NYS COMMON CORE MATHEMATICS CURRICULUM Lesson 12 M1 ALGEBRA I Lesson 12: Solving Equations Student Outcomes ο§ Students are introduced to the formal process of solving an equation: starting from the assumption that the original equation has a solution. Students explain each step as following from the properties of equality. Students identify equations that have the same solution set. Classwork Opening Exercise (4 minutes) Opening Exercise Answer the following questions. a. Why should the equations (π β π)(π + π) = ππ + π and (π β π)(π + π) = π + ππ have the same solution set? The commutative property b. Why should the equations (π β π)(π + π) = ππ + π and (π + π)(π β π) = ππ + π have the same solution set? The commutative property c. Do you think the equations (π β π)(π + π) = ππ + π and (π β π)(π + π) + πππ = πππ + π should have the same solution set? Why? Yes. πππ was added to both sides. d. Do you think the equations (π β π)(π + π) = ππ + π and π(π β π)(π + π) = ππ + ππ should have the same solution set? Explain why. Yes. Both sides were multiplied by π. Discussion (4 minutes) Allow students to attempt to justify their answers for parts (a) and (b) above. Then, summarize with the following: ο§ We know that the commutative and associative properties hold for all real numbers. We also know that variables are placeholders for real numbers, and the value(s) assigned to a variable that make an equation true is the βsolution.β If we apply the commutative and associative properties of real numbers to an expression, we obtain an equivalent expression. Therefore, equations created this way (by applying the commutative and associative properties to one or both expressions) consist of expressions equivalent to those in the original equation. ο§ In other words, if π₯ is a solution to an equation, then it will also be a solution to any new equation we make by applying the commutative and associative properties to the expression in that equation. Lesson 12: Solving Equations This work is derived from Eureka Math β’ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from ALG I-M1-TE-1.3.0-07.2015 160 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. Lesson 12 NYS COMMON CORE MATHEMATICS CURRICULUM M1 ALGEBRA I Exercise 1 (3 minutes) Exercise 1 a. Use the commutative property to write an equation that has the same solution set as ππ β ππ + π = (π + π)(π β ππ)(π). βππ + ππ + π = (π + π)(π)(π β ππ) b. Use the associative property to write an equation that has the same solution set as ππ β ππ + π = (π + π)(π β ππ)(π). (ππ β ππ) + π = ((π + π)(π β ππ))(π) c. Does this reasoning apply to the distributive property as well? Yes, it does apply to the distributive property. Discussion (3 minutes) ο§ Parts (c) and (d) of the Opening Exercise rely on key properties of equality. What are they? Call on students to articulate and compare their thoughts as a class discussion. In middle school, these properties are simply referred to as the if-then moves. Introduce their formal names to the class: the additive and multiplicative properties of equality. While writing it on the board, summarize with the following: ο§ So, whenever π = π is true, then π + π = π + π will also be true for all real numbers π. ο§ What if π = π is false? οΊ ο§ Then π + π = π + π will also be false. Is it also okay to subtract a number from both sides of the equation? οΊ Yes. This is the same operation as adding the opposite of that number. ο§ Whenever π = π is true, then ππ = ππ will also be true, and whenever π = π is false, ππ = ππ will also be false for all nonzero real numbers π. ο§ So, we have said earlier that applying the distributive, associative, and commutative properties does not change the solution set, and now we see that applying the additive and multiplicative properties of equality also preserves the solution set (does not change it). ο§ Suppose I see the equation |π₯| + 5 = 2. (Write the equation on the board.) ο§ Is it true, then, that |π₯| + 5 β 5 = 2 β 5? (Write the equation on the board.) Allow students to verbalize their answers and challenge each other if they disagree. If they all say yes, prompt students with, βAre you sure?