{"id":85293,"date":"2022-05-30T14:12:34","date_gmt":"2022-05-30T08:42:34","guid":{"rendered":"https:\/\/ccssanswers.com\/?p=85293"},"modified":"2022-05-10T14:19:54","modified_gmt":"2022-05-10T08:49:54","slug":"eureka-math-grade-8-module-4-lesson-8","status":"publish","type":"post","link":"https:\/\/ccssanswers.com\/eureka-math-grade-8-module-4-lesson-8\/","title":{"rendered":"Eureka Math Grade 8 Module 4 Lesson 8 Answer Key"},"content":{"rendered":"

Engage NY Eureka Math 8th Grade Module 4 Lesson 8 Answer Key<\/h2>\n

Eureka Math Grade 8 Module 4 Lesson 8 Example Answer Key<\/h3>\n

Example 1.
\n\u2192 Given a linear equation in disguise, we will try to solve it. To do so, we must first assume that the following equation is true for some number x.
\n\\(\\frac{x-1}{2}\\) = \\(\\frac{x+\\frac{1}{3}}{4}\\)
\nWe want to make this equation look like the linear equations we are used to. For that reason, we will multiply both sides of the equation by 2 and 4, as we normally do with proportions:
\n2(x+\\(\\frac{1}{3}\\))=4(x-1).
\n\u2192 Is this a linear equation? How do you know?
\n\u2192 Yes, this is a linear equation because the expressions on the left and right of the equal sign are linear expressions.
\n\u2192 Notice that the expressions that contained more than one term were put in parentheses. We do that so we do not make a mistake and forget to use the distributive property.
\n\u2192 Now that we have a linear equation, we will use the distributive property and solve as usual.
\n2(x+\\(\\frac{1}{3}\\))=4(x-1)
\n2x+\\(\\frac{2}{3}\\)=4x-4
\n2x-2x+\\(\\frac{2}{3}\\)=4x-2x-4
\n\\(\\frac{2}{3}\\)=2x-4
\n\\(\\frac{2}{3}\\)+4=2x-4+4
\n\\(\\frac{14}{3}\\)=2x
\n\\(\\frac{1}{2}\\)\u2219\\(\\frac{14}{3}\\)=\\(\\frac{1}{2}\\)\u22192x
\n\\(\\frac{7}{3}\\)=x
\n\u2192 How can we verify that \\(\\frac{7}{3}\\) is the solution to the equation?
\n\u2192 We can replace x with \\(\\frac{7}{3}\\) in the original equation.
\n\"Engage
\nSince \\(\\frac{7}{3}\\) made the equation true, we know it is a solution to the equation.<\/p>\n

Example 2.
\n\u2192 Can we solve the following equation? Explain.
\n\\(\\frac{\\frac{1}{5}-x}{7}\\) = \\(\\frac{2 x+9}{3}\\)
\n\u2192 We need to multiply each numerator with the other fraction\u2019s denominator.
\n\u2192 So,
\n\\(\\frac{\\frac{1}{5}-x}{7}\\) = \\(\\frac{2 x+9}{3}\\)
\n7(2x+9)=3(\\(\\frac{1}{5}\\)-x).
\n\u2192 What would be the next step?
\n\u2192 Use the distributive property.
\n\u2192 Now we have
\n7(2x+9)=3(\\(\\frac{1}{5}\\)-x)
\n14x+63=\\(\\frac{3}{5}\\)-3x
\n14x+3x+63=\\(\\frac{3}{5}\\)-3x+3x
\n17x+63=\\(\\frac{3}{5}\\)
\n17x+63-63=\\(\\frac{3}{5}\\)-63
\n17x=\\(\\frac{3}{5}\\)–\\(\\frac{315}{5}\\)
\n17x=-\\(\\frac{312}{5}\\)
\n\\(\\frac{1}{17}\\) (17x)=(-\\(\\frac{312}{5}\\))\\(\\frac{1}{17}\\)
\nx=-\\(\\frac{312}{85}\\).
\n\u2192 Is this a linear equation? How do you know?
\n\u2192 Yes, this is a linear equation because the left and right side are linear expressions.<\/p>\n

