3.2.2 Warm-up, P.57 Benito is a waiter. He earns a base salary of $1,500 a month, plus 20% of the price of the meals he serves. Write an equation to predict the amount of money Benito will earn if he serves $350 in meals. Benito earned $1,650 in one month. What was the total price of the meals that Benito served? Create a graph to show the possible amount of money Benito could earn each month. Common Core State Standard: A–REI.6 3.2.2: Solving Systems of Linear Equations

Let x represent the total cost of the meals Benito serves. Write an equation to predict the amount of money Benito will earn if he serves $350 in meals. Translate the verbal description of Benito’s pay into an algebraic equation. Let x represent the total cost of the meals Benito serves. Let y represent the amount of money Benito earns. Benito earns a base salary of $1,500, plus 20% of the price of the meals. 20% of the price of the meals is 0.20x. y = 0.20x + 1500 3.2.2: Solving Systems of Linear Equations

y = 0.20x + 1500 Original equation Benito served $350 in food. Substitute $350 for x and solve the equation for y, the amount of money Benito will earn. y = 0.20x + 1500 Original equation y = 1500 + 0.20(350) Substitute 350 for x. y = 1500 + 70 Simplify. y = 1570 Benito earns $1,570 if he serves $350 worth of food. 3.2.2: Solving Systems of Linear Equations

Substitute $1,650 for y, the total amount of money Benito earned. Benito earned $1,650 in one month. What was the total price of the meals that Benito served? Use your equation from part 1 to find the price of the meals Benito served. Substitute $1,650 for y, the total amount of money Benito earned. 1650 = 0.20x + 1500 3.2.2: Solving Systems of Linear Equations

Solve for x. Benito served $750 worth of meals. Original equation 150 = 0.20x Subtract 1,500 from both sides. 750 = x Divide both sides by 0.20. 3.2.2: Solving Systems of Linear Equations

The slope of the equation is 0.20. The y-intercept is 1,500. Create a graph to show the possible amount of money Benito could earn each month. The equation y = 0.20x + 1500 can be graphed using slope-intercept form. The slope of the equation is 0.20. The y-intercept is 1,500. 3.2.2: Solving Systems of Linear Equations

3.2.2: Solving Systems of Linear Equations Connection to the Lesson • Students will be asked to analyze similar situations and go a step further by finding one solution for both scenarios. • Students will be asked to use graphs to determine solutions to scenarios. 3.2.2: Solving Systems of Linear Equations

3.2.2: Introduction Point of intersection: The solution to a system of equations that make both equations true. This is the point at which two lines cross or meet. Systems of equations can have one solution, no solutions, or an infinite number of solutions. On a graph, the solution to a system of equations can be easily seen. 3.2.2: Solving Systems of Linear Equations

Solving Systems of Equations by Graphing: 3.2.2: Key Concepts There are various methods to solving a system of equations. One is the graphing method. Solving Systems of Equations by Graphing: Graphing a system of equations on the same coordinate plane allows you to visualize the solution to the system. Use a table of values, the slope-intercept form of the equations (y = mx + b), or a graphing calculator. Creating a table of values can be time consuming depending on the equations, but will work for all equations. 3.2.2: Solving Systems of Linear Equations

Key Concepts, continued Equations not written in slope-intercept form will need to be rewritten in order to determine the slope and y-intercept. Once graphed, you can determine the number of solutions the system has. Graphs of systems with one solution have two intersecting lines. The point of intersection is the solution to the system. These systems are considered consistent(at least 1 solutiuon) or independent(exactly 1 solution). 3.2.2: Solving Systems of Linear Equations

Key Concepts, continued Inconsistent Systems: Graphs of systems with no solution have parallel lines, which means that there are no points that satisfy both of the equations. Dependent Systems: with an infinite number of solutions are equations of the same line—the lines for the equations in the system overlap. Graphing the system of equations can sometimes be inaccurate, but solving the system algebraically will always give an exact answer. 3.2.2: Solving Systems of Linear Equations

Key Concepts, continued Intersecting Lines Parallel Lines Same Line One solution No solutions Infinitely many solutions Consistent, Independent Inconsistent Dependent 3.2.2: Solving Systems of Linear Equations

Common Errors/Misconceptions incorrectly graphing each equation misidentifying the point of intersection 3.2.2: Solving Systems of Linear Equations

Guided Practice Example #1: Graph the system of equations. Then determine whether the system has one solution, no solution, or infinitely many solutions. If the system has a solution, name it. 3.2.2: Solving Systems of Linear Equations

Guided Practice: Example #1, continued Solve each equation for y. The first equation needs to be solved for y. The second equation (y = –x + 3) is already in slope-intercept form. 4x – 6y = 12 Original equation –6y = -4x + 12 Subtract 4x from both sides. Divide both sides by –6. 3.2.2: Solving Systems of Linear Equations

