n\\(\\frac{4}{1}\\) (\\(\\frac{3}{5}\\))
\n\\(\\frac{12}{5}\\)
\n2 \\(\\frac{2}{5}\\)<\/p>\n

e. Mr. Jackson bought 1 \\(\\frac{3}{5}\\) lb. of beef. He cooked \\(\\frac{3}{4}\\) of it for lunch. How much does he have left?
\nAnswer:
\nIf he cooked \\(\\frac{3}{4}\\) of it for lunch, he had \\(\\frac{1}{4}\\) of the original amount left. Since (1 \\(\\frac{3}{5}\\))(\\(\\frac{1}{4}\\))=\\(\\frac{8}{5}\\)\u22c5\\(\\frac{1}{4}\\)=\\(\\frac{2}{5}\\), he had \\(\\frac{2}{5}\\) lb. left. Teachers: You can also show your students how to write the answer as one expression.
\n(1\\(\\frac{3}{5}\\))(1-\\(\\frac{3}{4}\\))<\/p>\n

\u2192 How is the process of writing equivalent expressions by combining like terms in this Opening Exercise different from the previous lesson?
\n\u2192 There are additional steps to find common denominators, convert mixed numbers to improper numbers (in some cases), and convert back.<\/p>\n

Eureka Math Grade 7 Module 3 Lesson 6 Exercise Answer Key<\/h3>\n

Exercise 1.
\nFor the following exercises, predict how many terms the resulting expression will have after collecting like terms. Then, write the expression in standard form by collecting like terms.
\na. \\(\\frac{2}{5}\\) g-\\(\\frac{1}{6}\\)-g+\\(\\frac{3}{10}\\) g-\\(\\frac{4}{5}\\)
\nAnswer:
\nThere will be two terms.
\n\\(\\frac{2}{5}\\) g-1g+\\(\\frac{3}{10}\\) g-\\(\\frac{1}{6}\\)–\\(\\frac{4}{5}\\)
\n(\\(\\frac{2}{5}\\)-1+\\(\\frac{3}{10}\\))g-(\\(\\frac{1}{6}\\)+\\(\\frac{4}{5}\\))
\n–\\(\\frac{3}{10}\\) g-\\(\\frac{29}{30}\\)<\/p>\n

b. i+6i-\\(\\frac{3}{7}\\) i+\\(\\frac{1}{3}\\) h+\\(\\frac{1}{2}\\) i-h+\\(\\frac{1}{4}\\) h
\nAnswer:
\ni+6i-\\(\\frac{3}{7}\\) i+\\(\\frac{1}{3}\\) h+\\(\\frac{1}{2}\\) i-h+\\(\\frac{1}{4}\\) h
\nThere will be two terms.
\n\\(\\frac{1}{3}\\) h+\\(\\frac{1}{4}\\) h-h+i-\\(\\frac{3}{7}\\) i+6i+\\(\\frac{1}{2}\\) i
\n(\\(\\frac{1}{3}\\)+\\(\\frac{1}{4}\\)+(-1))h+(1-\\(\\frac{3}{7}\\)+6+\\(\\frac{1}{2}\\))i
\n–\\(\\frac{5}{12}\\) h+7 \\(\\frac{1}{14}\\) i<\/p>\n

Exercise 2.
\nWrite two different expressions that represent the total cost of the items if tax was \\(\\frac{1}{10}\\) of the original price. Explain the different information each shows.
\nFor the cost of a shirt:
\n\\(\\frac{1}{2}\\) (s-10)+\\(\\frac{1}{10}\\) s The cost of each shirt is \\(\\frac{1}{2}\\) of the quantity of the original cost of the shirt, minus 10, plus \\(\\frac{1}{10}\\) of the cost of the shirt.
\n\\(\\frac{3}{5}\\) s-5 The cost of each shirt is \\(\\frac{3}{5}\\) of the original price (because it is \\(\\frac{1}{2}\\) s+\\(\\frac{1}{10}\\) s=\\(\\frac{6}{10}\\) s), minus 5, since half of 10 is 5.
\nFor the cost of a pair of pants:
\n\\(\\frac{1}{2}\\) (2p)+\\(\\frac{1}{10}\\) p The cost of each pair of pants is half off double the price plus \\(\\frac{1}{10}\\) of the cost of a pair of pants.
\n1 \\(\\frac{1}{10}\\) p The cost of each pair of pants is 1 \\(\\frac{1}{10}\\) (because 1p+\\(\\frac{1}{10}\\) p=1 \\(\\frac{1}{10}\\) p) times the number of pair of pants.<\/p>\n