Example 3.
\nCan this equation be solved?
\n\\(\\frac{6+x}{7 x+\\frac{2}{3}}\\)=\\(\\frac{3}{8}\\)
\nGive students a few minutes to work. Provide support to students as needed.
\n\u2192 Yes, we can solve the equation because we can multiply each numerator with the other fraction\u2019s denominator and then use the distributive property to begin solving it.
\nAnswer:
\n\\(\\frac{6+x}{7 x+\\frac{2}{3}}\\)=\\(\\frac{3}{8}\\)
\n(6+x)8=(7x+\\(\\frac{2}{3}\\))3
\n48+8x=21x+2
\n48+8x-8x=21x-8x+2
\n48=13x+2
\n48-2=13x+2-2
\n46=13x
\n\\(\\frac{46}{13}\\)=x<\/p>\n

Example 4.
\nCan this equation be solved?
\n\\(\\frac{7}{3 x+9}\\)=\\(\\frac{1}{8}\\)
\nGive students a few minutes to work. Provide support to students as needed.
\n\u2192 Yes, we can solve the equation because we can multiply each numerator with the other fraction\u2019s denominator and then use the distributive property to begin solving it.
\nAnswer:
\n\\(\\frac{7}{3 x+9}\\)=\\(\\frac{1}{8}\\)
\n7(8)=(3x+9)1
\n56=3x+9
\n56-9=3x+9-9
\n47=3x
\n\\(\\frac{47}{3}\\)=x<\/p>\n

Example 5.
\nIn the diagram below, \u25b3ABC~ \u25b3A’ B’ C’. Using what we know about similar triangles, we can determine the value of x.
\n\"Engage
\n\u2192 Begin by writing the ratios that represent the corresponding sides.
\nAnswer:
\n\\(\\frac{x-2}{9.5}\\) = \\(\\frac{x+2}{12}\\)<\/p>\n

It is possible to write several different proportions in this case. If time, discuss this fact with students.
\n\u2192 Now that we have the ratios, solve for x and find the lengths of \\(\\overline{A B}\\) and \\(\\overline{A C}\\).
\nAnswer:
\n\\(\\frac{x-2}{9.5}\\) = \\(\\frac{x+2}{12}\\)
\n(x-2)12=9.5(x+2)
\n12x-24=9.5x+19
\n12x-24+24=9.5x+19+24
\n12x=9.5x+43
\n12x-9.5x=9.5x-9.5x+43
\n2.5x=43
\nx=\\(\\frac{43}{2.5}\\)
\nx=17.2
\n|AB|=15.2 cm, and |AC|=19.2 cm.<\/p>\n

Eureka Math Grade 8 Module 4 Lesson 8 Exercise Answer Key<\/h3>\n

Solve the following equations of rational expressions, if possible.<\/p>\n

Question 1.
\n\\(\\frac{2 x+1}{9}\\)=\\(\\frac{1-x}{6}\\)
\nAnswer:
\n\\(\\frac{2 x+1}{9}\\)=\\(\\frac{1-x}{6}\\)
\n9(1-x)=(2x+1)6
\n9-9x=12x+6
\n9-9x+9x=12x+9x+6
\n9=21x+6
\n9-6=21x+6-6
\n3=21x
\n\\(\\frac{3}{21}\\)=\\(\\frac{21}{2}\\)1 x
\n\\(\\frac{1}{7}\\)=x<\/p>\n

Question 2.
\n\\(\\frac{5+2 x}{3 x-1}\\)=\\(\\frac{6}{7}\\)
\nAnswer:
\n\\(\\frac{5+2 x}{3 x-1}\\)=\\(\\frac{6}{7}\\)
\n(5+2x)7=(3x-1)6
\n35+14x=18x-6
\n35-35+14x=18x-6-35
\n14x=18x-41
\n14x-18x=18x-18x-41
\n-4x=-41
\n\\(\\frac{-4}{-4}\\) x=\\(\\frac{-41}{-4}\\)
\nx=\\(\\frac{41}{4}\\)<\/p>\n

Question 3.
\n\\(\\frac{x+9}{12}\\)=\\(\\frac{-2 x-\\frac{1}{2}}{3}\\)
\nAnswer:
\n\\(\\frac{x+9}{12}\\)=\\(\\frac{-2 x-\\frac{1}{2}}{3}\\)
\n12(-2x-\\(\\frac{1}{2}\\))=(x+9)3
\n-24x-6=3x+27
\n-24x+24x-6=3x+24x+27
\n-6=27x+27
\n-6-27=27x+27-27
\n-33=27x
\n\\(\\frac{-33}{27}\\)=\\(\\frac{27}{27}\\) x
\n–\\(\\frac{11}{9}\\)=x<\/p>\n