Guided Practice: Example #1, continued Graph both equations using the slope-intercept method. The y-intercept of is –2. The slope is . The y-intercept of y = –x + 3 is 3. The slope is –1. 3.2.2: Solving Systems of Linear Equations

Guided Practice: Example #1, continued y = -x + 3 (3, 0) y = ⅔x - 2 3.2.2: Solving Systems of Linear Equations

Guided Practice: Example #1, continued The system has one solution: (3, 0) ✔ 3.2.2: Solving Systems of Linear Equations

Dependent Solutions Guided Practice Example #2: Graph the system of equations. Then determine whether the system has one solution, no solution, or infinitely many solutions. If the system has a solution, name it. –8x + 4y = 4 4y = 8x + 4 y = 2x + 1 Both lines have the same equation, y = 2x + 1, so there are infinite solutions because the lines are the same. Dependent Solutions 3.2.2: Solving Systems of Linear Equations

Guided Practice: Example #2, continued y =2x + 1 3.2.2: Solving Systems of Linear Equations

Guided Practice Example #3: Graph the system of equations. Then determine whether the system has one solution, no solution, or infinitely many solutions. If the system has a solution, name it. 3.2.2: Solving Systems of Linear Equations

Guided Practice: Example #3, continued Solve each equation for y. The first equation needs to be solved for y. The second equation (y = 3x – 5) is already in slope-intercept form. –6x + 2y = 8 Original equation 2y = 6x + 8 Add 6x to both sides. y = 3x + 4 Divide both sides by 2. 3.2.2: Solving Systems of Linear Equations

Guided Practice: Example #3, continued Graph both equations using the slope-intercept method. The y-intercept of y = 3x + 4 is 4. The slope is 3. The y-intercept of y = 3x – 5 is –5. The slope is 3. 3.2.2: Solving Systems of Linear Equations

Guided Practice: Example #3, continued y = 3x + 4 y = 3x - 5 3.2.2: Solving Systems of Linear Equations

Guided Practice: Example #3, continued The system has no solutions, the lines are parallel. ✔ 3.2.2: Solving Systems of Linear Equations

Bonus Example #4(Use P.58): Guided Practice Bonus Example #4(Use P.58): Write a system of equations for #4 and then use the graphing method to solve it. Copy the following problem: #4) A test is worth 100 points and has 26 questions. True/False questions are worth 2 points and short answers are worth 5 points. How many 2-point questions are on the test? How many 5-point questions are on the test?? How many points is the test worth? b) How many total questions are there? c) How many 2-point questions are there? d) How many 5-point questions are there? 100 26 ?? ?? 3.2.2: Solving Systems of Linear Equations

Guided Practice: Example #4, continued Set up an equation for each scenario(2) in order to create the system of equations needed to solve. Create an equation for the total number of questions on the test: Identify what each variable represents: Create an equation for the total number of each question on the test in relation to the total number of points for the test: x + y = 26 x = 2-point ?’s y = 5-point ?’s 2x + 5y = 100 3.2.2: Solving Systems of Linear Equations

Guided Practice: Example #4, continued Set up an equation for each scenario(2) in order to create the system of equations needed to solve. 3.2.2: Solving Systems of Linear Equations

Guided Practice: Example #4, continued 2. Solve for the variables by either substitution or elimination. Then graph the equations to confirm the answer. 2x + 5y = 100 2x + 5y = 100 x + y = 26 x + y = 26 2x + 5y = 100 2 (x + y = 26) x + y = 26 2x + 5y = 100 x + 16 = 26 - 2x + 2y = 52 x = 10 3y = 48 (10, 16) y = 16 3.2.2: Solving Systems of Linear Equations

Guided Practice: Example #4, continued 3. Solve each equation for y in order to graph each equation The first equation needs to be solved for y. 2x + 5y = 100 Original equation 5y = -2x + 100 Subtract -2x from both sides. y = -⅖x + 20 Divide both sides by 5. 3.2.2: Solving Systems of Linear Equations

Guided Practice: Example #4, continued 3. Solve each equation for y. The second equation needs to be solved for y. x + y = 26 Original equation y = -x + 26 Subtract x from both sides. 3.2.2: Solving Systems of Linear Equations

Guided Practice: Example #4, continued 4. Graph both equations using the slope- intercept method. The y-intercept of y = -⅖x + 20 is 20. The slope is -⅖. The y-intercept of y = -x + 26 is 26. The slope is -1. 3.2.2: Solving Systems of Linear Equations

Show all work(Use P.64 or P.65)!! Use Example #4 to help with #5 Homework Workbook(3.2.2): P.63 # 1-5 Show all work(Use P.64 or P.65)!! Use Example #4 to help with #5 Use the graphs in the back of the workbook!! http://www.walch.com/ei/00004

In Class, 10/17: Problem Based Task, P.51 Homework/Classwork Study for Unit 3 TEST Will cover the following sections: 3.1.1, 3.1.2, 3.1.3, 3.1.4, 3.2.1, 3.2.2 http://www.walch.com/ei/00004