Lesson 12 NYS COMMON CORE MATHEMATICS CURRICULUM 8β’4 Lesson 12: Linear Equations in Two Variables Student Outcomes ο§ Students use a table to find solutions to a given linear equation and plot the solutions on a coordinate plane. Lesson Notes In this lesson, students find solutions to a linear equation in two variables using a table and then plot the solutions as points on the coordinate plane. Students need graph paper in order to complete the Exercises and the Problem Set. Classwork Opening Exercise (5 minutes) Students complete the Opening Exercise independently in preparation for the discussion about standard form and the solutions that follow. Opening Exercise Emily tells you that she scored ππ points in a basketball game. Write down all the possible ways she could have scored ππ with only two- and three-point baskets. Use the table below to organize your work. Number of Two-Pointers Number of Three-Pointers ππ π ππ π ππ π π π π π π ππ Let π be the number of two-pointers and π be the number of three-pointers that Emily scored. Write an equation to represent the situation. ππ + ππ = ππ Discussion (10 minutes) ο§ An equation in the form of ππ₯ + ππ¦ = π is called a linear equation in two variables, where π, π, and π are constants, and at least one of π and π are not zero. In this lesson, neither π nor π will be equal to zero. In the Opening Exercise, what equation did you write to represent Emilyβs score at the basketball game? οΊ ο§ 2π₯ + 3π¦ = 32 The equation 2π₯ + 3π¦ = 32 is an example of a linear equation in two variables. Lesson 12: Linear Equations in Two Variables 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 140 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. Lesson 12 NYS COMMON CORE MATHEMATICS CURRICULUM ο§ An equation of this form, ππ₯ + ππ¦ = π, is also referred to as an equation in standard form. Is the equation you wrote in the Opening Exercise in standard form? οΊ ο§ No, because β50(1) + 2 = β50 + 2 = β48 β 15. What pairs of numbers did you find that worked for Emilyβs basketball score? Did just any pair of numbers work? Explain. οΊ ο§ The symbols π₯ and π¦ are numbers. Since they are not constants, it means they are unknown numbers, typically called variables, in the equation ππ₯ + ππ¦ = π. For example, β50π₯ + π¦ = 15 is a linear equation in π₯ and π¦. As you can easily see, not just any pair of numbers π₯ and π¦ will make the equation true. Consider π₯ = 1 and π¦ = 2. Does it make the equation true? οΊ ο§ Yes. It is in the same form as ππ₯ + ππ¦ = π. In the equation ππ₯ + ππ¦ = π, the symbols π, π, and π are constants. What, then, are π₯ and π¦? οΊ ο§ 8β’4 Students should identify the pairs of numbers in the table of the Opening Exercise. No, not just any pair of numbers worked. For example, I couldnβt say that Emily scored 15 two-pointers and 1 three-pointer because that would mean she scored 33 points in the game, and she only scored 32 points. A solution to the linear equation in two variables is an ordered pair of numbers (π₯, π¦) so that π₯ and π¦ makes the equation a true statement. The pairs of numbers that you wrote in the table for Emily are solutions to the equation 2π₯ + 3π¦ = 32 because they are pairs of numbers that make the equation true. The question becomes, how do we find an unlimited number of solutions to a given linear equation? οΊ Guess numbers until you find a pair that makes the equation true. ο§ A strategy that will help us find solutions to a linear equation in two variables is as follows: We fix a number for π₯. That means we pick any number we want and call it π₯. Since we know how to solve a linear equation in one variable, then we solve for π¦. The number we picked for π₯ and the number we get when we solve for π¦ is the ordered pair (π₯, π¦), which is a solution to the two-variable linear equation. ο§ For example, let π₯ = 5. Then, in the equation β50π₯ + π¦ = 15, we have β50(5) + π¦ β250 + π¦ β250 + 250 + π¦ π¦ = 15 = 15 = 15 + 250 = 265. Therefore, (5, 265) is a solution to the equation β50π₯ + π¦ = 15. ο§ Similarly, we can fix a number for π¦ and solve for π₯. Let π¦ = 10; then β50π₯ + 10 = 15 β50π₯ + 10 β 10 = 15 β 10 β50π₯ = 5 β50 5 π₯= β50 β50 1 π₯=β . 