Example 1.
\nSubtracting Expressions
\na. Subtract: (40 + 9) – (30 + 2).
\nAnswer:
\nThe opposite of a sum is the sum of its opposites. Order of operations<\/p>\n
40 + 9 + (- (30 + 2))
\n40 + 9 + (- 30) + (- 2)
\n49 + (- 30) + (- 2)
\n19 + (- 2)
\n17<\/p>\n
(40 + 9)- (30 + 2)
\n(49)- (32)
\n17<\/p>\n
b. Subtract: (3x + 5y – 4) – (4x + 11).
\nAnswer:
\n3x + 5y + ( – 4) + ( – (4x + 11)) Subtraction as adding the opposite
\n3x + 5y + ( – 4) + ( – 4x) + ( – 11) The opposite of a sum is the sum of its opposites.
\n3x + ( – 4x) + 5y + ( – 4) + ( – 11) Any order, any grouping
\n– x + 5y + ( – 15) Combining like terms
\n– x + 5y – 15 Subtraction replaces adding the opposite.
\nHave students check the equivalency of the expressions by substituting 2 for x and 6 for y.
\n(3x + 5y – 4) – (4x + 11)
\n(3(2) + 5(6) – 4) – (4(2) + 11)
\n(6 + 30 – 4) – (8 + 11)
\n(36 – 4) – (19)
\n32 – 19
\n13
\n– x + 5y – 15
\n– (2) + 5(6) – 15
\n– 2 + 30 + ( – 15)
\n28 + ( – 15)
\n13
\n\u2192 When writing the difference as adding the expression\u2019s opposite in Example 1(b), what happens to the grouped terms that are being subtracted?
\n\u2192 When the subtraction is changed to addition, every term in the parentheses that follows must be converted to its opposite.<\/p>\n
Example 2.
\nCombining Expressions Vertically
\na. Find the sum by aligning the expressions vertically.
\n(5a + 3b – 6c) + (2a – 4b + 13c)
\nAnswer:
\n(5a + 3b + ( – 6c)) + (2a + ( – 4b) + 13c) Subtraction as adding the opposite
\n![\"Engage](\"https:\/\/ccssanswers.com\/wp-content\/uploads\/2021\/03\/Engage-NY-Math-7th-Grade-Module-3-Lesson-2-Example-Answer-Key-30.png\")
\n7a – b + 7c Adding the opposite is equivalent to subtraction.<\/p>\n
b. Find the difference by aligning the expressions vertically.
\n(2x + 3y – 4) – (5x + 2)
\nAnswer:
\n(2x + 3y + ( – 4)) + ( – 5x + ( – 2)) Subtraction as adding the opposite
\n![\"Engage](\"https:\/\/ccssanswers.com\/wp-content\/uploads\/2021\/03\/Engage-NY-Math-7th-Grade-Module-3-Lesson-2-Example-Answer-Key-31.png\")
\n– 3x + 3y – 6 Adding the opposite is equivalent to subtraction.
\nStudents should recognize that the subtracted expression in Example 1(b) did not include a term containing the variable y, so the 3y from the first grouped expression remains unchanged in the answer.<\/p>\n
Example 3.
\nUsing Expressions to Solve Problems
\nA stick is x meters long. A string is 4 times as long as the stick.
\na. Express the length of the string in terms of x.
\nAnswer:
\nThe length of the stick in meters is x meters, so the string is 4\u2219x, or 4x, meters long.<\/p>\n
b. If the total length of the string and the stick is 15 meters long, how long is the string?
\nAnswer:
\nThe length of the stick and the string together in meters can be represented by x + 4x, or 5x. If the length of the stick and string together is 15 meters, the length of the stick is 3 meters, and the length of the string is 12 meters.<\/p>\n
Example 4.
\nExpressions from Word Problems
\nIt costs Margo a processing fee of $3 to rent a storage unit, plus $17 per month to keep her belongings in the unit. Her friend Carissa wants to store a box of her belongings in Margo\u2019s storage unit and tells her that she will pay her $1 toward the processing fee and $3 for every month that she keeps the box in storage. Write an expression in standard form that represents how much Margo will have to pay for the storage unit if Carissa contributes. Then, determine how much Margo will pay if she uses the storage unit for 6 months.
\nAnswer:
\nLet m represent the number of months that the storage unit is rented.
