Objectives

<p>Students will go through linear equation systems in this unit. Students will:&nbsp;<br>- find solutions to real-world systems of linear equations.&nbsp;<br>- create real-world problems using linear equation systems.</p>

Core Questions

<p>- How can graphing help students get a conceptual knowledge of algebraic and/or number theory topics?&nbsp;<br>&nbsp;</p>

Vocabulary

<p>- Inconsistent System: Two or more equations that are not satisfied by any one set of values for the variables.&nbsp;<br>- Solution to a Linear System: The process of finding a set of values of the unknowns which satisfy all the equations.&nbsp;<br>- System of Linear Equations: A set of two or more linear equations, which can be solved to find one particular value for a variable.</p>

Materials

<p>- copies of Lesson 3 Exit Ticket (M-A1-3-3_Lesson 3 Exit Ticket and KEY)&nbsp;<br>- a graphing calculator&nbsp;<br>- GeoGebra (optional)</p>

Assignment

<p>- <span style="background-color:rgb(255,255,255);color:rgb(8,42,61);">While graphing calculators or computer software are highly useful teaching tools, their use requires adherence to procedures and routines.</span> Assign individual presentations to the class to assess the student's understanding of both methods and concepts. Students are expected to explain topics and show that they understand procedures and routines throughout the presentations. Particularly helpful are calculators and computer screens that may be displayed for the entire class.&nbsp;<br>- Examine problems 2 and 3 of the Lesson 3 Exit Ticket to find the solutions that match the proper ordered pairs for each equation. Utilize a more comprehensive evaluation of the graphs, taking into account the approximate locations of the two graphs' intersections as well as the roughly accurate <i>x</i>- and <i>y</i>-intercepts and slopes (29, 17).&nbsp;<br>&nbsp;</p>

Supports

<p>Explicit Instruction, Modeling, Scaffolding, and Active Engagement<br>W: The lesson focuses on solving system-of-linear-equation word problems by graphing. Along with five systems-of-equation word problems that the students designed and a PowerPoint presentation, the lesson also contains an Exit Ticket evaluation that has additional system-of-linear-equation word problems.<br>H: Students will probably become interested in the presentation of instances of linear systems of equations and the modeling of each word problem. Students' attention will probably be maintained by the assignment to create five more system-of-linear-equation word problems, complete with a PowerPoint presentation and a solution component. Every student will have a fellow student help them solve their tasks, which should encourage ingenuity and precision.<br>E: The lesson is both concrete and abstract; at the start of the lesson, concrete modeling is given, and at the end, abstract thinking is needed.<br>R: Students revisit, rethink, contemplate, and revise during the class discussion and the PowerPoint presentation, as well as during the class review time.<br>E: Due to the requirement of creating five original word problems with solution keys including systems of linear equations, students must assess their&nbsp;understanding.<br>T: Learning for all learning styles will be facilitated by&nbsp;discussion, graphs, and explicit modeling. Support for individual students might be provided as required. Students will have the chance to question other students as they work through difficulties that other students have created.<br>O: After showing students how to solve problems, the lesson allows them&nbsp;to "teach" other students using the PowerPoint presentation. An approach known as "train-the-trainer" is used in this lesson.</p>

