Lesson 5 NYS COMMON CORE MATHEMATICS CURRICULUM 8β’4 Lesson 5: Writing and Solving Linear Equations Student Outcomes ο§ Students apply knowledge of geometry to writing and solving linear equations. Lesson Notes All of the problems in this lesson relate to what students have learned about geometry in recent modules and previous years. The purpose is twofold: first, to reinforce what students have learned about geometry, and second, to demonstrate a need for writing and solving an equation in a context that is familiar. Throughout the lesson, students solve mathematical problems that relate directly to what students have learned about angle relationships, congruence, and the triangle sum theorem. Encourage students to draw diagrams to represent the situations presented in the word problems. Classwork Example 1 (5 minutes) ο§ Solve the following problem: Example 1 One angle is five degrees less than three times the measure of another angle. Together, the angle measures have a sum of πππ°. What is the measure of each angle? MP.1 Provide students with time to make sense of the problem and persevere in solving it. They could begin their work by guessing and checking, drawing a diagram, or other methods as appropriate. Then, move to the algebraic method shown below. ο§ What do we need to do first to solve this problem? οΊ ο§ If π₯ is the measure of the first angle, how do you represent the measure of the second angle? οΊ ο§ The second angle is 3π₯ β 5. What is the equation that represents this situation? οΊ ο§ First, we need to define our variable (symbol). Let π₯ be the measure of the first angle in degrees. The equation is π₯ + 3π₯ β 5 = 143. Scaffolding: Model for students how to use diagrams to help make sense of the problems throughout this lesson. Encourage students to use diagrams to help them understand the situation. The equation that represents this situation is π₯ + 3π₯ β 5 = 143. Solve for π₯, and then determine the measure of each angle. Lesson 5: Writing and Solving Linear Equations This work is derived from Eureka Math β’ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from G8-M4-TE-1.3.0-09.2015 53 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. Lesson 5 NYS COMMON CORE MATHEMATICS CURRICULUM 8β’4 As students share their solutions for this and subsequent problems, ask them a variety of questions to reinforce the concepts of the last few lessons. For example, ask students to discuss whether or not this is a linear equation and how they know, to justify their steps and explain why they chose their particular first step, to explain what the solution means, or to justify how they know their answer is correct. π + ππ β π = πππ (π + π)π β π = πππ ππ β π = πππ ππ β π + π = πππ + π ππ = πππ π = ππ The measure of the first angle is ππ°. The second angle is π(ππ°) β π° = πππ° β π° = πππ°. ο§ Compare the method you tried at the beginning of the problem with the algebraic method. What advantage does writing and solving an equation have? οΊ ο§ Could we have defined π₯ to be the measure of the second angle? If so, what, if anything, would change? οΊ ο§ ο§ If we let π₯ be the measure of the second angle, then the equation would change, but the answers for the measures of the angles should remain the same. If π₯ is the measure of the second angle, how would we write the measure of the first angle? οΊ ο§ Writing and solving an equation is a more direct method than the one I tried before. It allows me to find the answer more quickly. The first angle would be π₯+5 3 . The equation that represents the situation is π₯ + π₯+5 = 143. How should we solve this equation? 3 οΊ We could add the fractions together and then solve for π₯. οΊ We could multiply every term by 3 to change the fraction to a whole number. Using either method, solve the equation. Verify that the measures of the angles are the same as before. οΊ π₯+5 = 143 3 3π₯ π₯ + 5 + = 143 3 3 3π₯ + π₯ + 5 = 143 3 3π₯ + π₯ + 5 = 143(3) π₯+ (3 + 1)π₯ + 5 = 429 Scaffolding: It may be necessary to remind students how to add fractions by rewriting the term or terms as equivalent fractions and then adding the numerators. Provide support as needed. 