10 Therefore, (β 1 , 10) is a solution to the equation β50π₯ + π¦ = 15. 10 Ask students to provide a number for π₯ or π¦ and demonstrate how to find a solution. This can be done more than once in order to prove to students that they can find a solution no matter which number they choose to fix for π₯ or π¦. Once they are convinced, allow them to work on the Exploratory Challenge. Lesson 12: Linear Equations in Two Variables 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 141 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. Lesson 12 NYS COMMON CORE MATHEMATICS CURRICULUM 8β’4 Exploratory Challenge/Exercises (20 minutes) Students can work independently or in pairs to complete the exercises. Every few minutes, have students share their tables and graphs with the class. Make suggestions to students as they work as to which values for π₯ or π¦ they could choose. For example, in Exercises 1 and 2, small numbers would ease the mental math work. Exercise 3 may be made easier if they choose a number for π¦ and solve for π₯. Exercise 4 can be made easier if students choose values for π₯ that are multiples of 5. While making suggestions, ask students why the suggestions would make the work easier. Exploratory Challenge/Exercises 1. Find five solutions for the linear equation π + π = π, and plot the solutions as points on a coordinate plane. Lesson 12: π Linear Equation: π+π=π π π π+π =π π π π+π =π π π π+π =π π π π+π =π βπ π π+π =π βπ Linear Equations in Two Variables 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 142 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. Lesson 12 NYS COMMON CORE MATHEMATICS CURRICULUM 2. 8β’4 Find five solutions for the linear equation ππ β π = ππ, and plot the solutions as points on a coordinate plane. π Linear Equation: ππ β π = ππ π π(π) β π = ππ π β π = ππ π π β π β π = ππ β π βπ βπ = π π = βπ π(π) β π = ππ π β π = ππ π π β π β π = ππ β π βπ βπ = π π = βπ π(π) β π = ππ π β π = ππ π π β π β π = ππ β π βπ βπ = π π = βπ π(π) β π = ππ π β π = ππ π π β π β π = ππ β π βπ βπ = π π = βπ π(π) β π = ππ π ππ β π = ππ π π=π Lesson 12: Linear Equations in Two Variables 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 143 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. Lesson 12 NYS COMMON CORE MATHEMATICS CURRICULUM 3. 8β’4 Find five solutions for the linear equation π + ππ = ππ, and plot the solutions as points on a coordinate plane. Linear Equation: π + ππ = ππ π π π + π(π) = ππ π + π = ππ ππ π π = ππ π + π(π) = ππ π + ππ = ππ ππ π π = ππ π + π(π) = ππ π + ππ = ππ π π π=π π + π(π) = ππ π + ππ = ππ π π π=π π + π(π) = ππ π + ππ = ππ βπ π π = βπ 4. Consider the linear equation a. π π π + π = ππ. Will you choose to fix values for π or π? Explain. If I fix values for π, it will make the computations easier. Solving for π can be done in one step. b. Are there specific numbers that would make your computational work easier? Explain. Values for π that are multiples of π will make the computations easier. When I multiply π π by a multiple of π, I will get an integer. Lesson 12: Linear Equations in Two Variables 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 144 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. Lesson 12 NYS COMMON CORE MATHEMATICS CURRICULUM c. Find five solutions to the linear equation π π 8β’4 π + π = ππ, and plot the solutions as points on a coordinate plane. π π Linear Equation: π π + π = ππ π π (π) + π = ππ π π + π = ππ π π π=π ππ π (ππ) + π = ππ π π + π = ππ π π=π ππ π (ππ) + π = ππ π π + π = ππ π π=π ππ π (ππ) + π = ππ π π + π = ππ π π=π ππ π (ππ) + π = ππ π ππ + π = ππ π π=π Lesson 12: Linear Equations in Two Variables 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 145 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. Lesson 12 NYS COMMON CORE MATHEMATICS CURRICULUM 5. 