β until one or more students articulate that we would get the false statement: |π₯| = β3. Then, summarize with the following points. Give these points great emphasis, perhaps adding grand gestures or voice MP.1 inflection to recognize the importance of this moment, as it addresses a major part of A-REI.A.1: ο§ So, our idea that adding the same number to both sides gives us another true statement depends on the idea that the first equation has a value of π₯ that makes it true to begin with. Lesson 12: Solving Equations This work is derived from Eureka Math β’ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from ALG I-M1-TE-1.3.0-07.2015 161 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. Lesson 12 NYS COMMON CORE MATHEMATICS CURRICULUM M1 ALGEBRA I MP.1 ο§ It is a big assumption that we make when we start to solve equations using properties of equality. We are assuming there is some value for the variable that makes the equation true. IF there is, then it makes sense that applying the properties of equality will give another true statement. But we must be cognizant of that big IF. ο§ What if there was no value of π₯ to make the equation true? What is the effect of adding a number to both sides of the equation or multiplying both sides by a nonzero number? οΊ ο§ There still will be no value of π₯ that makes the equation true. The solution set is still preserved; it will be the empty set. Create another equation that initially seems like a reasonable equation to solve but in fact has no possible solution. Exercise 2 (7 minutes) Exercise 2 Consider the equation ππ + π = π β π. a. Verify that this has the solution set {βπ, π}. Draw this solution set as a graph on the number line. We will later learn how to show that these happen to be the ONLY solutions to this equation. ππ + π = π β π True. (βπ)π + π = π β (βπ) True. b. Letβs add 4 to both sides of the equation and consider the new equation ππ + π = ππ β π. Verify π and βπ are still solutions. ππ + π = ππ β π True. (βπ)π + π = ππ β (βπ) True. They are still solutions. c. Letβs now add π to both sides of the equation and consider the new equation ππ + π + π = ππ. Are π and βπ still solutions? ππ + π + π = ππ True. (βπ)π + π + βπ = ππ True. π and βπ are still solutions. d. Letβs add βπ to both sides of the equation and consider the new equation ππ + π = π. Are π and βπ still solutions? ππ + π = π True. (βπ)π + βπ = π True. π and βπ are still solutions. e. π ππ +π π π Letβs multiply both sides by to get ππ +π π = π True. Lesson 12: (βπ)π +(βπ) π = π. Are π and βπ still solutions? = π True. π and βπ are still solutions. Solving Equations This work is derived from Eureka Math β’ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from ALG I-M1-TE-1.3.0-07.2015 162 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. Lesson 12 NYS COMMON CORE MATHEMATICS CURRICULUM M1 ALGEBRA I f. Letβs go back to part (d) and add πππ to both sides of the equation and consider the new equation ππ + π + πππ = π + πππ . Are π and βπ still solutions? ππ + π + π(π)π = π + π(π)π (βπ)π + βπ + π(βπ)π = π + π(βπ)π π + π + ππ = π + ππ True. π β π β ππ = π β ππ True. π and βπ are still solutions. Discussion (4 minutes) ο§ In addition to applying the commutative, associative, and distributive properties to equations, according to the exercises above, what else can be done to equations that does not change the solution set? οΊ Adding a number to or subtracting a number from both sides. οΊ Multiplying or dividing by a nonzero number. ο§ What we discussed in Example 1 can be rewritten slightly to reflect what we have just seen: If π₯ is a solution to an equation, it will also be a solution to the new equation formed when the same number is added to (or subtracted from) each side of the original equation or when the two sides of the original equation are multiplied by the same number or divided by the same nonzero number. These are referred to as the properties of equality. This now gives us a strategy for finding solution sets. ο§ Is π₯ = 5 an equation? If so, what is its solution set? οΊ Yes. Its solution set is 5. ο§ This example is so simple that it is hard to wrap your brain around, but it points out that if ever we have an equation that is this simple, we know its solution set. ο§ We also know the solution sets to some other simple equations, such as (a) π€ 2 = 64 ο§ (b) 7 + π = 5 (c) 3π½ = 10 Hereβs the strategy: If we are faced