4π₯ + 5 = 429 4π₯ + 5 β 5 = 429 β 5 4π₯ = 424 π₯ = 106 Lesson 5: Writing and Solving Linear Equations This work is derived from Eureka Math β’ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from G8-M4-TE-1.3.0-09.2015 54 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. Lesson 5 NYS COMMON CORE MATHEMATICS CURRICULUM 8β’4 OR π₯+5 = 143 3 π₯+5 3 (π₯ + = 143) 3 3π₯ + π₯ + 5 = 429 π₯+ (3 + 1)π₯ + 5 = 429 4π₯ + 5 = 429 4π₯ + 5 β 5 = 429 β 5 4π₯ = 424 π₯ = 106 So, the measure of the second angle is 106°. π₯ + 5 106 + 5 111 = = = 37 3 3 3 The measure of the first angle is 37°. ο§ Whether we let π₯ represent the measure of the first angle or the second angle does not change our answers. Whether we solve the equation using the first or second method does not change our answers. What matters is that we accurately write the information in the problem and correctly use the properties of equality. You may solve a problem differently from your classmates or teachers. Again, what matters most is that what you do is accurate and correct. Example 2 (12 minutes) ο§ Solve the following problem: Example 2 Given a right triangle, find the degree measure of the angles if one angle is ten degrees more than four times the degree measure of the other angle and the third angle is the right angle. Give students time to work. As they work, walk around and identify students who are writing and solving the problem in different ways. The instructional goal of this example is to make clear that there are different ways to solve a linear equation as opposed to one βright way.β Select students to share their work with the class. If students do not come up with different ways of solving the equation, talk them through the following student work samples. Again, as students share their solutions, ask them a variety of questions to reinforce the concepts of the last few lessons. For example, ask students to discuss whether or not this is a linear equation and how they know, to justify their steps and explain why they chose their particular first step, to explain what the solution means, or to justify how they know their answer is correct. Lesson 5: Writing and Solving Linear Equations This work is derived from Eureka Math β’ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from G8-M4-TE-1.3.0-09.2015 55 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. Lesson 5 NYS COMMON CORE MATHEMATICS CURRICULUM 8β’4 Solution One Let π be the measure of the first angle. Then, the second angle is ππ + ππ. The sum of the measures for the angles for this right triangle is π + ππ + ππ + ππ = πππ. π + ππ + ππ + ππ = πππ (π + π)π + πππ = πππ ππ + πππ = πππ ππ + πππ β πππ = πππ β πππ ππ = ππ π = ππ The measure of the first angle is ππ°, the measure of the second angle is π(ππ°) + ππ° = ππ° + ππ° = ππ°, and the measure of the third angle is ππ°. Solution Two Let π be the measure of the first angle. Then, the second angle is ππ + ππ. Since we have a right triangle, we already know that one angle is ππ°, which means that the sum of the other two angles is ππ: π + ππ + ππ = ππ. π + ππ + ππ = ππ (π + π)π + ππ = ππ ππ + ππ = ππ ππ + ππ β ππ = ππ β ππ ππ = ππ π = ππ The measure of the first angle is ππ°, the measure of the second angle is π(ππ°) + ππ° = ππ° + ππ° = ππ°, and the measure of the third angle is ππ°. Solution Three Let π be the measure of the second angle. Then, the first angle is πβππ π . Since we have a right triangle, we already know that one angle is ππ°, which means that the sum of the other two angles is ππ: π + πβππ = ππ. π π β ππ = ππ π π β ππ π (π + = ππ) π ππ + π β ππ = πππ π+ (π + π)π β ππ = πππ ππ β ππ = πππ ππ β ππ + ππ = πππ + ππ ππ = πππ π = ππ The measure of the second angle is ππ°, the measure of the first angle is ππ°βππ° π = ππ° π = ππ°, and the measure of the third angle is ππ°. Lesson 5: Writing and Solving Linear Equations This work is derived from Eureka Math β’ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from G8-M4-TE-1.3.0-09.2015 56 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. Lesson 5 NYS COMMON CORE MATHEMATICS CURRICULUM 8β’4 Solution Four Let π be the measure of the second angle. Then, the first angle is πβππ πβππ π+ + ππ = πππ. π π . The sum of the three angles is π β ππ + ππ = πππ π π β ππ π+ + ππ β ππ = πππ β ππ π π β ππ π+ = ππ π ππ π β ππ + = ππ π π ππ + π β ππ = ππ π ππ + π β ππ = πππ π+ ππ β ππ + ππ = πππ + ππ ππ = πππ π = ππ The measure of the second angle is ππ°, the measure of the first angle is third angle is ππ°. ππ°βππ° π = ππ° π = ππ°, and the measure of the Make sure students see at least four different methods of solving the problem. Conclude this example with the statements below. ο§ Each method is slightly different either in terms of how the variable is defined or how the properties of equality are used to solve the equation. The way you find the answer may be different from your classmatesβ or your teacherβs. ο§ As long as you are accurate and do what is mathematically correct, you will find the correct answer. Example 3 (4 minutes) ο§ A pair of alternate interior angles are described as follows. One angle measure, in degrees, is fourteen more than half a number. The other angle measure, in degrees, is six less than half that number. Are the angles congruent? ο§ We begin by assuming that the angles are congruent. If the angles are congruent, what does that mean about their measures? οΊ ο§ It means that they are equal in measure. Write an expression that describes each angle. οΊ One angle measure is Lesson 5: π₯ 2 + 14, and the other angle measure is Writing and Solving Linear Equations This work is derived from Eureka Math β’ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from G8-M4-TE-1.3.0-09.2015 π₯ 2 β 6. 