Procedures

<p>Present systems of linear equations as a new topic. Ask, <strong>"What is a system of linear equation? A </strong><i><strong>system of linear equations</strong></i><strong> is a collection of two or more linear equations that can be resolved so that a variable has a single defined value. Let's check a few instances."</strong> Write these down on the board:&nbsp;<br><br><i>y</i> = 6<i>x</i><br><br>3<i>x</i> - <i>y</i> = 1<br><br>8<i>x</i> - 9<i>y</i> = 3<br><br>-4<i>x</i> + 2<i>y</i> = -1<br><br><i>x</i> + 3<i>y</i> = 2<br><br>-<i>x</i> + <i>y</i> = 4<br><br>Using substitution, elimination, and graphing, the three systems displayed can be solved for the <i>x</i>- or <i>y</i>-variable. This lesson's primary goal is to teach you how to solve systems of linear equations graphically. <strong>"Let's now work on converting word problems into linear equation systems and solving them."</strong><br><br><strong>Problem 1</strong><br><br><strong>"We'll begin with the well-known Pigs and Chickens system of equations problem:"</strong><br><br>Anna has a certain number of pigs and chickens on her farm. She counts 36 heads and 126 feet . How many chickens and pigs does she have on her farm?<br><br><strong>"The following set of equations can be derived from this problem:</strong><br><br><i><strong>p</strong></i><strong> + </strong><i><strong>c</strong></i><strong> = 36</strong><br><br><strong>4</strong><i><strong>p</strong></i><strong> + 2</strong><i><strong>c</strong></i><strong> = 126&nbsp;</strong><br><br><strong>To solve the system, we'll use a graphical method. How will the resolution appear?" </strong><i>(The intersection of the two graphs will serve as the solution.)</i><br><br><strong>"We will rewrite each equation in slope-intercept form and input it into the graphing calculator. We will simply substitute </strong><i><strong>x</strong></i><strong> and </strong><i><strong>y</strong></i><strong> for </strong><i><strong>c</strong></i><strong> and </strong><i><strong>p</strong></i><strong>. To start, we will solve for just one variable. By doing this, you gain:</strong><br><br><i><strong>p</strong></i><strong> = -</strong><i><strong>c</strong></i><strong> + 36</strong><br><br><i><strong>p</strong></i><strong> = -\(1 \over 2\)&nbsp;</strong><i><strong>c</strong></i><strong> + \(63 \over 2\)&nbsp;</strong><br><br><strong>Substituting </strong><i><strong>y</strong></i><strong> for </strong><i><strong>p</strong></i><strong> and </strong><i><strong>x</strong></i><strong> for </strong><i><strong>c</strong></i><strong>, and we have:</strong><br><br><i><strong>y</strong></i><strong> = -</strong><i><strong>x</strong></i><strong> + 36</strong><br><br><i><strong>y</strong></i><strong> = -\(1 \over 2\)</strong><i><strong>x</strong></i><strong> + \(63 \over 2\)</strong><br><br><strong>Graphing calculators like GeoGebra or our own will be used to enter each equation. It is necessary to consult the table to confirm the point of intersection because it is not always obvious. We know that the intersection happens somewhere close&nbsp;to an </strong><i><strong>x</strong></i><strong>-value of 8 or 9. We discover that at the point (9, 27), the </strong><i><strong>y</strong></i><strong>-value is the same for every line. Therefore, our precise point of intersection, or the answer, is (9, 27).</strong></p><figure class="image"><img style="aspect-ratio:390/333;" src="https://storage.googleapis.com/worksheetzone/images/Screenshot_83.png" width="390" height="333"></figure><p><strong>"What can we infer from this? We can determine that Anna has 9 chickens and 27 pigs on her farm, as </strong><i><strong>x</strong></i><strong> stands for the number of chickens (</strong><i><strong>c</strong></i><strong>) and </strong><i><strong>y</strong></i><strong> for the number of pigs (</strong><i><strong>p</strong></i><strong>)."</strong><br><br><strong>Problem 2</strong><br><br><strong>"Another real-world problem that involves the total number of attendees at an event and the fee per admission, where the number of people for each admission type must be determined:</strong><br><br><strong>It costs $5.25 for kids under 12 to see a movie. A movie ticket costs $7.50 for adults (or children over 12). The manager reports making $1473.75 on Friday night from 210 ticket sales. How many children and adults bought tickets?"</strong><br><br><strong>"This problem can be expressed as the following set of equations:</strong><br><br><i><strong>a</strong></i><strong> + </strong><i><strong>c</strong></i><strong> = 210</strong><br><br><strong>7.50</strong><i><strong>a</strong></i><strong> + 5.25</strong><i><strong>c</strong></i><strong> = 1473.75</strong><br><br><strong>where </strong><i><strong>a</strong></i><strong> represent the number of adults and </strong><i><strong>c</strong></i><strong> represents the number of children. Again, the solution will be the intersection of the two lines."</strong><br><br><strong>"We will rewrite each equation in slope-intercept form and input it into the graphing calculator. We'll use </strong><i><strong>y</strong></i><strong> and </strong><i><strong>x</strong></i><strong> in place of a and c, respectively. First, we'll solve for just one variable. We'll work out the value of </strong><i><strong>a</strong></i><strong>. By doing this, we get:</strong><br><br><i><strong>a</strong></i><strong> = -</strong><i><strong>c</strong></i><strong> + 210</strong><br><br><i><strong>a</strong></i><strong> = -.7</strong><i><strong>c</strong></i><strong> + 196.50</strong></p><p><strong>Substituting </strong><i><strong>y</strong></i><strong> for </strong><i><strong>a</strong></i><strong>, and </strong><i><strong>x</strong></i><strong> for </strong><i><strong>c</strong></i><strong>, we get:&nbsp;</strong><br><br><i><strong>y</strong></i><strong> = -</strong><i><strong>x</strong></i><strong> + 210</strong><br><br><i><strong>y</strong></i><strong> = -.7</strong><i><strong>x</strong></i><strong> + 196.50</strong><br><br><strong>We'll enter each equation into our own graphing calculator or a graphing calculator like GeoGebra. The point of intersection is difficult to determine, so consult the table. We know that the point of intersection is somewhere between 43 and 46 </strong><i><strong>x</strong></i><strong>-values. Using the table, we can see that the </strong><i><strong>y</strong></i><strong>-value for each line at (45, 165) is the same. Thus, (45, 165) is our precise point of intersection, and it is the solution."</strong></p><figure class="image"><img style="aspect-ratio:508/380;" src="https://storage.googleapis.com/worksheetzone/images/Screenshot_84.png" width="508" height="380"></figure><p><strong>"What is this telling us? Because </strong><i><strong>x</strong></i><strong> represents </strong><i><strong>c</strong></i><strong> (the number of children) and </strong><i><strong>y</strong></i><strong> represents </strong><i><strong>a</strong></i><strong> (the number of adults), we know 45 children's tickets and 165 adult tickets were sold on Friday night."