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<\/p>\nExamples 1\u20132
\nWork through the examples as a whole class.<\/p>\n
Example 1.
\nConsider the equation \\(\\frac{1}{x}\\)=\\(\\frac{3}{x-2}\\).
\na. Rewrite the equation into a system of equations.
\nAnswer:
\n\\(\\frac{1}{x}\\)=\\(\\frac{3}{x-2}\\) and x\u22600 and x\u22602<\/p>\n
b. Solve the equation for x, excluding the value(s) of x that lead to a denominator of zero.
\nAnswer:
\nx=-1 and x\u22600 and x\u22602 Solution set: {-1}<\/p>\n
Example 2.
\nConsider the equation \\(\\frac{x+3}{x-2}\\)=\\(\\frac{5}{x-2}\\).
\na. Rewrite the equation into a system of equations.
\nAnswer:
\n\\(\\frac{x+3}{x-2}\\)=\\(\\frac{5}{x-2}\\) and x\u22602<\/p>\n
b. Solve the equation for x, excluding the value(s) of x that lead to a denominator of zero.
\nAnswer:
\nx=2 and x\u22602 Solution set: \u0278<\/p>\n
Emphasize the process of recognizing a rational equation as a system composed of the equation itself and the excluded value(s) of x. For Example 1, this is really the compound statement:
\n\\(\\frac{1}{x}\\)=\\(\\frac{3}{x-2}\\) and x\u22600 and x-2\u22600
\n\u2192 By the properties of equality, we can multiply through by nonzero quantities. Within this compound statement, x and x-2 are nonzero, so we may write x-2=3x and x\u22600 and x-2\u22600, which is equivalent to -2=2x and x\u22600 and x\u22602.
\n\u2192 All three declarations in this compound statement are true for x=-1. This is the solution set.
\nIn Example 2, remind students of the previous lesson on solving equations involving factored expressions. Students will need to factor out the common factor and then apply the zero-product property.
\n\u2192 What happens in Example 2 when we have x=2 and x\u22602? Both declarations cannot be true. What can we say about the solution set of the equation?
\n\u2192 There is no solution.<\/p>\n
<\/p>\n
Opening Exercise
\nNolan says that he checks the answer to a division problem by performing multiplication. For example, he says that
\n20\u00f74=5 is correct because 5\u00d74 is 20, and \\(\\frac{3}{\\frac{1}{2}}\\)=6 is correct because 6\u00d7\\(\\frac{1}{2}\\) is 3.
\na. Using Nolan\u2019s reasoning, explain why there is no real number that is the answer to the division problem
\n5\u00f70.
\nAnswer:
\nThere is no number n such that 0\u00d7n=5.<\/p>\n
b. Quentin says that \\(\\frac{0}{0}\\)=17. What do you think?
\nAnswer:
\nWhile it is true that 0\u00d717=0, the problem is that by that principle \\(\\frac{0}{0}\\) could equal any number.<\/p>\n
c. Mavis says that the expression \\(\\frac{3 x-6}{x-2}\\) has a meaningful value for whatever value one chooses to assign to x. Do you agree?
\nAnswer:
\nNo. The expression does not have a meaningful value when x=-2.<\/p>\n
d. Bernoit says that the expression \\(\\frac{3 x-6}{x-2}\\) always has the value 3 for whichever value one assigns to x. Do you agree?
\nAnswer:
\nThe expression does equal 3 for all values of x except x=2.<\/p>\n
Note that the problem with \\(\\frac{0}{0}\\) is that too many numbers pass Nolan\u2019s criterion! Have students change 17 to a different number. It still passes Nolan\u2019s multiplication check. Like \\(\\frac{5}{0}\\), it is a problematic notion. For this reason, we want to disallow the possibility of ever dividing by zero.
\nPoint out that an expression like \\(\\frac{5}{x+2}\\) is really accompanied with the clause \u201cunder the assumption the denominator is not zero.\u201d So, \\(\\frac{5}{x+2}\\) should be read as a compound statement:
\n\\(\\frac{5}{x+2}\\) and x+2\u22600 OR \\(\\frac{5}{x+2}\\) and x\u2260-2<\/p>\n
Exercises 1\u20132
\nGive students a few minutes to complete the problems individually. Then, elicit answers from students.<\/p>\n
Exercise 1.
\nRewrite \\(\\frac{10}{x+5}\\) as a compound statement.
\nAnswer:
\n\\(\\frac{10}{x+5}\\) and x\u2260-5<\/p>\n
Exercise 2.