\n(17m + 3) – (3m + 1) Original expression
\n17m + 3 + ( – (3m + 1)) Subtraction as adding the opposite
\n17m + 3 + ( – 3m) + ( – 1) The opposite of the sum is the sum of its opposites.
\n17m + ( – 3m) + 3 + ( – 1) Any order, any grouping
\n14m + 2 Combined like terms
\nThis means that Margo will have to pay only $2 of the processing fee and $14 per month that the storage unit is used.
\n14(6) + 2
\n84 + 2
\n86
\nMargo will pay $86 toward the storage unit rental for 6 months of use.<\/p>\n
Example 5.
\nExtending Use of the Inverse to Division
\n![\"Engage](\"https:\/\/ccssanswers.com\/wp-content\/uploads\/2021\/03\/Engage-NY-Math-7th-Grade-Module-3-Lesson-2-Example-Answer-Key-20.png\")
\nAnswer:
\n![\"Engage](\"https:\/\/ccssanswers.com\/wp-content\/uploads\/2021\/03\/Engage-NY-Math-7th-Grade-Module-3-Lesson-2-Example-Answer-Key-21.png\")
<\/p>\n
Opening Exercise
\nAdditive inverses have a sum of zero. Fill in the center column of the table with the opposite of the given number or expression, then show the proof that they are opposites. The first row is completed for you.
\n![\"Eureka](\"https:\/\/ccssanswers.com\/wp-content\/uploads\/2021\/03\/Eureka-Math-Grade-7-Module-3-Lesson-2-Opening-Exercise-Answer-Key-1.png\")
\nAnswer:
\n![\"Eureka](\"https:\/\/ccssanswers.com\/wp-content\/uploads\/2021\/03\/Eureka-Math-Grade-7-Module-3-Lesson-2-Opening-Exercise-Answer-Key-2.png\")
<\/p>\n
Question 1.
\nWrite each expression in standard form. Verify that your expression is equivalent to the one given by evaluating each expression using x=5.<\/p>\n
a. 3x + (2 – 4x)
\nAnswer:
\n3x + (2 – 4x)
\n– x + 2
\n– 5 + 2
\n– 3<\/p>\n
3(5) + (2 – 4(5))
\n15 + (2 + ( – 20))
\n15 + ( – 18)
\n– 3<\/p>\n
b. 3x + ( – 2 + 4x)
\nAnswer:
\n3x + ( – 2 + 4x)
\n7x – 2
\n7(5) – 2
\n35 – 2
\n33<\/p>\n
3(5) + ( – 2 + 4(5))
\n15 + ( – 2 + 20)
\n15 + 18
\n33<\/p>\n
c. – 3x + (2 + 4x)
\nAnswer:
\n– 3x + (2 + 4x)
\nx + 2
\n5 + 2
\n7<\/p>\n
– 3(5) + (2 + 4(5))
\n– 15 + (2 + 20)
\n– 15 + 22
\n7<\/p>\n
d. 3x + ( – 2 – 4x)
\nAnswer:
\n3x + ( – 2 – 4x)
\n– x – 2
\n– 5 – 2
\n– 7<\/p>\n
3(5) + ( – 2 – 4(5))
\n15 + ( – 2 + ( – 4(5)))
\n15 + ( – 2 + ( – 20))
\n15 + ( – 22)
\n– 7<\/p>\n
e. 3x – (2 + 4x)
\nAnswer:
\n3x – (2 + 4x)
\n– x – 2
\n– 5 – 2
\n– 7<\/p>\n
3(5) – (2 + 4(5))
\n15 – (2 + 20)
\n15 – 22
\n15 + ( – 22)
\n– 7<\/p>\n
f. 3x – ( – 2 + 4x)
\nAnswer:
\n3x – ( – 2 + 4x)
\n– x + 2
\n– 5 + 2
\n– 3<\/p>\n
3(5) – ( – 2 + 4(5))
\n15 – ( – 2 + 20)
\n15 – (18)
\n15 + ( – 18)
\n– 3<\/p>\n
g. 3x – ( – 2 – 4x)
\nAnswer:
\n3x – ( – 2 – 4x)
\n7x + 2
\n7(5) + 2
\n35 + 2
\n37<\/p>\n
3(5) – ( – 2 – 4(5))
\n15 – ( – 2 + ( – 4(5)))
\n15 – ( – 2 + ( – 20))
\n15 – ( – 22)
\n15 + 22
\n37<\/p>\n
h. 3x – (2 – 4x)
\nAnswer:
\n3x – (2 – 4x)
\n7x – 2
\n7(5) – 2
\n35 – 2
\n33<\/p>\n
3(5) – (2 – 4(5))
\n15 – (2 + ( – 4(5)))
\n15 – (2 + ( – 20))
\n15 – ( – 18)
\n15 + 18
\n33<\/p>\n
i. – 3x – ( – 2 – 4x)
\nAnswer:
\n– 3x – ( – 2 – 4x)
\nx + 2
\n5 + 2
\n7<\/p>\n
– 3(5) – ( – 2 – 4(5))
\n– 15 – ( – 2 + ( – 4(5)))
\n– 15 – ( – 2 + ( – 20))
\n– 15 – ( – 22)
\n– 15 + 22
\n7<\/p>\n
j. In problems (a)\u2013(d) above, what effect does addition have on the terms in parentheses when you removed the parentheses?
