Lesson 23 NYS COMMON CORE MATHEMATICS CURRICULUM 7β’2 Lesson 23: Solving Equations Using Algebra Student Outcomes ο§ Students use algebra to solve equations (of the form ππ₯ + π = π and π(π₯ + π) = π, where π, π, and π are specific rational numbers) using techniques of making zero (adding the additive inverse) and making one (multiplying by the multiplicative inverse) to solve for the variable. ο§ Students identify and compare the sequence of operations used to find the solution to an equation algebraically, with the sequence of operations used to solve the equation with tape diagrams. They recognize the steps as being the same. ο§ Students solve equations for the value of the variable using inverse operations by making zero (adding the additive inverse) and making one (multiplying by the multiplicative inverse). Classwork As in Lesson 22, students continue solving equations using properties of equality and inverse operations to relate their MP.1 steps to the steps taken when solving problems algebraically. In this lesson, students decontextualize word problems to MP.2 & create equations that model given situations. Students justify their solutions by comparing their algebraic steps to the MP.3 steps taken when using a tape diagram. Have students work in cooperative groups and share out their solutions on chart paper. Use the class discussion as a way to have students view the differences in problem-solving approaches. Exercises (35 minutes) Exercises 1. Youth Group Trip The youth group is going on a trip to an amusement park in another part of the state. The trip costs each group member $πππ, which includes $ππ for the hotel and two one-day combination entrance and meal plan passes. a. Write an equation representing the cost of the trip. Let π· be the cost of the park pass. Provide a review card showing examples of fraction multiplication and division for students who do not have adequate prerequisite skills. ππ + ππ· = πππ Lesson 23: Scaffolding: Solving Equations Using Algebra This work is derived from Eureka Math β’ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from G7-M2-TE-1.3.0-08.2015 273 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. Lesson 23 NYS COMMON CORE MATHEMATICS CURRICULUM b. 7β’2 Solve the equation algebraically to find the cost of the park pass. Then write the reason that justifies each step using if-then statements. If: ππ + ππ· = πππ, Then: ππ β ππ + ππ· = πππ β ππ Subtraction property of equality for the additive inverse of ππ If: π + ππ· = ππ Then: ππ· = ππ Additive identity If: ππ· = ππ π π π π Then: ( ) ππ· = ( ) ππ Multiplication property of equality using the multiplicative inverse of π If: ππ· = ππ. π Then: π· = ππ. π Multiplicative identity The park pass costs $ππ. ππ. c. Model the problem using a tape diagram to check your work. $πππ πππ β ππ = ππ $ππ Hotel Fee ππ ÷ π = ππ. ππ Park Pass Park Pass Suppose you want to buy your favorite ice cream bar while at the amusement park, and it costs $π. ππ. If you purchase the ice cream bar and π bottles of water, pay with a $ππ bill, and receive no change, then how much did each bottle of water cost? d. Write an equation to model this situation. πΎ: the cost of one bottle of water π. ππ + ππΎ = ππ e. Solve the equation to determine the cost of one water bottle. Then write the reason that justifies each step using if-then statements. If: π. ππ + ππΎ = ππ Then: π. ππ β π . ππ + ππΎ = ππ β π. ππ Subtraction property of equality for the additive inverse of π. ππ If: π + ππΎ = π. ππ Then: ππΎ = π. ππ Additive identity If: ππΎ = π. ππ Then: π π π (ππΎ) = (π. ππ) π Multiplication property of equality using the multiplicative inverse of π If: ππΎ = π. ππ Then: πΎ = π. ππ Multiplicative identity A bottle of water costs $π. ππ. Lesson 23: Solving Equations Using Algebra This work is derived from Eureka Math β’ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from G7-M2-TE-1.3.0-08.2015 274 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. Lesson 23 NYS COMMON CORE MATHEMATICS CURRICULUM f. 7β’2 Model the problem using a tape diagram to check your work. ππ β π. ππ = π. ππ $π. ππ Ice cream πΎ πΎ πΎ π. ππ = π. ππ π $ππ 2. Weekly Allowance Charlotte receives a weekly allowance from her parents. She spent half of this