with the task of solving an equation, that is, finding the solution set of the equation: Use the commutative, associative, distributive properties AND Use the properties of equality (adding, subtracting, multiplying by nonzeros, dividing by nonzeros) to keep rewriting the equation into one whose solution set you easily recognize. (We observed that the solution set will not change under these operations.) ο§ This usually means rewriting the equation so that all the terms with the variable appear on one side of the equation. Lesson 12: Solving Equations This work is derived from Eureka Math β’ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from ALG I-M1-TE-1.3.0-07.2015 163 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. Lesson 12 NYS COMMON CORE MATHEMATICS CURRICULUM M1 ALGEBRA I Exercise 3 (5 minutes) Exercise 3 Solve for π: a. π ππ = π π π=π Solve for π: ππ + π = ππ b. π = π, π = βπ Solve for π: ππ β π = ππ β π c. π=π Exercise 4 (5 minutes) ο§ Does it matter which step happens first? Let's see what happens with the following example. Do a quick count-off, or separate the class into quadrants. Give groups their starting points. Have each group designate a presenter so the whole class can see the results. Exercise 4 Consider the equation ππ + π = ππ β ππ. Solve for π using the given starting point. MP.3 ο§ Group 1 Subtract ππ from both sides Group 2 Subtract π from both sides ππ + π β ππ = ππ β ππ β ππ π = ππ β ππ π + ππ = ππ β ππ + ππ ππ = ππ ππ ππ = π π π=π ππ + π β π = ππ β ππ β π ππ = ππ β ππ ππ β ππ = ππ β ππ β ππ βππ = βππ βππ βππ = βπ βπ π=π {π} {π} Group 4 Add ππ to both sides ππ + π β ππ = ππ β ππ β ππ ππ + π + ππ = ππ β ππ + ππ βππ + π = βππ ππ + ππ = ππ βππ + π β π = βππ β π ππ + ππ β ππ = ππ β ππ βππ = βππ ππ = ππ βππ βππ ππ ππ = = βπ βπ π π π=π π=π {π} {π} Therefore, according to this exercise, does it matter which step happens first? Explain why or why not. οΊ ο§ Group 3 Subtract ππ from both sides No, because the properties of equality produce equivalent expressions, no matter the order in which they happen. How does one know βhow muchβ to add/subtract/multiply/divide? What is the goal of using the properties, and how do they allow equations to be solved? Encourage students to verbalize their strategies to the class and to question each otherβs reasoning and question the precision of each otherβs description of their reasoning. From middle school, students recall that the goal is to isolate the variable by making 0s and 1s. Add/subtract numbers to make the zeros, and multiply/divide numbers to make the 1s. The properties say any numbers will work, which is true, but with the 0s and 1s goal in mind, equations can be solved very efficiently. ο§ The ability to pick the most efficient solution method comes with practice. Lesson 12: Solving Equations This work is derived from Eureka Math β’ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from ALG I-M1-TE-1.3.0-07.2015 164 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. Lesson 12 NYS COMMON CORE MATHEMATICS CURRICULUM M1 ALGEBRA I Closing (5 minutes) Answers will vary. As time permits, share several examples of student responses. Closing Consider the equation πππ + π = (π β π)(π + π)π. a. Use the commutative property to create an equation with the same solution set. π + πππ = (π + π)(π β π)π b. Using the result from part (a), use the associative property to create an equation with the same solution set. (π + πππ ) = ((π + π)(π β π))π c. Using the result from part (b), use the distributive property to create an equation with the same solution set. π + πππ = ππ + πππ β πππ d. MP.2 Using the result from part (c), add a number to both sides of the equation. π + πππ + π = ππ + πππ β πππ + π e. Using the result from part (d), subtract a number from both sides of the equation. π + πππ + π β π = ππ + πππ β πππ + π β π f. Using the result from part (e), multiply both sides of the equation by a number. π(π + πππ + π) = π(ππ + πππ β πππ + π) g. Using the result from part (f), divide both sides of the equation by a number. π + πππ + π = ππ + πππ β πππ + π h. What do all seven equations have in common? Justify your answer. They will all have the same solution set. Lesson 12: Solving Equations This work is derived from Eureka Math β’ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from