Exercise 3.
\nRewrite the following expressions in standard form by finding the product and collecting like terms.
\na. -6 \\(\\frac{1}{3}\\)–\\(\\frac{1}{2}\\) (\\(\\frac{1}{2}\\)+y)
\nAnswer:
\n-6 \\(\\frac{1}{3}\\)+(-\\(\\frac{1}{2}\\))(\\(\\frac{1}{2}\\))+(-\\(\\frac{1}{2}\\))y
\n-6 \\(\\frac{1}{3}\\)+(-\\(\\frac{1}{4}\\))+(-\\(\\frac{1}{2}\\) y)
\n–\\(\\frac{1}{2}\\) y-(6 \\(\\frac{1}{3}\\)+\\(\\frac{1}{4}\\))
\n–\\(\\frac{1}{2}\\) y-(6 \\(\\frac{4}{12}\\)+\\(\\frac{3}{12}\\))
\n–\\(\\frac{1}{2}\\) y-6 \\(\\frac{7}{12}\\)<\/p>\n

b. \\(\\frac{2}{3}\\)+\\(\\frac{1}{3}\\) (\\(\\frac{1}{4}\\) f-1 \\(\\frac{1}{3}\\))
\nAnswer:
\n\\(\\frac{2}{3}\\)+\\(\\frac{1}{3}\\) (\\(\\frac{1}{4}\\) f)+\\(\\frac{1}{3}\\) (-\\(\\frac{4}{3}\\))
\n\\(\\frac{2}{3}\\)+\\(\\frac{1}{12}\\) f-\\(\\frac{4}{9}\\)
\n\\(\\frac{1}{12}\\) f+(\\(\\frac{6}{9}\\)–\\(\\frac{4}{9}\\))
\n\\(\\frac{1}{12}\\) f+\\(\\frac{2}{9}\\)<\/p>\n

Exercise 4.
\nRewrite the following expression in standard form by finding common denominators and collecting like terms.
\n\\(\\frac{2h}{3}\\)–\\(\\frac{h}{9}\\)+\\(\\frac{h-4}{6}\\)
\nAnswer:
\n\\(\\frac{6(2h)}{18}\\)–\\(\\frac{2(h)}{18}\\)+\\(\\frac{3(h-4)}{18}\\)
\n\\(\\frac{12 h-2 h+3 h-12}{18}\\)
\n\\(\\frac{13 h-12}{18}\\)
\n\\(\\frac{13}{18}\\)h – \\(\\frac{2}{3}\\)<\/p>\n

Exercise 5.
\nWrite the following expression in standard form.
\n\\(\\frac{2 x-11}{4}-\\frac{3(x-2)}{10}\\)
\nAnswer:
\n\\(\\frac{5(2 x-11)}{20}-\\frac{2 \\cdot 3(x-2)}{20}\\)
\n\\(\\frac{(10 x-55)-6(x-2)}{20}\\)
\n\\(\\frac{10 x-55-6 x+12}{20}\\)
\n\\(\\frac{4 x-43}{20}\\)
\n\\(\\frac{1}{5} x-2 \\frac{3}{20}\\)<\/p>\n

Eureka Math Grade 7 Module 3 Lesson 6 Exit Ticket Answer Key<\/h3>\n

For the problem \\(\\frac{1}{5}\\) g-\\(\\frac{1}{10}\\)-g+1 \\(\\frac{3}{10}\\) g-\\(\\frac{1}{10}\\) , Tyson created an equivalent expression using the following steps.
\nAnswer:
\n\\(\\frac{1}{5}\\) g+-1g+1 \\(\\frac{3}{10}\\) g+-\\(\\frac{1}{10}\\)+-\\(\\frac{1}{10}\\)
\n–\\(\\frac{4}{5}\\) g+1 \\(\\frac{1}{10}\\)
\nIs his final expression equivalent to the initial expression? Show how you know. If the two expressions are not equivalent, find Tyson\u2019s mistake and correct it.<\/p>\n

