Solve Systems of Equations Algebraically

Part 1: Introduction Solve Systems of Equations Algebraically Develop Skills and Strategies CCSS 8.EE.C.8b You know that solutions to systems of linear equations can be shown in graphs. Now you will learn about other ways to find the solutions. Take a look at this problem. Sienna wrote these equations to help solve a number riddle. y 5 x 2 20 x 1 y 5 84 What values for x and y solve both equations? Explore It Use math you already know to solve the problem. What does y 5 x 2 20 tell you about the relationship between x and y? What does x 1 y 5 84 tell you about the relationship between x and y? You can guess and check to solve the problem. Try 44 for x and 40 for y. Do these numbers solve both equations? Now try 50 for x. If x 5 50, what is y when y 5 x 2 20? Does that work with the other equation? Try x 5 52. What do you find? Explain how you could find values for x and y that solve both equations. 138

Part 1: Introduction Find Out More In Lesson 15, you learned that without actually solving, you can tell if a system of equations will have exactly one solution, no solution, or infinitely many solutions. Here are some examples. x 1 y 5 6 2x 1 2y 5 12 5x 1 y 5 3 x 5 4 2 5x The second equation is double the first one, so they are the same equation with the same graph and solution set. This system has infinitely many solutions. If you write both equations in slope-intercept form, you find that y 5 25x 1 4 and y 5 25x 1 3. The lines have the same slope and different intercepts so they are parallel. This system has no solutions. If a system of equations has exactly one solution, like the problem on the previous page, there are different ways you can find the solution. You could guess and check, but that is usually not an efficient way to solve a system of equations. You could graph each equation, but sometimes you can t read an exact answer from the graph. Here is one way to solve the problem algebraically. y 5 x 2 20 x 1 y 5 84 Substitute x 2 20 for y in the second equation and solve for x. x 1 (x 2 20) 5 84 2x 2 20 5 84 2x 5 104 x 5 52, so y 5 32 You will learn more about algebraic methods later in the lesson. Reflect 1 How does knowing x 5 52 help you find the value of y? 139

Part 2: Modeled Instruction Read the problem below. Then explore how to use substitution to solve systems of equations. Solve this system of equations. y 5 x 1 2 y 1 1 5 24x Graph It You can graph the equations and estimate the solution. Find the point of intersection. It looks like the solution is close to ( 2 1 2, 1 1 2 ). Model It 5 4 3 2 1 2524232221 0 21 22 1 2 3 4 5 23 24 25 You can use substitution to solve for the first variable. Notice that one of the equations tells you that y 5 x 1 2. This allows you to use substitution to solve the system of equations. Substitute x 1 2 for y in the second equation. y 5 x 1 2 y 1 1 5 24x (x 1 2) 1 1 5 24x Now you can solve for x. x 1 2 1 1 5 24x x 1 3 5 24x 3 5 25x x 5 2 3 5 140

Part 2: Guided Instruction Connect It Now you will solve for the second variable and analyze the solution. 2 What is the value of x? How can you find the value of y if you know the value of x? 3 Substitute the value of x in the equation y 5 x 1 2 to find the value of y. 4 Now substitute the value of x in the equation y 5 24x 2 1 to find the value of y. 5 What is the ordered pair that solves both equations? Where is this ordered pair located on the graph? 6 Look back at Model It. How does substituting x 1 2 for y in the second equation give you an equation that you can solve? 7 How does substitution help you to solve systems of equations? Try It Use what you just learned to solve these systems of equations. Show your work on a separate sheet of paper. 8 y 2 3 5 2x y 5 4x 2 2 9 y 5 1.4x 1 2 y 2 3.4x 5 22 141