</strong><br><br><strong>Problem 3</strong><br><br><strong>"The graph sometimes offers us an approximation of the intersection point, and even the table, when set to its default setting, does not always supply the exact answer we require. In this situation, we just alter the table settings. Consider the case below:</strong><br><br><strong>On Monday, Elizabeth paid $4.58 for a mix of 5 apples and 4 bananas. On Tuesday, she paid $2.49 for&nbsp;2 apples and 3 bananas. What was the cost of an apple? What was the cost of a banana?"</strong><br><br><strong>"This problem can be expressed as the following set of equations:</strong><br><br><strong>5</strong><i><strong>a</strong></i><strong> + 4</strong><i><strong>b</strong></i><strong> = 4.58</strong><br><br><strong>2</strong><i><strong>a</strong></i><strong> + 3</strong><i><strong>b</strong></i><strong> = 2.49</strong><br><br><strong>where </strong><i><strong>a</strong></i><strong>&nbsp;stands for the price of one apple and </strong><i><strong>b</strong></i><strong> for the price of one banana."</strong><br><br><strong>"The point where the two lines intersect will be the solution."</strong><br><br><strong>"We'll enter every equation into the graphing calculator and rewrite it using the slope-intercept method. We will simply substitute </strong><i><strong>x</strong></i><strong> and </strong><i><strong>y</strong></i><strong> for </strong><i><strong>a</strong></i><strong> and </strong><i><strong>b </strong></i><strong>respectively. As a result, </strong><i><strong>x</strong></i><strong> will reflect the cost of one apple, while </strong><i><strong>y</strong></i><strong> will represent the cost of a banana. We'll start by solving for one variable. Doing so gives:</strong><br><br><i>a</i> = -\(4 \over 5\)<i>b</i> + .916<br><br><i>a</i> = -\(3 \over 2\)<i>b</i> + 1.245&nbsp;<br><br><strong>Substituting </strong><i><strong>y</strong></i><strong> for </strong><i><strong>a</strong></i><strong> and </strong><i><strong>x</strong></i><strong> for </strong><i><strong>b</strong></i><strong>, we get:</strong><br><br><i>y</i> = -\(4 \over 5\)<i>x</i> + .916<br><br><i>y</i> = (-3/2)<i>x</i> + 1.245&nbsp;&nbsp;<br><br><strong>“We'll type each equation into our own graphing calculator or a graphing calculator like GeoGebra. The graph for the two equations is provided below.”</strong></p><figure class="image"><img style="aspect-ratio:319/281;" src="https://storage.googleapis.com/worksheetzone/images/Screenshot_85.png" width="319" height="281"></figure><p><strong>"The point of intersection cannot be determined only from the graph; hence, a reference to the table is required. The program's default increment for x is commonly set to 1. However, the default choice of 1 for each succeeding x-value in these two equations does not offer the degree of precision required to display the point of intersection. We can tweak the table setting by performing the following program adjustment: "</strong><br><br>1. Press \(2^{nd}\) Window (TBLSET).<br><br>2. Change the interval Δ<i>Tbl</i> to another value.<br><br>Just clarify<strong>: "'Change </strong><i><strong>ΔTbl'</strong></i><strong> refers to either the increment value for </strong><i><strong>x</strong></i><strong> or the change in </strong><i><strong>x</strong></i><strong>-values. As we work with money quantities to the hundredth place, we will adjust our </strong><i><strong>x</strong></i><strong>-value increment to .01."</strong><br><br><strong>"Using the table, we determine that the </strong><i><strong>y</strong></i><strong>-value is the same for each line at the point (0.47, 0.54), where </strong><i><strong>x</strong></i><strong> represents the cost per banana and </strong><i><strong>y</strong></i><strong> represents the cost per apple. So each apple costs $0.54, and each banana costs $0.47."</strong> Please provide as many instances as necessary.<br><br>Students should do the Lesson 3 Exit Ticket as an activity (M-A1-3-3_Lesson 3 Exit Ticket and KEY).<br><br><strong>Review Activity</strong><br><br>Create five linear-system-of-equations word problems. Prepare a solution key, including a graph and any table processes that were used. Use PowerPoint to show the word problems. To show the solutions, create a separate "answer key and solution" PowerPoint presentation. Each PowerPoint presentation will be switched with a classmate at random using a numbered system. Each student will complete five different word problems. The solutions will be compared to the author's techniques and solutions. There will be an opportunity for discussion and questions.<br><br>To recap the lesson, hold a class discussion about any questions, concerns, or problems that arose throughout the session. In addition, have students generate a list of at least ten "scenarios" in which systems of linear equations are required to solve a problem. For example, students could present an example of needing to know how many red rose bushes and yellow rose bushes were purchased, as well as the amount of money spent per rose bush type, given the total number of rose bushes and money spent.<br><br><strong>Extension:</strong><br><br>Ask students to develop a list of consumer products they are acquainted with, along with an estimate of the product's cost that they can best recall. Assign them to examine sales data at quarterly or annual intervals and form conclusions regarding the relationships they can discern between changes in the cost of each item and its sales.<br><br>(As an academic field, economics teaches the law of supply and demand, which states that when an item or service's price rises, the demand for buying it falls, and vice versa, when its price falls, the demand for buying it rises. The law is also expressed in reverse: as demand for an item or service rises, so does the price of the item or service; as demand falls, so does the price of the thing. This is a dynamic system in which producers sell more when costs are low and sell less when they are high, while consumers buy more when costs are low and less when costs are high. Despite its crudeness, the supply and demand model accurately predicts how a large number of people will purchase and sell a huge number of commodities and services.)<br><br>Assign students to create lists of additional elements (variables) that influence the purchase and sale of products and services. A good place to start is to inquire whether purchasing particular types of things is discretionary or mandatory. (<i>Answers will vary, but some examples include:</i> food, clothing, housing, medical expenses, advertising, controlling legislation, and seasonal shopping.)</p>
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Solutions of Systems of Linear Equations