\nAnswer:
\nBy the any grouping property, the terms remained the same with or without the parentheses.<\/p>\n
k. In problems (e)\u2013(i), what effect does subtraction have on the terms in parentheses when you removed the parentheses?
\nAnswer:
\nThe opposite of a sum is the sum of the opposites; each term within the parentheses is changed to its opposite.<\/p>\n
Question 2.
\nWrite each expression in standard form. Verify that your expression is equivalent to the one given by evaluating each expression for the given value of the variable.
\na. 4y – (3 + y); y=2
\nAnswer:
\n4y – (3 + y)
\n3y – 3
\n3(2) – 3
\n6 – 3
\n3<\/p>\n
4(2) – (3 + 2)
\n8 – 5
\n8 + ( – 5)
\n3<\/p>\n
b. (2b + 1) – b; b= – 4
\nAnswer:
\n(2b + 1) – b
\nb + 1
\n– 4 + 1
\n– 3<\/p>\n
(2( – 4) + 1) – ( – 4)
\n( – 8 + 1) + 4
\n( – 7) + 4
\n– 3<\/p>\n
c. (6c – 4) – (c – 3); c= – 7
\nAnswer:
\n(6c – 4) – (c – 3)
\n5c – 1
\n5( – 7) – 1
\n– 35 – 1
\n– 36<\/p>\n
(6( – 7) – 4) – ( – 7 – 3)
\n( – 42 – 4) – ( – 10)
\n– 42 + ( – 4) + (10)
\n– 46 + 10
\n– 36<\/p>\n
d. (d + 3d) – ( – d + 2); d=3
\nAnswer:
\n(d + 3d) – ( – d + 2)
\n5d – 2
\n5(3) – 2
\n15 – 2
\n13<\/p>\n
(3 + 3(3)) – ( – 3 + 2)
\n(3 + 9) – ( – 1)
\n12 + 1
\n13<\/p>\n
e. ( – 5x – 4) – ( – 2 – 5x); x=3
\nAnswer:
\n– 2
\n( – 5(3) – 4) – ( – 2 – 5(3))
\n( – 15 – 4) – ( – 2 – 15)
\n( – 19) – ( – 17)
\n( – 19) + 17
\n– 2<\/p>\n
f. 11f – ( – 2f + 2); f=\\(\\frac{1}{2}\\)
\n13f – 2
\n13(\\(\\frac{1}{2}\\) ) – 2
\n\\(\\frac{13}{2}\\) – 2
\n6 \\(\\frac{1}{2}\\) – 2
\n4 \\(\\frac{1}{2}\\)<\/p>\n
11(\\(\\frac{1}{2}\\) ) – ( – 2(\\(\\frac{1}{2}\\) ) + 2)
\n1\\(\\frac{1}{2}\\) – ( – 1 + 2)
\n1\\(\\frac{1}{2}\\) – 1
\n1\\(\\frac{1}{2}\\) + ( – \\(\\frac{2}{2}\\) )
\n\\(\\frac{9}{2}\\)
\n4 \\(\\frac{1}{2}\\)<\/p>\n
g. – 5g + (6g – 4); g= – 2
\nAnswer:
\n– 5g + (6g – 4)
\ng – 4
\n– 2 – 4
\n– 6<\/p>\n
– 5( – 2) + (6( – 2) – 4)
\n10 + ( – 12 – 4)
\n10 + ( – 12 + ( – 4))
\n10 + ( – 16)
\n– 6<\/p>\n
h. (8h – 1) – (h + 3); h= – 3
\nAnswer:
\n(8h – 1) – (h + 3)
\n7h – 4
\n7( – 3) – 4
\n– 21 – 4
\n– 25<\/p>\n