Part 3: Modeled Instruction Read the problem below. Then explore how to solve systems of equations using elimination. Solve this system of equations. 2x 2 2y 5 4 3y 5 20.5x 1 2 Model It You can use elimination to solve for one variable. First, write both equations so that like terms are in the same position. Then try to eliminate one of the variables, so you are left with one variable. To do this, look for a way to get opposite coefficients for one variable in the two equations. 22y 5 x 1 4 3y 5 20.5x 1 2 2(3y 5 20.5x 1 2) 6y 5 2x 1 4 22y 5 x 1 4 6y 5 2x 1 4 4y 5 8 y 5 2 22(2) 5 x 1 4 2x 2 4 5 4 2x 5 8 x 5 28 Check: 3(2) 5 20.5(28) 1 2 6 5 4 1 2 Multiply the second equation by 2. Now you have opposite terms: x in the first equation and 2x in the second equation. Add the like terms in the two equations. The result is an equation in just one variable. Divide each side by 4 to solve the equation for y. Substitute the value of y into one of the original equations and solve for x. Substitute your solution in the other original equation. 142

Part 3: Guided Instruction Connect It Now analyze the solution and compare methods for solving systems of equations. 10 What happens when you add opposites? Why do you want to get opposite coefficients for one of the variables? 11 How did you get opposite coefficients for x in the solution on the previous page? 12 Why does the equation stay balanced when you add the values on each side of the equal sign? 13 Which equation in the system was used to find the value of x? Can you use the other equation? Explain. 14 How is elimination like substitution? How is it different? 15 How can you check your answer? Try It Use what you just learned about elimination to solve this problem. Show your work on a separate sheet of paper. 16 2x 1 y 5 9 3x 2 y 5 16 143

Part 4: Guided Practice The student divided each term in the equation 2y 5 6x 2 2 by 2 to get an expression equal to y. Study the model below. Then solve problems 17 19. Solve this system of equations. 3y 5 x 1 1 2y 5 6x 2 2 Student Model Look at how you could use substitution to solve a system of equations. 2y 5 6x 2 2 2 y 5 3x 2 1 Since y 5 3x 2 1, I can substitute 3x 2 1 for y in the first equation. 3(3x 2 1) 5 x 1 1 9x 2 3 5 x 1 1 8x 5 4 x 5 1 2 Solve the problem using elimination. 3y 5 Solution: 1 1; y 5 1 2 1 2 1 1 2, 1 2 2 Can it help to write both equations in the same form? 17 What ordered pair is a solution to y 5 x 1 5 and x 2 5y 5 29? Show your work. Discuss your solution methods. Do you prefer using substitution or elimination? Solution: 144

Part 4: Guided Practice 18 Graph the equations. What is your estimate of the solution of this system of equations? y 5 3x 2 2 y 5 22x Show your work. Do the equations have the same or different slopes? Solution: 5 4 3 2 1 2524232221 0 21 22 1 2 3 4 5 23 24 25 Solve the problem algebraically and compare the solution to your estimate. 19 Which of these systems of equations has no solution? A y 5 x 4 1 2 B y 5 4x 2 1 y 5 2x 3 2 3 y 5 2x 2 3 What do you know about the lines in a system of equations with no solution? C y 5 4x y 5 4x 2 5 D x 1 y 5 3 2y 5 22x 1 6 Sheila chose D as the correct answer. How did she get that answer? Graph the solution to verify your answer. 145