8β’4 At the store, you see that you can buy a bag of candy for $π and a drink for $π. Assume you have a total of $ππ to spend. You are feeling generous and want to buy some snacks for you and your friends. a. Write an equation in standard form to represent the number of bags of candy, π, and the number of drinks, π, that you can buy with $ππ. ππ + π = ππ b. Find five solutions to the linear equation from part (a), and plot the solutions as points on a coordinate plane. Linear Equation: π ππ + π = ππ π π(π) + π = ππ π π + π = ππ ππ π = ππ π(π) + π = ππ π ππ + π = ππ ππ π = ππ π(π) + π = ππ π ππ + π = ππ ππ π = ππ π(ππ) + π = ππ ππ ππ + π = ππ ππ π = ππ π(ππ) + π = ππ ππ ππ + π = ππ π π=π Lesson 12: Linear Equations in Two Variables 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 146 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. NYS COMMON CORE MATHEMATICS CURRICULUM Lesson 12 8β’4 Closing (5 minutes) Summarize, or ask students to summarize, the main points from the lesson: ο§ A two-variable equation in the form of ππ₯ + ππ¦ = π is known as a linear equation in standard form. ο§ A solution to a linear equation in two variables is an ordered pair (π₯, π¦) that makes the given equation true. ο§ We can find solutions by fixing a number for π₯ or π¦ and then solving for the other variable. Our work can be made easier by thinking about the computations we will need to make before fixing a number for π₯ or π¦. For 1 example, if π₯ has a coefficient of , we should select values for π₯ that are multiples of 3. 3 Lesson Summary A linear equation in two-variables π and π is in standard form if it is the form ππ + ππ = π for numbers π, π, and π, where π and π are both not zero. The numbers π, π, and π are called constants. A solution to a linear equation in two variables is the ordered pair (π, π) that makes the given equation true. Solutions can be found by fixing a number for π and solving for π or fixing a number for π and solving for π. Exit Ticket (5 minutes) Lesson 12: Linear Equations in Two Variables 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 147 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. Lesson 12 NYS COMMON CORE MATHEMATICS CURRICULUM Name 8β’4 Date Lesson 12: Linear Equations in Two Variables Exit Ticket 1. Is the point (1, 3) a solution to the linear equation 5π₯ β 9π¦ = 32? Explain. 2. Find three solutions for the linear equation 4π₯ β 3π¦ = 1, and plot the solutions as points on a coordinate plane. π Linear Equation: 4π₯ β 3π¦ = 1 Lesson 12: π Linear Equations in Two Variables 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 148 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. Lesson 12 NYS COMMON CORE MATHEMATICS CURRICULUM 8β’4 Exit Ticket Sample Solutions 1. Is the point (π, π) a solution to the linear equation ππ β ππ = ππ? Explain. No, (π, π) is not a solution to ππ β ππ = ππ because π(π) β π(π) = π β ππ = βππ, and βππ β ππ. 2. Find three solutions for the linear equation ππ β ππ = π, and plot the solutions as points on a coordinate plane. Linear Equation: ππ β ππ = π π π π(π) β ππ = π π β ππ = π π βππ = βπ π π=π ππ β π(π) = π ππ β ππ = π π ππ = ππ π π=π π(π) β ππ = π ππ β ππ = π π βππ = βππ π π=π Problem Set Sample Solutions Students practice finding and graphing solutions for linear equations that are in standard form. 1. π π Consider the linear equation π β π = βπ. a. Will you choose to fix values for π or π? Explain. If I fix values for π, it