\nConsider \\(\\frac{x^{2}-25}{\\left(x^{2}-9\\right)(x+4)}\\).<\/p>\n
a. Is it permissible to let x=5 in this expression?
\nAnswer:
\nYes. \\(\\frac{0}{144}\\)=0<\/p>\n
b. Is it permissible to let x=3 in this expression?
\nAnswer:
\nNo. –\\(\\frac{16}{0}\\) is not permissible.<\/p>\n
c. Give all the values of x that are not permissible in this expression.
\nAnswer:
\nx\u2260-3, 3, -4<\/p>\n
Exercises 3\u201311
\nAllow students time to complete the problems individually. Then, have students compare their work with another student. Make sure that students are setting up a system of equations as part of their solution.
\nWork through Exercises 11 as a class.<\/p>\n
Exercises 3\u201311
\nRewrite each equation into a system of equations excluding the value(s) of x that lead to a denominator of zero; then, solve the equation for x.<\/p>\n
Exercise 3.
\n\\(\\frac{5}{x}\\)=1
\nAnswer:
\n\\(\\frac{5}{x}\\)=1 and x \u2260 0
\n{5}<\/p>\n
Exercise 4.
\n\\(\\frac{1}{x-5}\\)=3
\nAnswer:
\n\\(\\frac{1}{x-5}\\)=3 and x \u2260 5
\n{\\(\\frac{16}{3}\\)}<\/p>\n
Exercise 5.
\n\\(\\frac{x}{x+1}\\)=4
\nAnswer:
\n\\(\\frac{x}{x+1}\\)=4 and x \u2260 -1
\n{-\\(\\frac{4}{3}\\)}<\/p>\n
Exercise 6.
\n\\(\\frac{2}{x}\\)=\\(\\frac{3}{x-4}\\)
\nAnswer:
\n\\(\\frac{2}{x}\\)=\\(\\frac{3}{x-4}\\) and x\u22600 and x\u22604
\n{-8}<\/p>\n
Exercise 7.
\n\\(\\frac{x}{x+6}\\)=-\\(\\frac{6}{x+6}\\)
\nAnswer:
\n\\(\\frac{x}{x+6}\\)=-\\(\\frac{6}{x+6}\\) and x\u2260-6
\nNo solution<\/p>\n
Exercise 8.
\n\\(\\frac{x-3}{x+2}\\)=0
\nAnswer:
\n\\(\\frac{x-3}{x+2}\\)=0 and x\u2260-2
\n{3}<\/p>\n
Exercise 9.
\n\\(\\frac{x+3}{x+3}\\)=5
\nAnswer:
\n\\(\\frac{x+3}{x+3}\\)=5 and x\u2260-3
\nNo solution<\/p>\n
Exercise 10.
\n\\(\\frac{x+3}{x+3}\\)=1
\nAnswer:
\n\\(\\frac{x+3}{x+3}\\)=1 and x\u2260-3
\nAll real numbers except -3<\/p>\n
Exercise 11.
\nA baseball player\u2019s batting average is calculated by dividing the number of times a player got a hit by the total number of times the player was at bat. It is expressed as a decimal rounded to three places. After the first 10 games of the season, Samuel had 12 hits off of 33 at bats.
\na. What is his batting average after the first 10 games?
\nAnswer:
\n\\(\\frac{12}{33}\\)\u22480 .364<\/p>\n
b. How many hits in a row would he need to get to raise his batting average to above 0.500?
\nAnswer:
\n\\(\\frac{12+x}{33+x}\\)>0.500
\nx>9
\nHe would need 10 hits in a row to be above 0.500.<\/p>\n
c. How many at bats in a row without a hit would result in his batting average dropping below 0.300?
\nAnswer:
\n\\(\\frac{12}{33+x}\\)<0.300 x > 7
\nIf he went 8 at bats in a row without a hit, he would be below 0.300.<\/p>\n
Ask:
\n\u2192 What was the difference between Exercises 9 and 10? How did that affect the solution set?<\/p>\n
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<\/p>\n
Question 1.
\nConsider the equation \\(\\frac{10\\left(x^{2}-49\\right)}{3 x\\left(x^{2}-4\\right)(x+1)}\\) =0. Is x=7 permissible? Which values of x are excluded? Rewrite as a system of equations.
\nAnswer:
\nYes, x=7 is permissible. The excluded values are 0, \u00b12, and -1. The system is \\(\\frac{10\\left(x^{2}-49\\right)}{3 x\\left(x^{2}-4\\right)(x+1)}\\) =0 and x\u22600 and x\u2260-2 and x\u2260-1 and x\u22602.<\/p>\n
Question 2.