Part 1: Introduction AT A GLANCE Students explore a problem involving a system of equations. They use guess-and-check to find a solution. Part 1: Introduction Solve Systems of Equations Algebraically Develop Skills and Strategies CCSS 8.EE.C.8b STEP BY STEP Tell students that this page models solving a system of equations. Have students read the problem at the top of the page. Work through Explore It as a class. When the class reaches the third bullet point, tell students that any solution to the system has to work for both equations, not just one. Ask student pairs or groups to explain their answers to the last three questions. Discuss what might make certain methods difficult. For example, guess and check can involve many trials and these equations graph with larger numbers to seeing an exact point might be tricky. ELL Support Students may be unfamiliar with this usage of the term system. Tell them that in this context it means that a problem contains two or more equations. 138 You know that solutions to systems of linear equations can be shown in graphs. Now you will learn about other ways to find the solutions. Take a look at this problem. Sienna wrote these equations to help solve a number riddle. y 5 x 2 20 x 1 y 5 84 What values for x and y solve both equations? Explore It Use math you already know to solve the problem. What does y 5 x 2 20 tell you about the relationship between x and y? y is 20 less than x. What does x 1 y 5 84 tell you about the relationship between x and y? x and y have a sum of 84. You can guess and check to solve the problem. Try 44 for x and 40 for y. Do these numbers solve both equations? 40 1 44 5 84, but 40 Þ 44 2 20 so the numbers don t solve both equations. Now try 50 for x. If x 5 50, what is y when y 5 x 2 20? Does that work with the other equation? If x 5 50, y 5 50 2 20, or 30. But, 50 1 30 Þ 84 so it doesn t work in the other equation. Try x 5 52. What do you find? If x 5 52, then y 5 52 2 20, or 32. 52 1 32 5 84. So, the values x 5 52 and y 5 32 solve both equations. Explain how you could find values for x and y that solve both equations. You can guess and check or graph both equations to find an intersection point. Mathematical Discourse Could you solve just one of the equations? Why or why not? Some students may say you cannot solve one of the equations because you have more than one variable. Other students may say that you can find lots of solutions, and that any pair of x and y that makes the equation true is a solution. Why would someone want to set up a system of equations? Students may be unsure. Some may have seen number riddles similar to the one described in the book. They may see how the system of equations could help you solve the riddles. Some may propose real-life situations in which it is helpful or necessary to use two or more equations to describe and solve a problem. 154

AT A GLANCE Part 1: Introduction Students discuss how systems of equations can have infinite solutions, no solution, or exactly one solution. They learn of algebraic methods that can solve systems of equations. Part 1: Introduction Find Out More In Lesson 15, you learned that without actually solving, you can tell if a system of equations will have exactly one solution, no solution, or infinitely many solutions. Here are some examples. STEP BY STEP x 1 y 5 6 2x 1 2y 5 12 The second equation is double the first one, so they are the same equation with the same graph and solution set. This system has infinitely many solutions. Read Find Out More as a class. Point out how it is much easier to compare the two equations if they are in the same form. Note that it is easy to tell when a system has infinite solutions or no solutions. Solving for one solution, however, takes more work. Ask, How did you solve for y? [substitute x 5 52 in y 5 x 2 20] Ask students to share their thoughts in Reflect. SMP Tip: Students persevere in solving the problem when they use guess and check. They make sense of the problem when they wonder if there might be a more efficient method (SMP 1). Tell students to think carefully about which methods might be best suited to each situation. Hands-On Activity 5x 1 y 5 3 x 5 4 2 5x If you write both equations in slope-intercept form, you find that y 5 25x 1 4 and y 5 25x 1 3. The lines have the same slope and different intercepts so they are parallel. This system has no solutions. If a system of equations has exactly one solution, like the problem on the previous page, there are different ways you can find the solution. You could guess and check, but that is usually not an efficient way to solve a system of equations. You could graph each equation, but sometimes you can t read an exact answer from the graph. Here is one way to solve the problem algebraically. y 5 x 2 20 x 1 y 5 84 Substitute x 2 20 for y in the second equation and solve for x. x 1 (x 2 20) 5 84 2x 2 20 5 84 2x 5 104 x 5 52, so y 5 32 You will learn more about algebraic methods later in the lesson. Reflect 1 How does knowing x 5 52 help you find the value of y? You can replace x with 52 in one of the equations to find the value of y. Real-World Connection 139 Practice substitution using algebra tiles. Material: a set of algebra tiles for each student pair Tell students they will explore this system of equations: 2x 1 y 5 7 y 5 3x 1 2 Have students model each equation with algebra tiles. Because y equals 3x 1 2, students should replace the y in the first equation with 3x 1 2. Have them physically do this with the tiles and then solve the equation for x. [x 5 1] Now have students model one of the equations again, replacing the x tiles with 11 tiles. Have them solve for y. [y 5 5] The solution is (1, 5). Have students model their own system of equations and trade with other student pairs to solve. Equations describe various constraints on variables. It may be that in a recipe, whether baking, or making cement, or some other compound, you need a certain ratio for the ingredients, and you also need the recipe to produce a certain amount. Those are two different constraints that can be described by two different equations. Solving that system of equations tells how much you need of each ingredient to get the desired result. In another situation, a company might source some of its products from two different vendors. They might use a system of equations to describe how much total product they need and describe the different charges they will incur from each vendor. Ask students to think of other situations where two or more variables might be affected by more than one constraint. 155