will make the computations easier. Solving for π can be done in one step. b. Are there specific numbers that would make your computational work easier? Explain. Values for π that are multiples of π will make the computations easier. When I multiply π π by a multiple of π, I will get a whole number. Lesson 12: Linear Equations in Two Variables 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 149 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. Lesson 12 NYS COMMON CORE MATHEMATICS CURRICULUM c. 8β’4 π π Find five solutions to the linear equation π β π = βπ, and plot the solutions as points on a coordinate plane. Lesson 12: π Linear Equation: π π β π = βπ π π π π π β (π) = βπ π π β π = βπ π β π + π = βπ + π π=π π π π π β (π) = βπ π π β π = βπ π β π + π = βπ + π π=π π π π π β (π) = βπ π π β π = βπ π β π + π = βπ + π π=π π ππ π π β (π) = βπ π π β ππ = βπ π β ππ + ππ = βπ + ππ π = ππ π ππ π π β (ππ) = βπ π π β ππ = βπ π β ππ + ππ = βπ + ππ π = ππ ππ Linear Equations in Two Variables 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 150 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. Lesson 12 NYS COMMON CORE MATHEMATICS CURRICULUM 2. Find five solutions for the linear equation π π π π 8β’4 π + π = ππ, and plot the solutions as points on a coordinate plane. Linear Equation: π π + π = ππ π π (π) + π = ππ π π + π = ππ π ππ π = ππ π π (π) + π = ππ π π + π = ππ ππ π = ππ π π (π) + π = ππ π π + π = ππ π π=π ππ π (ππ) + π = ππ π π + π = ππ π π=π ππ π (ππ) + π = ππ π π + π = ππ π π=π Lesson 12: Linear Equations in Two Variables 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 151 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. Lesson 12 NYS COMMON CORE MATHEMATICS CURRICULUM 3. 8β’4 π π Find five solutions for the linear equation βπ + π = βπ, and plot the solutions as points on a coordinate plane. π π Lesson 12: Linear Equation: π βπ + π = βπ π π βπ + (π) = βπ π βπ + π = βπ βπ + π + π = βπ + π π = βπ + π π + π = βπ + π + π π=π π π ππ π βπ + (π) = βπ π βπ + π = βπ βπ + π + π = βπ + π π = βπ + π π + π = βπ + π + π ππ = π π ππ π βπ + (ππ) = βπ π βπ + π = βπ βπ + π + π = βπ + π π = βπ + π π + π = βπ + π + π ππ = π ππ ππ π βπ + (ππ) = βπ π βπ + ππ = βπ βπ + π + ππ = βπ + π ππ = βπ + π ππ + π = βπ + π + π ππ = π ππ ππ π βπ + (ππ) = βπ π βπ + ππ = βπ βπ + π + ππ = βπ + π ππ = βπ + π ππ + π = βπ + π + π ππ = π ππ Linear Equations in Two Variables 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 152 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. Lesson 12 NYS COMMON CORE MATHEMATICS CURRICULUM 4. 8β’4 Find five solutions for the linear equation ππ + π = π, and plot the solutions as points on a coordinate plane. π Linear Equation: ππ + π = π π π(π) + π = π π π+π= π π π=π π(π) + π = π π π+π= π π π=π π(π) + π = π π π+π =π βπ π = βπ π(π) + π = π π π+π =π βπ π = βπ π(π) + π = π π ππ + π = π βπ π = βπ Lesson 12: Linear Equations in Two Variables 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 153 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. Lesson 12 NYS COMMON CORE MATHEMATICS CURRICULUM 5. 8β’4 Find five solutions for the linear equation ππ β ππ = ππ, and plot the solutions as points on a coordinate plane. π ππ π ππ π ππ ππ π ππ π Lesson 12: Linear Equation: ππ β ππ = ππ ππ β π(π) = ππ ππ β π = ππ ππ β π + π = ππ + π ππ = ππ π ππ π= π π ππ π= π ππ β π(π) = ππ ππ β ππ = ππ ππ β ππ + ππ = ππ + ππ ππ = ππ π ππ π= π π ππ π= π ππ β π(π) = ππ ππ β ππ = ππ ππ β ππ + ππ = ππ + ππ ππ = ππ π = ππ ππ β π(π) = ππ ππ β ππ = ππ ππ β ππ + ππ = ππ + ππ ππ = ππ π ππ π= π π ππ π= π ππ β π(π) = ππ ππ β ππ = ππ ππ β ππ + ππ = ππ + ππ ππ = ππ π ππ π= π π ππ π= π Linear Equations in Two Variables 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 π π π π π 154 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License.
© Copyright 2026 Paperzz