(8( – 3) – 1) – ( – 3 + 3)
\n( – 24 – 1) – (0)
\n( – 25) – 0
\n– 25<\/p>\n
i. (7 + w) – (w + 7); w= – 4
\nAnswer
\n(7 + w) – (w + 7)
\n0
\n(7 + ( – 4)) – ( – 4 + 7)
\n3 – 3
\n3 + ( – 3)
\n0<\/p>\n
j. (2g + 9h – 5) – (6g – 4h + 2); g= – 2 and h=5
\nAnswer:
\n(2g + 9h – 5) – (6g – 4h + 2)
\n– 4g + 13h – 7
\n– 4( – 2) + 13(5) – 7
\n8 + 65 + ( – 7)
\n73 + ( – 7)
\n66<\/p>\n
(2( – 2) + 9(5) – 5) – (6( – 2) – 4(5) + 2)
\n( – 4 + 45 – 5) – ( – 12 + ( – 4(5)) + 2)
\n(41 – 5) – ( – 12 + ( – 20) + 2)
\n(41 + ( – 5)) – ( – 32 + 2)
\n36 – ( – 30)
\n36 + 30
\n66<\/p>\n
Question 3.
\nWrite each expression in standard form. Verify that your expression is equivalent to the one given by evaluating both expressions for the given value of the variable.<\/p>\n
a. – 3(8x); x=\\(\\frac{1}{4}\\)
\nAnswer:
\n– 3(8x)
\n– 24x
\n– 24(\\(\\frac{1}{4}\\) )
\n– 24\/4
\n– 6<\/p>\n
– 3(8(\\(\\frac{1}{4}\\) ))
\n– 3(2)
\n– 6<\/p>\n
b. 5\u2219k\u2219( – 7); k=\\(\\frac{3}{5}\\)
\nAnswer:
\n5\u2219k\u2219( – 7)
\n– 35k
\n– 35(\\(\\frac{3}{5}\\) )
\n– 105\/5
\n– 21<\/p>\n
5(\\(\\frac{3}{5}\\) )( – 7)
\n3( – 7)
\n– 21<\/p>\n
c. 2( – 6x)\u22192; x=\\(\\frac{3}{4}\\)
\nAnswer:
\n2( – 6x)\u22192
\n– 24x
\n– 24(\\(\\frac{3}{4}\\) )
\n– \\(\\frac{72}{4}\\)
\n– 18<\/p>\n
2( – 6(\\(\\frac{3}{4}\\) ))\u22192
\n2( – 3(\\(\\frac{3}{2}\\) ))\u22192
\n2( – 3)(\\(\\frac{3}{2}\\) )(2)
\n– 6(3)
\n– 18<\/p>\n
d. – 3(8x) + 6(4x); x=2
\nAnswer:
\n– 3(8x) + 6(4x)
\n0<\/p>\n
– 3(8(2)) + 6(4(2))
\n– 3(16) + 6(8)
\n– 48 + 48
\n0<\/p>\n
e. 8(5m) + 2(3m); m= – 2
\nAnswer:
\n8(5m) + 2(3m)
\n46m
\n46( – 2)
\n– 92<\/p>\n
8(5( – 2)) + 2(3( – 2))
\n8( – 10) + 2( – 6)
\n– 80 + ( – 12)
\n– 92<\/p>\n
f. – 6(2v) + 3a(3); v=\\(\\frac{1}{3}\\) ; a=\\(\\frac{2}{3}\\)
\nAnswer:
\n– 6(2v) + 3a(3); v=\\(\\frac{1}{3}\\)
\n– 6(2v) + 3a(3)
\n– 12v + 9a
\n– 12(\\(\\frac{1}{3}\\) ) + 9(\\(\\frac{2}{3}\\) )
\n– 1\\(\\frac{2}{3}\\) + \\(\\frac{18}{3}\\)
\n– 4 + 6
\n2<\/p>\n
– 6(2(\\(\\frac{1}{3}\\) )) + 3(\\(\\frac{2}{3}\\) )(3)
\n– 6(\\(\\frac{2}{3}\\) ) + 2(3)
\n– 4 + 6
\n2<\/p>\n
Question 4.