Question 8.
\n\\(\\frac{8}{3-4 x}\\) = \\(\\frac{5}{2 x+\\frac{1}{4}}\\)
\nAnswer:
\n\\(\\frac{8}{3-4 x}\\) = \\(\\frac{5}{2 x+\\frac{1}{4}}\\)
\n8(2x+\\(\\frac{1}{4}\\))=(3-4x)5
\n16x+2=15-20x
\n16x+2-2=15-2-20x
\n16x=13-20x
\n16x+20x=13-20x+20x
\n36x=13
\n\\(\\frac{36}{36}\\) x=\\(\\frac{13}{36}\\)
\nx=\\(\\frac{13}{36}\\)<\/p>\n

Eureka Math Grade 8 Module 4 Lesson 8 Problem Set Answer Key<\/h3>\n

Students practice solving equations with rational expressions, if a solution is possible.<\/p>\n

Solve the following equations of rational expressions, if possible. If an equation cannot be solved, explain why.<\/p>\n

Question 1.
\n\\(\\frac{5}{6 x-2}\\) = \\(\\frac{-1}{x+1}\\)
\nAnswer:
\n\\(\\frac{5}{6 x-2}\\) = \\(\\frac{-1}{x+1}\\)
\n5(x+1)=-1(6x-2)
\n5x+5=-6x+2
\n5x+5-5=-6x+2-5
\n5x=-6x-3
\n5x+6x=-6x+6x-3
\n11x=-3
\nx=-\\(\\frac{3}{11}\\)<\/p>\n

Question 2.
\n\\(\\frac{4-x}{8}\\) = \\(\\frac{7 x-1}{3}\\)
\nAnswer:
\n\\(\\frac{4-x}{8}\\) = \\(\\frac{7 x-1}{3}\\)
\n8(7x-1)=(4-x)3
\n56x-8=12-3x
\n56x-8+8=12+8-3x
\n56x=20-3x
\n56x+3x=20-3x+3x
\n59x=20
\n\\(\\frac{59}{59}\\) x=\\(\\frac{20}{59}\\)
\nx=\\(\\frac{20}{59}\\)<\/p>\n

Question 3.
\n\\(\\frac{3 x}{x+2}\\) = \\(\\frac{5}{9}\\)
\nAnswer:
\n\\(\\frac{3 x}{x+2}\\) = \\(\\frac{5}{9}\\)
\n9(3x)=(x+2)5
\n27x=5x+10
\n27x-5x=5x-5x+10
\n22x=10
\n\\(\\frac{22}{22}\\) x=\\(\\frac{10}{22}\\)
\nx=\\(\\frac{5}{11}\\)<\/p>\n

Question 4.
\n\\(\\frac{\\frac{1}{2} x+6}{3}\\) = \\(\\frac{x-3}{2}\\)
\nAnswer:
\n\\(\\frac{\\frac{1}{2} x+6}{3}\\) = \\(\\frac{x-3}{2}\\)
\n3(x-3)=2(\\(\\frac{1}{2}\\) x+6)
\n3x-9=x+12
\n3x-9+9=x+12+9
\n3x=x+21
\n3x-x=x-x+21
\n2x=21
\nx=\\(\\frac{21}{2}\\)<\/p>\n

Question 5.
\n\\(\\frac{7-2 x}{6}\\) = \\(\\frac{x-5}{1}\\)
\nAnswer:
\n\\(\\frac{7-2 x}{6}\\) = \\(\\frac{x-5}{1}\\)
\n6(x-5)=(7-2x)1
\n6x-30=7-2x
\n6x-30+30=7+30-2x
\n6x=37-2x
\n6x+2x=37-2x+2x
\n8x=37
\n\\(\\frac{8}{8}\\) x=\\(\\frac{37}{8}\\)
\nx=\\(\\frac{37}{8}\\)<\/p>\n

Question 6.
\n\\(\\frac{2 x+5}{2}\\) = \\(\\frac{3 x-2}{6}\\)
\nAnswer:
\n\\(\\frac{2 x+5}{2}\\) = \\(\\frac{3 x-2}{6}\\)
\n2(3x-2)=6(2x+5)
\n6x-4=12x+30
\n6x-4+4=12x+30+4
\n6x=12x+34
\n6x-12x=12x-12x+34
\n-6x=34
\nx=-\\(\\frac{34}{6}\\)
\nx=-\\(\\frac{17}{3}\\)<\/p>\n