\nRewrite each equation as a system of equations excluding the value(s) of x that lead to a denominator of zero. Then, solve the equation for x.
\na. 25x=\\(\\frac{1}{x}\\)
\nAnswer:
\nSystem: 25x=\\(\\frac{1}{x}\\) and x\u22600; solution set: {\u00b1\\(\\frac{1}{5}\\)}<\/p>\n
b. \\(\\frac{1}{5x}\\)=10
\nAnswer:
\nSystem: \\(\\frac{1}{5}\\)x=10 and x\u22600; solution set: {\\(\\frac{1}{5}\\)0}<\/p>\n
c. \\(\\frac{x}{7-x}\\)=2x
\nAnswer:
\nSystem: \\(\\frac{x}{7-x}\\)=2x and x\u22607; solution set: {0,\\(\\frac{13}{2}\\)}<\/p>\n
d. \\(\\frac{2}{x}\\)=\\(\\frac{5}{x+1}\\)
\nAnswer:
\nSystem: \\(\\frac{2}{x}\\)=\\(\\frac{5}{x+1}\\) and x\u2260-1 and x\u22600; solution set: {\\(\\frac{2}{3}\\)}<\/p>\n
e. \\(\\frac{3+x}{3-x}\\)=\\(\\frac{3+2x}{3-2x}\\)
\nAnswer:
\nSystem: \\(\\frac{3+x}{3-x}\\)=\\(\\frac{3+2x}{3-2x}\\) and x\u22603\/2 and x\u22603; solution set: {0}<\/p>\n
Question 3.
\nRoss wants to cut a 40-foot rope into two pieces so that the length of the first piece divided by the length of the second piece is 2.
\na. Let x represent the length of the first piece. Write an equation that represents the relationship between the pieces as stated above.
\nAnswer:
\n\\(\\frac{x}{40-x}\\)=2<\/p>\n
b. What values of x are not permissible in this equation? Describe within the context of the problem what situation is occurring if x were to equal this value(s). Rewrite as a system of equations to exclude the value(s).
\nAnswer:
\n40 is not a permissible value because it would mean the rope is still intact. System: \\(\\frac{x}{40-x}\\)=2 and x\u226040<\/p>\n
c. Solve the equation to obtain the lengths of the two pieces of rope. (Round to the nearest tenth if necessary.)
\nAnswer:
\nFirst piece is \\(\\frac{80}{3}\\)\u224826.7 feet long; second piece is 40\/3\u224813.3 feet long.<\/p>\n
Question 4.
\nWrite an equation with the restrictions x\u226014, x\u22602, and x\u22600.
\nAnswer:
\nAnswers will vary. Sample equation: \\(\\frac{1}{x}\\)(x-2)(x-14) =0<\/p>\n
Question 5.
\nWrite an equation that has no solution.
\nAnswer:
\nAnswers will vary. Sample equation: \\(\\frac{1}{x}\\)=0<\/p>\n
Question 1.
\nRewrite the equation \\(\\frac{x-2}{x-9}\\)=2 as a system of equations. Then, solve for x.
\nAnswer:
\n\\(\\frac{x-2}{x-9}\\)=2 and x \u2260 9
\nx-2=2(x-9)
\nx-2=2x-18
\n16=x
\n{16}<\/p>\n
Question 2.
\nWrite an equation that would have the restriction x\u2260-3.
\nAnswer:
\nSample answer: \\(\\frac{4}{x+3}\\)=2<\/p>\n","protected":false},"excerpt":{"rendered":"
Engage NY Eureka Math Algebra 1 Module 1 Lesson 18 Answer Key Eureka Math Algebra 1 Module 1 Lesson 18 Example Answer Key Examples 1\u20132 Work through the examples as a whole class. Example 1. Consider the equation =. a. Rewrite the equation into a system of equations. Answer: = and x\u22600 and x\u22602 b. … Read more<\/a><\/p>\n","protected":false},"author":3,"featured_media":171565,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[6],"tags":[],"jetpack_sharing_enabled":true,"jetpack_featured_media_url":"https:\/\/ccssanswers.com\/wp-content\/uploads\/2022\/05\/Eureka-Math-Algebra-1-Module-1-Lesson-18-Answer-Key.png","_links":{"self":[{"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/posts\/87654"}],"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=87654"}],"version-history":[{"count":3,"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/posts\/87654\/revisions"}],"predecessor-version":[{"id":171569,"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/posts\/87654\/revisions\/171569"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/media\/171565"}],"wp:attachment":[{"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/media?parent=87654"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/categories?post=87654"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/tags?post=87654"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}