weekβs allowance at the movies but earned an additional $π for performing extra chores. If she did not spend any additional money and finished the week with $ππ, what is Charlotteβs weekly allowance? a. Write an equation that can be used to find the original amount of Charlotteβs weekly allowance. Let π¨ be the value of Charlotteβs original weekly allowance. π π¨ + π = ππ π b. Solve the equation to find the original amount of allowance. Then write the reason that justifies each step using if-then statements. If: π π π¨ + π = ππ Then: If: π π π π¨ + π β π = ππ β π Subtraction property of equality for the additive inverse of π π¨+π=π Then: If : π π π π π π¨=π Additive identity π¨=π π π Then: (π) π¨ = (π)π Multiplication property of equality using the multiplicative inverse of π π If: ππ¨ = ππ Then: π¨ = ππ Multiplicative identity The original allowance was $ππ. c. Explain your answer in the context of this problem. Charlotteβs weekly allowance is $ππ. d. Charlotteβs goal is to save $πππ for her beach trip at the end of the summer. Use the amount of weekly allowance you found in part (c) to write an equation to determine the number of weeks that Charlotte must work to meet her goal. Let π represent the number of weeks. ππ π = πππ π π ( ) πππ = ( ) πππ ππ ππ ππ = π. ππ π = π. ππ Lesson 23: Solving Equations Using Algebra This work is derived from Eureka Math β’ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from G7-M2-TE-1.3.0-08.2015 275 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. Lesson 23 NYS COMMON CORE MATHEMATICS CURRICULUM e. 7β’2 In looking at your answer to part (d) and based on the story above, do you think it will take Charlotte that many weeks to meet her goal? Why or why not? Charlotte needs more than π weeksβ allowance, so she will need to save π weeksβ allowance (and not spend any of it). There are ππβππ weeks in the summer; so, yes, she can do it. 3. Travel Baseball Team Allen is very excited about joining a travel baseball team for the fall season. He wants to determine how much money he should save to pay for the expenses related to this new team. Players are required to pay for uniforms, travel expenses, and meals. a. If Allen buys π uniform shirts at one time, he gets a $ππ. ππ discount so that the total cost of π shirts would be $ππ. Write an algebraic equation that represents the regular price of one shirt. Solve the equation. Write the reason that justifies each step using if-then statements. π: the cost of one shirt If: ππ β ππ = ππ Then: ππ β ππ + ππ = ππ + ππ Addition property of equality using the additive inverse of βππ If: ππ + π = ππ Then: ππ = ππ Additive identity If: ππ = ππ π π π π Then: ( ) ππ = ( ) ππ Multiplication property of equality using multiplicative inverse of π If: ππ = ππ. ππ Then: π = ππ. ππ b. Multiplicative identity What is the cost of one shirt without the discount? The cost of one shirt is $ππ. ππ. c. What is the cost of one shirt with the discount? ππ = ππ π π ( ) ππ = ( ) ππ π π ππ = ππ π = ππ The cost of one shirt with the discount is $ππ. ππ. d. How much more do you pay per shirt if you buy them one at a time (rather than in bulk)? ππ. ππ β ππ. ππ = π. ππ One shirt costs $ππ if you buy them in bulk. So, Allen would pay $π. ππ more per shirt if he bought them one at a time. Allenβs team was also required to buy two pairs of uniform pants and two baseball caps, which total $ππ. A pair of pants costs $ππ more than a baseball cap. e. Write an equation that models this situation. Let π represent the cost of a baseball cap. π(cap + π pair of pants) = ππ π(π + π + ππ) = ππ Lesson 23: or π ( ππ + ππ) = ππ or Solving Equations Using Algebra This work is derived from Eureka Math β’ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from G7-M2-TE-1.3.0-08.2015 ππ + ππ = ππ 276 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. Lesson 23 NYS COMMON CORE MATHEMATICS CURRICULUM f. 7β’2 Solve the equation algebraically to find the cost of a baseball cap. Write the reason that justifies each step using if-then statements. If: π (π π + ππ) = ππ Then: π π π π ( ) (π )(π π + ππ) = ( ) ππ Multiplication property of equality using the multiplicative inverse of π If: π(ππ + ππ) = ππ Then: ππ + ππ = ππ Multiplicative identity If: ππ + ππ = ππ Then: ππ + ππ β ππ = ππ β ππ Subtraction property of equality for the additive inverse of ππ If: ππ + π = ππ Then: ππ = ππ Additive identity If: ππ = ππ Then: π π π π ( ) ππ = ( ) ππ Multiplication property of equality using the multiplicative inverse of π If: ππ = ππ Then: π = ππ g. Multiplicative identity Model the problem using a tape diagram in order to check your work from part (f). ππ π h. π ππ ππ β ππ = ππ π π ππ ππ = ππ π What is the cost of one cap? The cost of one cap is $ππ. i. What is the cost of one pair of pants? ππ + ππ = ππ The cost of one pair of pants is $ππ. Closing (5 minutes) ο§ How do we translate a word problem into an equation? For instance, in Exercise 1 about the youth group trip, what key words and statements helped you determine the operations and values used in the equation? οΊ ο§ Answers will vary, but students should talk about key words to determine the total value, the constant value, and the coefficient. How do we make sense of a word problem and model it with an equation? οΊ Answers will vary and may be similar to answers to the previous question. Lesson 23: Solving Equations Using Algebra This work is derived from Eureka Math β’ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from G7-M2-TE-1.3.0-08.2015 277 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. NYS COMMON CORE MATHEMATICS CURRICULUM Lesson 23 7β’2 Lesson Summary Equations are useful to model and solve real-world problems. The steps taken to solve an algebraic equation are the same steps used in an arithmetic solution. Exit Ticket (5 minutes) Lesson 23: Solving Equations Using Algebra This work is derived from Eureka Math β’ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from G7-M2-TE-1.3.0-08.2015 278 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. Lesson 23 NYS COMMON CORE MATHEMATICS CURRICULUM Name 7β’2 Date Lesson 23: Solving Equations Using Algebra Exit Ticket Andrewβs math teacher entered the seventh-grade students in a math competition. There was an enrollment fee of $30 and also an $11 charge for each packet of 10 tests. The total cost was $151. How many tests were purchased? Set up an equation to model this situation, solve it using if-then statements, and justify the reasons for each step in your solution. Lesson 23: Solving Equations Using Algebra This work is derived from Eureka Math β’ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from G7-M2-TE-1.3.0-08.2015 279 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. Lesson 23 NYS COMMON CORE MATHEMATICS CURRICULUM 7β’2 Exit Ticket Sample Solutions Andrewβs math teacher entered the seventh-grade students in a math competition. There was an enrollment fee of $ππ and also an $ππ charge for each packet of ππ tests. The total cost was $πππ. How many tests were purchased? Set up an equation to model this situation, solve it using if-then statements, and justify the reasons for each step in your solution. Let π represent the number of test packets. Enrollment fee + cost of test = πππ If: ππ + πππ = πππ Then: ππ β ππ + πππ = πππ β ππ Subtraction property of equality for the additive inverse of ππ If: π + πππ = πππ Then: πππ = πππ Additive identity If: πππ = πππ Then: π ππ (πππ) = π ππ (πππ) Multiplication property of equality using the multiplicative inverse of ππ If: ππ = ππ Then: π = ππ Multiplicative identity Andrewβs math teacher bought ππ packets of tests. There were ππ tests in each packet, and ππ × ππ = πππ. So, there were πππ tests purchased. Problem Set Sample Solutions For Exercises 1β4, solve each equation algebraically using if-then statements to justify your steps. 1. π π β π = ππ π π If: π π β π = ππ Then: If: π π π π π π β π + π = ππ + π Addition property of equality using the additive inverse of βπ π + π = ππ Then: If: π π π π = ππ Additive identity π = ππ π π π π π π Then: ( ) π = ( ) ππ Multiplication property of equality using the multiplicative inverse of π π If: ππ = ππ Then: π = ππ Lesson 23: Multiplicative identity Solving Equations Using Algebra This work is derived from Eureka Math β’ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from G7-M2-TE-1.3.0-08.2015 280 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. NYS COMMON CORE MATHEMATICS CURRICULUM 2. π= Lesson 23 7β’2 βπ+π π βπ+π π If: π = Then: π (π) = π ( βπ+π ) π Multiplication property of equality using the multiplicative inverse of π π If: π = π (βπ + π) Then: π = βπ + π Multiplicative identity If: π = βπ + π Then: π β (βπ) = βπ β (βπ) + π Subtraction property of equality for the additive inverse of βπ If: π = π + π Then: π = π 3. Additive identity ππ(π + π) = βπππ If: ππ(π + π) = βπππ Then: ( π π ) ππ(π + π) = ( ) (βπππ) ππ ππ Multiplication property of equality using the multiplicative inverse of ππ If: π (π + π) = βπ Then: π + π = βπ Multiplicative identity If: π + π = βπ Then: π + π β π = βπ β π Subtraction property of equality for the additive inverse of π If: π + π = βππ Then: π = βππ 4. Additive identity ππ + ππ = βπ If: ππ + ππ = βπ Then: ππ + ππ β ππ = βπ β ππ Subtraction property of equality for the additive inverse of ππ If: ππ + π = βππ Then: ππ = βππ Additive identity If: ππ = βππ π π π π Then: ( ) ππ = ( ) (βππ) Multiplication property of equality using the multiplicative inverse of π If: ππ = βπ. π Then: π = βπ. π Lesson 23: Multiplicative identity Solving Equations Using Algebra This work is derived from Eureka Math β’ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from G7-M2-TE-1.3.0-08.2015 281 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. Lesson 23 NYS COMMON CORE MATHEMATICS CURRICULUM 7β’2 For Exercises 5β7, write an equation to represent each word problem. Solve the equation showing the steps, and then state the value of the variable in the context of the situation. 5. A plumber has a very long piece of pipe that is used to run city water parallel to a major roadway. The pipe is cut into two sections. One section of pipe is ππ ππ. shorter than the other. If π π of the length of the shorter pipe is πππ ππ., how long is the longer piece of the pipe? Let π represent the longer piece of pipe. If: π π (π β ππ) = πππ Then: π π π ( ) (π β ππ) = (π) πππ π π Multiplication property of equality using the multiplicative inverse of π π If: π(π β ππ) = πππ Then: π β ππ = πππ Multiplicative identity If: π β ππ = πππ Then: π β ππ + ππ = πππ + ππ Addition property of equality for the additive inverse of βππ If: π + π = πππ Then: π = πππ Additive identity The longer piece of pipe is πππ ππ. 6. Bobβs monthly phone bill is made up of a $ππ fee plus $π. ππ per minute. Bobβs phone bill for July was $ππ. Write an equation to model the situation using π to represent the number of minutes. Solve the equation to determine the number of phone minutes Bob used in July. Let π represent the number of phone minutes Bob used. If: ππ + π. πππ = ππ Then: ππ β ππ + π. πππ = ππ β ππ Subtraction property of equality for the additive inverse of ππ If: π + π. πππ = ππ Then: π. πππ = ππ Additive identity If: π. πππ = ππ Then: ( π π ) π. πππ = ( ) ππ π.ππ π.ππ Multiplication property of equality using the multiplicative inverse of π. ππ If: ππ = πππ Then: π = πππ Multiplicative identity Bob used πππ phone minutes in July. Lesson 23: Solving Equations Using Algebra This work is derived from Eureka Math β’ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from G7-M2-TE-1.3.0-08.2015 282 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. Lesson 23 NYS COMMON CORE MATHEMATICS CURRICULUM 7. 7β’2 Kym switched cell phone plans. She signed up for a new plan that will save her $π. ππ per month compared to her old cell phone plan. The cost of the new phone plan for an entire year is $πππ. How much did Kym pay per month under her old phone plan? Let π represent the amount Kym paid per month for her old cell phone plan. If: πππ = ππ(π β π. ππ) Then: ( π π ) (πππ) = ( ) ππ(π β π. ππ) ππ ππ Multiplication property of equality using the multiplicative inverse of ππ If: ππ. π = π (π β π. ππ) Then: ππ. π = π β π. ππ Multiplicative identity If: ππ. π = π β π. ππ Then: ππ. π + π. ππ = π β π. ππ + π. ππ Addition property of equality for the additive inverse of βπ. ππ If: ππ = π + π Then: ππ = π Additive identity Kym paid $ππ per month for her old cell phone plan. Lesson 23: Solving Equations Using Algebra This work is derived from Eureka Math β’ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from G7-M2-TE-1.3.0-08.2015 283 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License.
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