\nWrite each expression in standard form. Verify that your expression is equivalent to the one given by evaluating both expressions for the given value of the variable<\/p>\n
a. 8x \u00f7 2; x= – \\(\\frac{1}{4}\\)
\nAnswer:
\n8x \u00f7 2
\n4x
\n4( – \\(\\frac{1}{4}\\) )
\n– 1<\/p>\n
8( – \\(\\frac{1}{4}\\) ) \u00f7 2
\n– 2 \u00f7 2
\n– 1<\/p>\n
b. 18w \u00f7 6; w=6
\nAnswer:
\n18w \u00f7 6
\n3w
\n3(6)
\n18<\/p>\n
18(6) \u00f7 6
\n108 \u00f7 6
\n18<\/p>\n
c. 25r \u00f7 5r; r= – 2
\nAnswer:
\n25r \u00f7 5r
\n5<\/p>\n
25( – 2) \u00f7 (5( – 2))
\n– 50 \u00f7 ( – 10)
\n5<\/p>\n
d. 33y \u00f7 11y; y= – 2
\nAnswer:
\n33y \u00f7 11y
\n3<\/p>\n
33( – 2) \u00f7 (11( – 2))
\n( – 66) \u00f7 ( – 22)
\n3<\/p>\n
e. 56k \u00f7 2k; k=3
\nAnswer:
\n56k \u00f7 2k
\n28<\/p>\n
56(3) \u00f7 (2(3))
\n168 \u00f7 6
\n28<\/p>\n
f. 24xy \u00f7 6y; x= – 2;y=3
\nAnswer:
\n24xy \u00f7 6y
\n4x
\n4( – 2)
\n– 8<\/p>\n
24( – 2)(3) \u00f7 (6(3))
\n– 48(3) \u00f7 18
\n– 144 \u00f7 18
\n– 8<\/p>\n
Question 5.
\nFor each problem (a)\u2013(g), write an expression in standard form.
\na. Find the sum of – 3x and 8x.
\nAnswer:
\n– 3x + 8x
\n5x<\/p>\n
b. Find the sum of – 7g and 4g + 2.
\nAnswer:
\n– 7g + (4g + 2)
\n– 3g + 2<\/p>\n
c. Find the difference when 6h is subtracted from 2h – 4.
\nAnswer:
\n(2h – 4) – 6h
\n– 4h – 4<\/p>\n
d. Find the difference when – 3n – 7 is subtracted from n + 4.
\nAnswer:
\n(n + 4) – ( – 3n – 7)
\n4n + 11<\/p>\n
e. Find the result when 13v + 2 is subtracted from 11 + 5v.
\nAnswer:
\n(11 + 5v) – (13v + 2)
\n– 8v + 9<\/p>\n
f. Find the result when – 18m – 4 is added to 4m – 14.
\nAnswer:
\n(4m – 14) + ( – 18m – 4)
\n– 14m – 18<\/p>\n
g. What is the result when – 2x + 9 is taken away from – 7x + 2?
\nAnswer:
\n( – 7x + 2) – ( – 2x + 9)
\n– 5x – 7<\/p>\n
Question 6.
\nMarty and Stewart are stuffing envelopes with index cards. They are putting x index cards in each envelope. When they are finished, Marty has 15 stuffed envelopes and 4 extra index cards, and Stewart has 12 stuffed envelopes and 6 extra index cards. Write an expression in standard form that represents the number of index cards the boys started with. Explain what your expression means.
\nAnswer:
\nThey inserted the same number of index cards in each envelope, but that number is unknown, x. An expression that represents Marty\u2019s index cards is 15x + 4 because he had 15 envelopes and 4 cards left over. An expression that represents Stewart\u2019s index cards is 12x + 6 because he had 12 envelopes and 6 left over cards. Their total number of cards together would be:
\n15x + 4 + 12x + 6
\n15x + 12x + 4 + 6
\n27x + 10
\nThis means that altogether, they have 27 envelopes with x index cards in each, plus another 10 leftover index cards.<\/p>\n
Question 7. Directions: Write each as an equivalent expression in standard form as quickly and as accurately as possible within the allotted time. Directions: Write each as an equivalent expression in standard form as quickly and as accurately as possible within the allotted time.<\/p>\n Question 1. Question 2. Question 3. Engage NY Eureka Math 7th Grade Module 3 Lesson 2 Answer Key Eureka Math Grade 7 Module 3 Lesson 2 Example Answer Key Example 1. Subtracting Expressions a. Subtract: (40 + 9) – (30 + 2). Answer: The opposite of a sum is the sum of its opposites. Order of operations 40 + 9 + … 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\/76697"}],"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=76697"}],"version-history":[{"count":1,"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/posts\/76697\/revisions"}],"predecessor-version":[{"id":137836,"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/posts\/76697\/revisions\/137836"}],"wp:attachment":[{"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/media?parent=76697"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/categories?post=76697"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/tags?post=76697"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
\nThe area of the pictured rectangle below is 24b ft2<\/sup>. Its width is 2b ft. Find the height of the rectangle and name any properties used with the appropriate step.