57 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. Lesson 5 NYS COMMON CORE MATHEMATICS CURRICULUM ο§ If the angles are congruent, we can write the equation as π₯ 2 + 14 = π₯ 2 8β’4 β 6. We know that our properties of equality allow us to transform equations while making sure that they remain true. π₯ π₯ + 14 = β 6 2 2 π₯ π₯ π₯ π₯ β + 14 = β β 6 2 2 2 2 14 β β6 Therefore, our assumption was not correct, and the angles are not congruent. Exercises (16 minutes) Students complete Exercises 1β6 independently or in pairs. Exercises For each of the following problems, write an equation and solve. 1. A pair of congruent angles are described as follows: The degree measure of one angle is three more than twice a number, and the other angleβs degree measure is ππ. π less than three times the number. Determine the measure of the angles in degrees. Let π be the number. Then, the measure of one angle is π + ππ, and the measure of the other angle is ππ β ππ. π. Because the angles are congruent, their measures are equal. Therefore, π + ππ = ππ β ππ. π π + ππ β ππ = ππ β ππ β ππ. π π = π β ππ. π π + ππ. π = π β ππ. π + ππ. π ππ. π = π Replacing π with ππ. π in π + ππ gives π + π(ππ. π) = π + πππ = πππ; therefore the measure of the angles is πππ°. 2. The measure of one angle is described as twelve more than four times a number. Its supplement is twice as large. Find the measure of each angle in degrees. Let π be the number. Then, the measure of one angle is ππ + ππ. The other angle is π(ππ + ππ) = ππ + ππ. Since the angles are supplementary, their sum must be πππ°. ππ + ππ + ππ + ππ = πππ πππ + ππ = πππ πππ + ππ β ππ = πππ β ππ πππ = πππ π = ππ Replacing π with ππ in ππ + ππ gives π(ππ) + ππ = ππ + ππ = ππ. Replacing π with ππ in π(ππ + ππ) gives π(π(ππ) + ππ) = π(ππ + ππ) = π(ππ) = πππ. Therefore, the measures of the angles are ππ° and πππ°. Lesson 5: Writing and Solving Linear Equations This work is derived from Eureka Math β’ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from G8-M4-TE-1.3.0-09.2015 58 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. Lesson 5 NYS COMMON CORE MATHEMATICS CURRICULUM 3. 8β’4 A triangle has angles described as follows: The measure of the first angle is four more than seven times a number, the measure of the second angle is four less than the first, and the measure of the third angle is twice as large as the first. What is the measure of each angle in degrees? Let π be the number. The measure of the first angle is ππ + π. The measure of the second angle is ππ + π β π = ππ. The measure of the third angle is π(ππ + π) = πππ + π. The sum of the angles of a triangle must be πππ°. ππ + π + ππ + πππ + π = πππ πππ + ππ = πππ πππ + ππ β ππ = πππ β ππ πππ = πππ π=π Replacing π with π in ππ + π gives π(π) + π = ππ + π = ππ. Replacing π with π in ππ gives π(π) = ππ. Replacing π with π in πππ + π gives ππ(π) + π = ππ + π = ππ. Therefore, the measures of the angles are ππ°, ππ°, and ππ°. 4. One angle measures nine more than six times a number. A sequence of rigid motions maps the angle onto another angle that is described as being thirty less than nine times the number. What is the measure of the angle in degrees? Let π be the number. Then, the measure of one angle is ππ + π. The measure of the other angle is ππ β ππ. Since rigid motions preserve the measures of angles, then the measure of these angles is equal. ππ + π = ππ β ππ ππ + π β π = ππ β ππ β π ππ = ππ β ππ ππ β ππ = ππ β ππ β ππ βππ = βππ π = ππ Replacing π with ππ in ππ + π gives π(ππ) + π = ππ + π = ππ. Therefore, the angle measure is ππ°. 5. A right triangle is described as having an angle of measure six less than negative two times a number, another angle measure that is three less than negative one-fourth the number, and a right angle. What are the measures of the angles in degrees? π π Let π be a number. Then, the measure of one angle is βππ β π. The measure of the other angle is β β π. The sum of the two angles must be ππ°. π βππ β π + (β ) β π = ππ π ππ π (β ) + (β ) β π = ππ π π ππ (β ) β π + π = ππ + π π ππ β = ππ π βππ = πππ π = βππ π π Replacing π with βππ gives βππ β π gives βπ(βππ) β π = ππ β π = ππ. Replacing π with βππ in β β π gives ππ β ππ = π. Therefore, the angle measures are ππ° and π°. Lesson 5: Writing and Solving Linear Equations This work is derived from Eureka Math β’ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from G8-M4-TE-1.3.0-09.2015 59 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. Lesson 5 NYS COMMON CORE MATHEMATICS CURRICULUM 6. 