ALG I-M1-TE-1.3.0-07.2015 165 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. Lesson 12 NYS COMMON CORE MATHEMATICS CURRICULUM M1 ALGEBRA I Lesson Summary MP.2 If π is a solution to an equation, it will also be a solution to the new equation formed when the same number is added to (or subtracted from) each side of the original equation or when the two sides of the original equation are multiplied by (or divided by) the same nonzero number. These are referred to as the properties of equality. If one is faced with the task of solving an equation, that is, finding the solution set of the equation: Use the commutative, associative, and distributive properties, AND use the properties of equality (adding, subtracting, multiplying by nonzeros, dividing by nonzeros) to keep rewriting the equation into one whose solution set you easily recognize. (We believe that the solution set will not change under these operations.) Exit Ticket (5 minutes) Lesson 12: Solving Equations This work is derived from Eureka Math β’ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from ALG I-M1-TE-1.3.0-07.2015 166 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. Lesson 12 NYS COMMON CORE MATHEMATICS CURRICULUM M1 ALGEBRA I Name Date Lesson 12: Solving Equations Exit Ticket Determine which of the following equations have the same solution set by recognizing properties rather than solving. a. 2π₯ + 3 = 13 β 5π₯ b. 6 + 4π₯ = β10π₯ + 26 d. 0.6 + 0.4π₯ = βπ₯ + 2.6 e. 3(2π₯ + 3) = g. 15(2π₯ + 3) = 13 β 5π₯ h. 15(2π₯ + 3) + 97 = 110 β 5π₯ Lesson 12: 13 βπ₯ 5 Solving Equations This work is derived from Eureka Math β’ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from ALG I-M1-TE-1.3.0-07.2015 13 βπ₯ 5 c. 6π₯ + 9 = f. 4π₯ = β10π₯ + 20 167 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. Lesson 12 NYS COMMON CORE MATHEMATICS CURRICULUM M1 ALGEBRA I Exit Ticket Sample Solutions Determine which of the following equations have the same solution set by recognizing properties rather than solving. a. ππ + π = ππ β ππ b. π + ππ = βπππ + ππ d. π. π + π. ππ = βπ + π. π e. π(ππ + π) = g. ππ(ππ + π) = ππ β ππ h. ππ(ππ + π) + ππ = πππ β ππ ππ βπ π ππ βπ π c. ππ + π = f. ππ = βπππ + ππ (a), (b), (d), and (f) have the same solution set. (c), (e), (g), and (h) have the same solution set. Problem Set Sample Solutions 1. Which of the following equations have the same solution set? Give reasons for your answers that do not depend on solving the equations. I. π β π = ππ + π II. ππ β π = ππ + π III. πππ β π = ππ + ππ IV. ππ β ππ = πππ + ππ V. ππ + ππ = ππ β ππ VI. βπ. ππ + π ππ = + π. ππ πππ πππ I, V, and VI all have the same solution set; V is the same as I after multiplying both sides by π and switching the left side with the right side; VI is the same as I after dividing both sides by πππ and using the commutative property to rearrange the terms on the left side of the equation. II and IV have the same solution set. IV is the same as II after multiplying both sides by π and subtracting π from both sides. III does not have the same solution set as any of the others. Solve the following equations, check your solutions, and then graph the solution sets. 2. βππ β ππ = βπ(ππ β π) 3. π(ππ + π) = π + ππ {βπ} {π} 5. π β ππ = π(π + ππ) 6. ππ β ππ = βπ(π + ππ) + ππ {π} Lesson 12: 4. {βπ} Solving Equations This work is derived from Eureka Math β’ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from ALG I-M1-TE-1.3.0-07.2015 ππ β ππ + π = π {π} 7. (π β π)(π + π) = ππ + ππ β π no solution 168 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. Lesson 12 NYS COMMON CORE MATHEMATICS CURRICULUM M1 ALGEBRA I 8. ππ β π = ππ β ππ β π 9. βπ β ππ + ππ = ππ β ππ 11. π(π β π) = π(π β π) β ππ 12. βπ(π β π) = βπ β ππ 14. βππ β ππ = ππ + π(π + π) 15. π π+π 13. βππ β ππ = βπ(π + π) π {β } π 18. π+π π {βπ} = {π} 16. π + =π {βπ} Lesson 12: {βπ} π {β } π {π} 17. βπ(βππ β π) = βππ β π 10. π β ππ = π β ππ + ππ {π} {π} π+π π π = βπ π π { 19. βπ(ππ β π. π) + π. π(ππ + π) = βππ π {βπ} Solving Equations This work is derived from Eureka Math β’ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from ALG I-M1-TE-1.3.0-07.2015 ππ } π { ππ } π 169 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License.
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