For the problem \\(\\frac{1}{5}\\) g-\\(\\frac{1}{10}\\)-g+1 \\(\\frac{3}{10}\\) g-\\(\\frac{1}{10}\\) , Tyson created an equivalent expression using the following steps.
\n\\(\\frac{1}{5}\\) g+-1g+1 \\(\\frac{3}{10}\\) g+-\\(\\frac{1}{10}\\)+-\\(\\frac{1}{10}\\)
\n–\\(\\frac{4}{5}\\) g+1 \\(\\frac{1}{10}\\)
\nIs his final expression equivalent to the initial expression? Show how you know. If the two expressions are not equivalent, find Tyson\u2019s mistake and correct it.
\nAnswer:
\nNo, he added the first two terms correctly, but he forgot the third term and added to the other like terms.
\nIf g=10,
\n\\(\\frac{1}{5}\\) g+-1g+1 \\(\\frac{3}{10}\\) g+-\\(\\frac{1}{10}\\)+-\\(\\frac{1}{10}\\)
\n\\(\\frac{1}{5}\\)(10)+-1(10)+1 \\(\\frac{3}{10}\\) (10)+-\\(\\frac{1}{10}\\)+-\\(\\frac{1}{10}\\)
\n2+(-10)+13+(-\\(\\frac{2}{10}\\))
\n4\\(\\frac{4}{5}\\)<\/p>\n

–\\(\\frac{4}{5}\\)g +1 \\(\\frac{1}{10}\\)
\n–\\(\\frac{4}{5}\\)(10)+1 \\(\\frac{1}{10}\\)
\n-8+1 \\(\\frac{1}{10}\\)
\n-6\\(\\frac{9}{10}\\)
\nThe expressions are not equal.
\nHe should factor out the g and place parentheses around the values using the distributive property in order to make it obvious which rational numbers need to be combined.
\n\\(\\frac{1}{5}\\) g+-1g+1 \\(\\frac{3}{10}\\) g+-\\(\\frac{1}{10}\\)+-\\(\\frac{1}{10}\\)
\n(\\(\\frac{1}{5}\\) g+-1g+1 \\(\\frac{3}{10}\\) g)+(-\\(\\frac{1}{10}\\)+-\\(\\frac{1}{10}\\))
\n(\\(\\frac{1}{5}\\)+-1+1 \\(\\frac{3}{10}\\))g+(-\\(\\frac{2}{10}\\))
\n(\\(\\frac{2}{10}\\)+\\(\\frac{3}{10}\\))g+(-\\(\\frac{1}{5}\\))
\n\\(\\frac{1}{2}\\) g-\\(\\frac{1}{5}\\)<\/p>\n

Eureka Math Grade 7 Module 3 Lesson 6 Problem Set Answer Key<\/h3>\n

Question 1.
\nWrite the indicated expressions.
\na. \\(\\frac{1}{2}\\) m inches in feet
\nAnswer:
\n\\(\\frac{1}{2}\\) m\u00d7\\(\\frac{1}{12}\\)=\\(\\frac{1}{24}\\)m. It is \\(\\frac{1}{24}\\)m ft.<\/p>\n

b. The perimeter of a square with \\(\\frac{2}{3}\\) g cm sides
\nAnswer:
\n4\u00d7\\(\\frac{2}{3}\\) g=\\(\\frac{8}{3}\\) g. The perimeter is \\(\\frac{8}{3}\\) g cm.<\/p>\n

c. The number of pounds in 9 oz.
\nAnswer:
\n9\u00d7\\(\\frac{1}{16}\\)=\\(\\frac{9}{16}\\). It is \\(\\frac{9}{16}\\) lb.<\/p>\n

d. The average speed of a train that travels x miles in \\(\\frac{3}{4}\\) hour
\nAnswer:
\n\"Engage. The average speed of the train is \\(\\frac{4}{3}\\) x miles per hour.<\/p>\n

e. Devin is 1 \\(\\frac{1}{4}\\) years younger than Eli. April is \\(\\frac{1}{5}\\) as old as Devin. Jill is 5 years older than April. If Eli is E years old, what is Jill\u2019s age in terms of E?
\nAnswer:
\nD=E-1 \\(\\frac{1}{4}\\), A=\\(\\frac{D}{5}\\), A+5=J, so J=(\\(\\frac{D}{5}\\))+5. J=\\(\\frac{1}{5}\\)(E-1 \\(\\frac{1}{4}\\))+5. J=\\(\\frac{E}{5}\\)+4 \\(\\frac{3}{4}\\).<\/p>\n