Part 2: Modeled Instruction AT A GLANCE Students review graphing to estimate a solution to a system of equations. Then they explore using substitution to find an exact solution. STEP BY STEP Read the problem at the top of the page as a class. In Graph It, review how to graph each equation. Remind students that the point of intersection is the solution to the system of equations. In Model It, discuss how both equations were solved for y. Make sure students understand where the equation (x 1 2) 1 1 5 24x comes from. Use the Mathematical Discourse to discuss alternative substitutions to method shown such as setting both equations equal to y and setting them equal. SMP Tip: Students model the problem when they use a graph to estimate the solution (SMP 4). Remind students that the solution to a system of equations is where the graphs of the equations intersect. 140 Part 2: Modeled Instruction Read the problem below. Then explore how to use substitution to solve systems of equations. Solve this system of equations. Graph It y 5 x 1 2 y 1 1 5 24x You can graph the equations and estimate the solution. Find the point of intersection. It looks like the solution is close to ( 2 1 2, 1 1 2 ). Model It You can use substitution to solve for the first variable. Notice that one of the equations tells you that y 5 x 1 2. This allows you to use substitution to solve the system of equations. Substitute x 1 2 for y in the second equation. y 5 x 1 2 y 1 1 5 24x (x 1 2) 1 1 5 24x Now you can solve for x. x 1 2 1 1 5 24x x 1 3 5 24x 3 5 25x x 5 2 3 5 5 4 3 2 1 2524232221 0 21 1 2 3 4 5 22 23 24 25 Hands-On Activity Review graphs of systems of equations. Tell students to look at these four equations: A y 5 x 1 3 B y 5 x 2 2 C 2y 5 2x 1 6 D y 5 2x Graph and label each equation. Ask, Will A and B ever meet? [No. They are parallel.] Do A and B have a solution in common? [No. There is no solution in common.] What do you notice about A and C? (When you solve C for y, you see they are the same equation. Their solution set would be every ordered pair on their graph.] What do you notice about A and D, or B and D? [They meet at one point and would have one solution.] Discuss how students can use the slope of each equation to learn about the solution set of any two equations. Mathematical Discourse Do you think graphing is a good way to find the solution to a system of equations? Why or why not? Some students might think graphing is a good way to estimate but realize it cannot accurately find fractional answers. Some may not like graphing and prefer an exact algebraic method. In the substitution method, why were the equations solved for y? Could it have been solved differently? Students may note that it was easy to solve both equations for y because one equation was already in that form. Some may suggest solving both equations for x. Listen for responses stating that a problem should have the same result no matter how it is solved. 156