\nAnswer:
\n24b \u00f7 2b
\n![\"Eureka](\"https:\/\/ccssanswers.com\/wp-content\/uploads\/2021\/03\/Eureka-Math-Grade-7-Module-3-Lesson-2-Problem-Set-Answer-Key-25.png\")
\n24b\u2219\\(\\frac{1}{2b}\\) Multiplying the reciprocal
\n\\(\\frac{27b}{2b}\\) Multiplication
\n\\(\\frac{24}{2}\\) \u2219\\(\\frac{b}{b}\\) Any order, any grouping in multiplication
\n12\u22191
\n12
\nThe height of the rectangle is 12 ft.<\/p>\nEureka Math Grade 7 Module 3 Lesson 2 Generating Equivalent Expressions\u2014Round 1 Answer Key<\/h3>\n
\n![\"Eureka](\"https:\/\/ccssanswers.com\/wp-content\/uploads\/2021\/03\/Eureka-Math-Grade-7-Module-3-Lesson-2-Generating-Equivalent-Expressions\u2014Round-1-Answer-Key-20.1.png\")
\nAnswer:
\n![\"Eureka](\"https:\/\/ccssanswers.com\/wp-content\/uploads\/2021\/03\/Eureka-Math-Grade-7-Module-3-Lesson-2-Generating-Equivalent-Expressions\u2014Round-1-Answer-Key-20.2.png\")
<\/p>\nEureka Math Grade 7 Module 3 Lesson 2 Generating Equivalent Expressions\u2014Round 2 Answer Key<\/h3>\n
![\"Eureka](\"https:\/\/ccssanswers.com\/wp-content\/uploads\/2021\/03\/Eureka-Math-Grade-7-Module-3-Lesson-2-Generating-Equivalent-Expressions\u2014Round-2-Answer-Key-25.png\")
\nAnswer:
\n![\"Eureka](\"https:\/\/ccssanswers.com\/wp-content\/uploads\/2021\/03\/Eureka-Math-Grade-7-Module-3-Lesson-2-Generating-Equivalent-Expressions\u2014Round-2-Answer-Key-26.png\")
<\/p>\nEureka Math Grade 7 Module 3 Lesson 2 Exit Ticket Answer Key<\/h3>\n
\nWrite the expression in standard form.
\nAnswer:
\n(4f – 3 + 2g) – ( – 4g + 2)
\n4f + ( – 3) + 2g + ( – ( – 4g + 2)) Subtraction as adding the opposite
\n4f + ( – 3) + 2g + 4g + ( – 2) The opposite of a sum is the sum of its opposites.
\n4f + 2g + 4g + ( – 3) + ( – 2) Any order, any grouping
\n4f + 6g + ( – 5) Combined like terms
\n4f + 6g – 5 Subtraction as adding the opposite<\/p>\n
\nFind the result when 5m + 2 is subtracted from 9m.
\nAnswer:
\n9m – (5m + 2) Original expression
\n9m + ( – (5m + 2)) Subtraction as adding the opposite
\n9m + ( – 5m) + ( – 2) The opposite of a sum is the sum of its opposites.
\n4m + ( – 2) Combined like terms
\n4m – 2 Subtraction as adding the opposite<\/p>\n
\nWrite the expression in standard form.
\n27h \u00f7 3h
\nAnswer:
\n27h\u2219\\(\\frac{1}{3 h}\\) Multiplying by the reciprocal
\n\\(\\frac{27h}{3 h}\\) Multiplication
\n\\(\\frac{27}{3}\\)\u2219\\(\\frac{h}{h}\\) Any order, any grouping
\n9\u22191
\n9<\/p>\n","protected":false},"excerpt":{"rendered":"