Question 2.
\nRewrite the expressions by collecting like terms.
\na. \\(\\frac{1}{2}\\) k-\\(\\frac{3}{8}\\) k
\nAnswer:
\n\\(\\frac{4}{8}\\) k-\\(\\frac{3}{8}\\) k
\n\\(\\frac{1}{8}\\) k<\/p>\n

b. \\(\\frac{2r}{5}\\)+\\(\\frac{7r}{15}\\)
\nAnswer:
\n\\(\\frac{6r}{15}\\)+\\(\\frac{7r}{15}\\)
\n\\(\\frac{13r}{15}\\)<\/p>\n

c. –\\(\\frac{1}{3}\\) a-\\(\\frac{1}{2}\\) b-\\(\\frac{3}{4}\\)+\\(\\frac{1}{2}\\) b-\\(\\frac{2}{3}\\) b+\\(\\frac{5}{6}\\) a
\nAnswer:
\n–\\(\\frac{1}{3}\\) a+\\(\\frac{5}{6}\\) a-\\(\\frac{1}{2}\\) b+\\(\\frac{1}{2}\\) b-\\(\\frac{2}{3}\\) b-\\(\\frac{3}{4}\\)
\n–\\(\\frac{2}{6}\\) a+\\(\\frac{5}{6}\\) a-\\(\\frac{2}{3}\\) b-\\(\\frac{3}{4}\\)
\n\\(\\frac{1}{2}\\) a-\\(\\frac{2}{3}\\) b-\\(\\frac{3}{4}\\)<\/p>\n

d. -p+\\(\\frac{3}{5}\\) q-\\(\\frac{1}{10}\\) q+\\(\\frac{1}{9}\\)–\\(\\frac{1}{9}\\) p+2 \\(\\frac{1}{3}\\) p
\nAnswer:
\n-p-\\(\\frac{1}{9}\\) p+2 \\(\\frac{1}{3}\\) p+\\(\\frac{3}{5}\\) q-\\(\\frac{1}{10}\\) q+\\(\\frac{1}{9}\\)
\n–\\(\\frac{9}{9}\\) p-\\(\\frac{1}{9}\\) p+2 \\(\\frac{3}{9}\\) p+\\(\\frac{6}{10}\\) q-\\(\\frac{1}{10}\\) q+\\(\\frac{1}{9}\\)
\n1\\(\\frac{1}{9}\\) p+\\(\\frac{5}{10}\\)q+\\(\\frac{1}{9}\\)
\n1 \\(\\frac{2}{9}\\) p+\\(\\frac{1}{2}\\) q+\\(\\frac{1}{9}\\)<\/p>\n

e. \\(\\frac{5}{7}\\) y-\\(\\frac{y}{14}\\)
\nAnswer:
\n\\(\\frac{10}{14}\\) y-\\(\\frac{1}{14}\\) y
\n\\(\\frac{9}{14}\\)y<\/p>\n

f. \\(\\frac{3n}{8}\\)–\\(\\frac{n}{4}\\)+2 \\(\\frac{n}{2}\\)
\nAnswer:
\n\\(\\frac{3n}{8}\\)–\\(\\frac{2n}{8}\\)+2\\(\\frac{4n}{8}\\)
\n2\\(\\frac{5n}{8}\\)<\/p>\n

Question 3.
\nRewrite the expressions by using the distributive property and collecting like terms.
\na. \\(\\frac{4}{5}\\) (15x-5)
\nAnswer:
\n12x-4<\/p>\n

b. \\(\\frac{4}{5}\\) (\\(\\frac{1}{4}\\) c-5)
\nAnswer:
\n\\(\\frac{1}{5}\\)c-4<\/p>\n

c. 2 \\(\\frac{4}{5}\\) v-\\(\\frac{2}{3}\\) (4v+1 \\(\\frac{1}{6}\\))
\nAnswer:
\n\\(\\frac{2}{5}\\) v-\\(\\frac{7}{9}\\)<\/p>\n

d. 8-4(\\(\\frac{1}{8}\\) r-3 \\(\\frac{1}{2}\\))
\nAnswer:
\n–\\(\\frac{1}{2}\\) r+22<\/p>\n