8β’4 One angle is one less than six times the measure of another. The two angles are complementary angles. Find the measure of each angle in degrees. Let π be the measure of the first angle. Then, the measure of the second angle is ππ β π. The sum of the measures will be ππ because the angles are complementary. π + ππ β π = ππ ππ β π = ππ ππ β π + π = ππ + π ππ = ππ π = ππ The first angle is π and therefore measures ππ°. Replacing π with ππ in ππ β π gives π(ππ) β π = ππ β π = ππ. Therefore, the second angle measure is ππ°. Closing (4 minutes) Summarize, or ask students to summarize, the main points from the lesson: ο§ We know that an algebraic method for solving equations is more efficient than guessing and checking. ο§ We know how to write and solve equations that relate to angles, triangles, and geometry, in general. ο§ We know that drawing a diagram can sometimes make it easier to understand a problem and that there is more than one way to solve an equation. Exit Ticket (4 minutes) Lesson 5: Writing and Solving Linear Equations This work is derived from Eureka Math β’ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from G8-M4-TE-1.3.0-09.2015 60 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. Lesson 5 NYS COMMON CORE MATHEMATICS CURRICULUM Name 8β’4 Date Lesson 5: Writing and Solving Linear Equations Exit Ticket For each of the following problems, write an equation and solve. 1. Given a right triangle, find the measures of all the angles, in degrees, if one angle is a right angle and the measure of the second angle is six less than seven times the measure of the third angle. 2. In a triangle, the measure of the first angle is six times a number. The measure of the second angle is nine less than the first angle. The measure of the third angle is three times the number more than the measure of the first angle. Determine the measure of each angle in degrees. Lesson 5: Writing and Solving Linear Equations This work is derived from Eureka Math β’ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from G8-M4-TE-1.3.0-09.2015 61 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. Lesson 5 NYS COMMON CORE MATHEMATICS CURRICULUM 8β’4 Exit Ticket Sample Solutions For each of the following problems, write an equation and solve. 1. Given a right triangle, find the measures of all of the angles, in degrees, if one angle is a right angle and the measure of the second angle is six less than seven times the measure of the third angle. Let π represent the measure of the third angle. Then, ππ β π can represent the measure of the second angle. The sum of the two angles in the right triangle will be ππ°. ππ β π + π = ππ ππ β π = ππ ππ β π + π = ππ + π ππ = ππ π ππ π= π π π = ππ The third angle is π and therefore measures ππ°. Replacing π with ππ in ππ β π gives π(ππ) β π = ππ β π = ππ. Therefore, the measure of the second angle is ππ°. The measure of the third angle is ππ°. 2. In a triangle, the measure of the first angle is six times a number. The measure of the second angle is nine less than the first angle. The measure of the third angle is three times the number more than the measure of the first angle. Determine the measure of each angle in degrees. Let π be the number. Then, the measure of the first angle is ππ, the measure of the second angle is ππ β π, and the measure of the third angle is ππ + ππ. The sum of the measures of the angles in a triangle is πππ°. ππ + ππ β π + ππ + ππ = πππ πππ β π = πππ πππ β π + π = πππ + π πππ = πππ ππ πππ π= ππ ππ π=π Replacing π with π in ππ gives π(π) = ππ. Replacing π with π in ππ β π gives π(π) β π = ππ β π = ππ. Replacing π with π in ππ + ππ gives ππ + π(π) = ππ + ππ = ππ. Therefore, the angle measures are ππ°, ππ°, and ππ°. Note to teacher: There are several ways to solve problems like these. For example, a student may let π₯ be the measure of the first angle and write the measure of the other angles accordingly. Either way, make sure that students are defining their symbols and correctly using the properties of equality to solve. Lesson 5: Writing and Solving Linear Equations This work is derived from Eureka Math β’ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from G8-M4-TE-1.3.0-09.2015 62 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. Lesson 5 NYS COMMON CORE MATHEMATICS CURRICULUM 8β’4 Problem Set Sample Solutions Students practice writing and solving linear equations. For each of the following problems, write an equation and solve. 