Question 7.
\n\\(\\frac{6 x+1}{3}\\) = \\(\\frac{9-x}{7}\\)
\nAnswer:
\n\\(\\frac{6 x+1}{3}\\) = \\(\\frac{9-x}{7}\\)
\n(6x+1)7=3(9-x)
\n42x+7=27-3x
\n42x+7-7=27-7-3x
\n42x=20-3x
\n42x+3x=20-3x+3x
\n45x=20
\n\\(\\frac{45}{45}\\)x=\\(\\frac{20}{45}\\)
\nx=\\(\\frac{4}{9}\\)<\/p>\n

Question 8.
\n\\(\\frac{\\frac{1}{3} x-8}{12}\\) = \\(\\frac{-2-x}{15}\\)
\nAnswer:
\n\\(\\frac{\\frac{1}{3} x-8}{12}\\) = \\(\\frac{-2-x}{15}\\)
\n12(-2-x)=(\\(\\frac{1}{3}\\) x-8)15
\n-24-12x=5x-120
\n-24-12x+12x=5x+12x-120
\n-24=17x-120
\n-24+120=17x-120+120
\n96=17x
\n\\(\\frac{96}{17}\\)=\\(\\frac{17}{17}\\) x
\n\\(\\frac{96}{17}\\)=x<\/p>\n

Question 9.
\n\\(\\frac{3-x}{1-x}\\)=\\(\\frac{3}{2}\\)
\nAnswer:
\n\\(\\frac{3-x}{1-x}\\)=\\(\\frac{3}{2}\\)
\n(3-x)2=(1-x)3
\n6-2x=3-3x
\n6-2x+2x=3-3x+2x
\n6=3-x
\n6-3=3-3-x
\n3=-x
\n-3=x<\/p>\n

Question 10.
\nIn the diagram below, \u25b3ABC~ \u25b3A’ B’ C’. Determine the lengths of \\(\\overline{A C}\\) and \\(\\overline{B C}\\).
\n\"Eureka
\nAnswer:
\n\\(\\frac{x+4}{4.5}\\) = \\(\\frac{3 x-2}{9}\\)
\n9(x+4)=4.5(3x-2)
\n9x+36=13.5x-9
\n9x+36+9=13.5x-9+9
\n9x+45=13.5x
\n9x-9x+45=13.5x-9x
\n45=4.5x
\n10=x
\n|AC|=14 cm, and |BC|=28 cm.<\/p>\n

Eureka Math Grade 8 Module 4 Lesson 8 Exit Ticket Answer Key<\/h3>\n

Solve the following equations for x.<\/p>\n

Question 1.
\n\\(\\frac{5 x-8}{3}\\) = \\(\\frac{11 x-9}{5}\\)
\nAnswer:
\n\\(\\frac{5 x-8}{3}\\) = \\(\\frac{11 x-9}{5}\\)
\n5(5x-8)=3(11x-9)
\n25x-40=33x-27
\n25x-25x-40=33x-25x-27
\n-40=8x-27
\n-40+27=8x-27+27
\n-13=8x
\n–\\(\\frac{13}{8}\\)=x<\/p>\n

Question 2.
\n\\(\\frac{x+11}{7}\\)=\\(\\frac{2 x+1}{-8}\\)
\nAnswer:
\n\\(\\frac{x+11}{7}\\)=\\(\\frac{2 x+1}{-8}\\)
\n7(2x+1)=-8(x+11)
\n14x+7=-8x-88
\n14x+7-7=-8x-88-7
\n14x=-8x-95
\n14x+8x=-8x+8x-95
\n22x=-95
\nx=-\\(\\frac{95}{22}\\)<\/p>\n

Question 3.
\n\\(\\frac{-x-2}{-4}\\) = \\(\\frac{3 x+6}{2}\\)
\nAnswer:
\n\\(\\frac{-x-2}{-4}\\) = \\(\\frac{3 x+6}{2}\\)
\n-4(3x+6)=2(-x-2)
\n-12x-24=-2x-4
\n-12x-24+24=-2x-4+24
\n-12x=-2x+20
\n-12x+2x=-2x+2x+20
\n-10x=20
\nx=-2<\/p>\n","protected":false},"excerpt":{"rendered":"

Engage NY Eureka Math 8th Grade Module 4 Lesson 8 Answer Key Eureka Math Grade 8 Module 4 Lesson 8 Example Answer Key Example 1. \u2192 Given a linear equation in disguise, we will try to solve it. To do so, we must first assume that the following equation is true for some number x. … 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\/85293"}],"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=85293"}],"version-history":[{"count":1,"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/posts\/85293\/revisions"}],"predecessor-version":[{"id":137164,"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/posts\/85293\/revisions\/137164"}],"wp:attachment":[{"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/media?parent=85293"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/categories?post=85293"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/tags?post=85293"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}