e. \\(\\frac{1}{7}\\) (14x+7)-5
\nAnswer:
\n2x-4<\/p>\n

f. \\(\\frac{1}{5}\\) (5x-15)-2x
\nAnswer:
\n-x-3<\/p>\n

g. \\(\\frac{1}{4}\\) (p+4)+\\(\\frac{3}{5}\\) (p-1)
\nAnswer:
\n\\(\\frac{17}{20}\\) p+\\(\\frac{2}{5}\\)<\/p>\n

h. \\(\\frac{7}{8}\\) (w+1)+\\(\\frac{5}{6}\\)(w-3)
\nAnswer:
\n\\(\\frac{41}{24}\\)w-\\(\\frac{39}{24}\\) or \\(\\frac{41}{24}\\) w-\\(\\frac{13}{8}\\)<\/p>\n

i. \\(\\frac{4}{5}\\) (c-1)-\\(\\frac{1}{8}\\)(2c+1)
\nAnswer:
\n\\(\\frac{11}{20}\\)c – \\(\\frac{37}{40}\\)<\/p>\n

j. \\(\\frac{2}{3}\\) (h+\\(\\frac{3}{4}\\))-\\(\\frac{1}{3}\\) (h+\\(\\frac{3}{4}\\))
\nAnswer:
\n\\(\\frac{1}{3}\\) h+\\(\\frac{1}{4}\\)<\/p>\n

k. \\(\\frac{2}{3}\\) (h+\\(\\frac{3}{4}\\))-\\(\\frac{2}{3}\\) (h-\\(\\frac{3}{4}\\))
\nAnswer:
\n1<\/p>\n

l. \\(\\frac{2}{3}\\) (h+\\(\\frac{3}{4}\\))+\\(\\frac{2}{3}\\) (h-\\(\\frac{3}{4}\\))
\nAnswer:
\n\\(\\frac{4}{3}\\) h<\/p>\n

m. \\(\\frac{k}{2}\\)–\\(\\frac{4k}{5}\\)-3
\nAnswer:
\n–\\(\\frac{3k}{10}\\)-3<\/p>\n

n. \\(\\frac{3t+2}{7}\\)+\\(\\frac{t-4}{14}\\)
\nAnswer:
\n\\(\\frac{1}{2}\\) t<\/p>\n

o. \\(\\frac{9x-4}{10}\\)+\\(\\frac{3x+2}{5}\\)
\nAnswer:
\n\\(\\frac{3x}{2}\\) or 1 \\(\\frac{1}{2}\\)x<\/p>\n

p. \\(\\frac{3(5 g-1)}{4}\\) – \\(\\frac{2 g+7}{6}\\)
\nAnswer:
\n3\\(\\frac{5}{12}\\)g – 1\\(\\frac{11}{12}\\)<\/p>\n

q. \\(\\frac{3 d+1}{5}\\) + \\(\\frac{d-5}{2}\\) + \\(\\frac{7}{10}\\)
\nAnswer:
\n\\(\\frac{-d}{10}\\)-2<\/p>\n

r. \\(\\frac{9 w}{6}+\\frac{2 w-7}{3}-\\frac{w-5}{4}\\)
\nAnswer:
\n\\(\\frac{23 w-13}{12}\\)
\n\\(\\frac{23}{12}\\)w-\\(\\frac{13}{12}\\)<\/p>\n

s. \\(\\frac{1+f}{5}-\\frac{1+f}{3}+\\frac{3-f}{6}\\)
\nAnswer:
\n\\(\\frac{11}{30}\\)–\\(\\frac{3}{10}\\)f<\/p>\n","protected":false},"excerpt":{"rendered":"

Engage NY Eureka Math 7th Grade Module 3 Lesson 6 Answer Key Eureka Math Grade 7 Module 3 Lesson 6 Example Answer Key Example 1. Rewrite the expression in standard form by collecting like terms. n- n+ n+2 n Answer: n- n+ n+2 n n- n+ n+2 n 2 n \u2192 What are various strategies … Read more<\/a><\/p>\n","protected":false},"author":3,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[6],"tags":[],"jetpack_sharing_enabled":true,"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/posts\/79092"}],"collection":[{"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/users\/3"}],"replies":[{"embeddable":true,"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/comments?post=79092"}],"version-history":[{"count":1,"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/posts\/79092\/revisions"}],"predecessor-version":[{"id":137641,"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/posts\/79092\/revisions\/137641"}],"wp:attachment":[{"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/media?parent=79092"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/categories?post=79092"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/tags?post=79092"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}