1. The measure of one angle is thirteen less than five times the measure of another angle. The sum of the measures of the two angles is πππ°. Determine the measure of each angle in degrees. Let π be the measure of the one angle. Then, the measure of the other angle is ππ β ππ. π + ππ β ππ = πππ ππ β ππ = πππ ππ β ππ + ππ = πππ + ππ ππ = πππ π = ππ. π Since one angle measure is π, it is ππ. π°. Replacing π with ππ. π in ππ β ππ gives π(ππ. π) β ππ = πππ β ππ. π = πππ. π. Therefore, the other angle measures πππ. π°. 2. An angle measures seventeen more than three times a number. Its supplement is three more than seven times the number. What is the measure of each angle in degrees? Let π be the number. Then, the measure of one angle is ππ + ππ. The measure of the other angle is ππ + π. Since the angles are supplementary, the sum of their measures will be πππ. ππ + ππ + ππ + π = πππ πππ + ππ = πππ πππ + ππ β ππ = πππ β ππ πππ = πππ π = ππ Replacing π with ππ in ππ + ππ gives π(ππ) + ππ = ππ. Replacing π with ππ in ππ + π gives (ππ) + π = πππ + π = πππ. Therefore, the angle measures are ππ° and πππ°. 3. The angles of a triangle are described as follows: β π¨ is the largest angle; its measure is twice the measure of β π©. The measure of β πͺ is π less than half the measure of β π©. Find the measures of the three angles in degrees. Let π be the measure of β π©. Then, the measure of β π¨ is ππ, and the measure of β πͺ is measures of the angles must be πππ°. π + ππ + ππ + π π β π. The sum of the π β π = πππ π π β π + π = πππ + π π π ππ + = πππ π ππ π + = πππ π π ππ = πππ π ππ = πππ π = ππ Since π is the measure of β π©, then β π© is ππ°. Replacing π with ππ in ππ gives π(ππ) = πππ. Therefore, the measure of β π¨ is πππ°. Replacing π with ππ in is ππ°. Lesson 5: π π β π gives ππ π Writing and Solving Linear Equations This work is derived from Eureka Math β’ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from G8-M4-TE-1.3.0-09.2015 β π = ππ β π = ππ. Therefore, the measure of β πͺ 63 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. Lesson 5 NYS COMMON CORE MATHEMATICS CURRICULUM 4. 8β’4 A pair of corresponding angles are described as follows: The measure of one angle is five less than seven times a number, and the measure of the other angle is eight more than seven times the number. Are the angles congruent? Why or why not? Let π be the number. Then, the measure of one angle is ππ β π, and the measure of the other angle is ππ + π. Assume they are congruent, which means their measures are equal. ππ β π = ππ + π ππ β ππ β π = ππ β ππ + π βπ β π Since βπ β π, the angles are not congruent. 5. The measure of one angle is eleven more than four times a number. Another angle is twice the first angleβs measure. The sum of the measures of the angles is πππ°. What is the measure of each angle in degrees? Let π be the number. The measure of one angle can be represented with ππ + ππ, and the other angleβs measure can be represented as π(ππ + ππ) = ππ + ππ. ππ + ππ + ππ + ππ = πππ πππ + ππ = πππ πππ + ππ β ππ = πππ β ππ πππ = πππ π = ππ. π Replacing π with ππ. π in ππ + ππ gives π(ππ. π) + ππ = ππ + ππ = ππ. Replacing π with ππ. π in π(ππ + ππ) gives π(π(ππ. π) + ππ) = π(ππ + ππ) = π(ππ) = πππ. Therefore, the measures of the angles are ππ° and πππ°. 6. Three angles are described as follows: β π© is half the size of β π¨. The measure of β πͺ is equal to one less than two times the measure of β π©. The sum of β π¨ and β π© is πππ°. Can the three angles form a triangle? Why or why not? π π π Let π represent the measure of β π¨. Then, the measure of β π© is , and the measure of β πͺ is π ( ) β π = π β π. π The sum of the measures of β π¨ and β π© is πππ. π = πππ π ππ = πππ π ππ = πππ π+ π = ππ Since π is the measure of β π¨, then β π¨ is ππ°. Replacing π with ππ in π π gives ππ π = ππ; therefore, the measure of β π© is ππ°. Replacing π with ππ in π β π gives ππ β π = ππ, therefore the measure of β πͺ is ππ°. The sum of the three angle measures is ππ° + ππ° + ππ° = πππ°. Since the sum of the measures of the interior angles of a triangle must equal πππ°, these angles do not form a triangle. The sum is too large. Lesson 5: Writing and Solving Linear Equations This work is derived from Eureka Math β’ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from G8-M4-TE-1.3.0-09.2015 64 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License.
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