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Description
What It Is:
This is a lesson plan worksheet focusing on 'Solutions of Systems of Linear Equations (M-A1-3-3)'. It outlines objectives such as finding solutions to real-world systems of linear equations and creating real-world problems using linear equation systems. The worksheet includes lesson core questions, vocabulary definitions (Inconsistent System, Solution to a Linear System, System of Linear Equations), a materials list (Lesson 3 Exit Ticket, graphing calculator, GeoGebra), and assessment guidelines for teachers. The assessment suggests using graphing calculators/software and examining specific problems from the Lesson 3 Exit Ticket.
Grade Level Suitability:
This worksheet is suitable for high school grades (9-12), specifically Algebra 1, as it deals with systems of linear equations, a topic typically covered in high school algebra courses. The concepts and vocabulary are appropriate for students at this level.
Why Use It:
This worksheet helps students understand systems of linear equations by providing clear objectives, defining key vocabulary, and suggesting assessment methods. It encourages the use of graphing calculators and software to visualize and solve problems, promoting a deeper understanding of the topic. It also helps teachers structure their lessons and assess student comprehension effectively.
How to Use It:
Teachers can use this worksheet as a guide for planning and delivering a lesson on systems of linear equations. They can use the core questions to initiate discussions, introduce the vocabulary, and utilize the suggested materials (Lesson 3 Exit Ticket, graphing calculator, GeoGebra). The assessment section provides guidance on evaluating student understanding through presentations and by examining specific problems.
Target Users:
The target users are high school math teachers teaching Algebra 1 or related courses, and high school students learning about systems of linear equations. The worksheet provides a framework for teachers to deliver the lesson and helps students understand the core concepts.
Lesson’s Materials
Teaching Progress
Solutions of Systems of Linear Equations
Solutions of Systems of Linear Equations
Solutions of Systems of Linear Equations
Solutions of Systems of Linear Equations
Solutions of Systems of Linear Equations
Solutions of Systems of Linear Equations
Solutions of Systems of Linear Equations
Solutions of Systems of Linear Equations