Part 2: Guided Instruction AT A GLANCE Students revisit the problem on page 140 and discuss how to finish solving it. They understand that the solution to this system of equations is an ordered pair. STEP BY STEP Remind students that Connect It refers to the problem on page 140. Ask students what it would mean if each equation produced a different value for y. [Getting a different value for y would indicate that something is wrong. When a system of linear equations has a single solution, the solution can have just one value for y. In such a case, it is possible that the value found for x is incorrect.] Ask students why they found the value for y two times. [to make certain the solution was true for both equations in the system] As needed, review and discuss different methods of solving equations involving fractions, such as multiplying by a common denominator or working with improper fractions instead of mixed numbers. Help students articulate their method or thinking as they answer each problem. Make sure they use the appropriate algebraic terms. Part 2: Guided Instruction Connect It Now you will solve for the second variable and analyze the solution. 2 What is the value of x? How can you find the value of y if you know the value of x? 3 Substitute the value of x in the equation y 5 x 1 2 to find the value of y. y 5 2 1 2; y 5 1 3 5 2 5 4 Now substitute the value of x in the equation y 5 24x 2 1 to find the value of y. 5 What is the ordered pair that solves both equations? Where is this ordered pair located (2 3 on the graph? 5, 1 2 ) It is the point at which the lines intersect. 5 TRY IT SOLUTIONS 8 Solution: 12 1 2, 8 2 ; Students may solve 2x 1 3 5 4x 2 2 to get x 5 5, then solve 2 y 5 21 5 2 2 1 3 to get y 5 8. ERROR ALERT: Students who wrote (2 1 2, 24) may have incorrectly solved for y 5 2x 2 3, then found the y value from one equation. Remind students to check their answer in both equations to avoid mistakes. x 5 2 3 ; I can substitute 2 for x in either equation and solve to find the value of y. 5 3 5 6 Look back at Model It. How does substituting x 1 2 for y in the second equation give you an equation that you can solve? After you substitute, the equation that results has 7 How does substitution help you to solve systems of equations? It allows you to write Try It y 5 24(2 3 5 Use what you just learned to solve these systems of equations. Show your work on a separate sheet of paper. 8 y 2 3 5 2x y 5 4x 2 2 (2 1 2, 8) 9 y 5 1.4x 1 2 y 2 3.4x 5 22 ) 2 1; y = 1 2 5 2 1, y = or 1 7 5 2 5 only one variable in it so you can solve for that variable. an equation with one variable instead of two and then solve for that variable. (2, 4.8) 141 9 Solution: (2, 4.8); Students may solve 1.4x 1 2 5 3.4x 2 2 to get x 5 2, then solve y 5 1.4(2) 1 2 to get y 5 4.8. 157

Part 3: Modeled Instruction AT A GLANCE Students explore solving a system of equations using the elimination method. STEP BY STEP Read the problem at the top of the page as a class. Ask students why they need to get the equations in the same form. [Putting the equations in the same form makes it easier to eliminate a variable and can reduce the chances of making a mistake.] Ask why multiplying the entire equation by 2 is a legitimate step. [When you perform the same operation on both sides of an equation, the equation remains true.] Ask students why they multiplied the equation 3y 5 20.5x 1 2 by 2. [Multiplying the equation by 2 made the coefficient of x equal to 21. This was a necessary step in getting the x terms in the two equations to cancel each other.] Ask students how they could eliminate the y terms. [First, multiply 22y 5 x 1 4 by 3 so that the coefficient for y becomes 26. Then, multiply 3y 5 20.5x 1 2 by 2 so that the coefficient for y becomes 6. The y terms cancel out, leaving only the x terms to solve for.] SMP Tip: Students make use of the structure in the equations when they determine how to eliminate a variable (SMP 7). Tell students to write equations in the same form to make the structure easier to see and to help avoid mistakes. 142 Part 3: Modeled Instruction Read the problem below. Then explore how to solve systems of equations using elimination. Solve this system of equations. 2x 2 2y 5 4 3y 5 20.5x 1 2 Model It You can use elimination to solve for one variable. First, write both equations so that like terms are in the same position. Then try to eliminate one of the variables, so you are left with one variable. To do this, look for a way to get opposite coefficients for one variable in the two equations. 22y 5 x 1 4 3y 5 20.5x 1 2 2(3y 5 20.5x 1 2) 6y 5 2x 1 4 22y 5 x 1 4 6y 5 2x 1 4 4y 5 8 y 5 2 22(2) 5 x 1 4 2x 2 4 5 4 2x 5 8 x 5 28 Check: 3(2) 5 20.5(28) 1 2 6 5 4 1 2 Multiply the second equation by 2. Now you have opposite terms: x in the first equation and 2x in the second equation. Add the like terms in the two equations. The result is an equation in just one variable. Divide each side by 4 to solve the equation for y. Substitute the value of y into one of the original equations and solve for x. Substitute your solution in the other original equation. Mathematical Discourse Do you prefer the substitution method or the elimination method? Explain your reasons. Students will give different reasons. Listen to see if students understand that the best method may depend on the types and numbers of equations in the system. Substitution might be well suited for solving one kind of problem, and elimination might be the simpler method for solving another. 158

Part 3: Guided Instruction AT A GLANCE Students revisit the problem on page 142. They analyze how to use the elimination method and why it works. Part 3: Guided Instruction Connect It STEP BY STEP Point out that Connect It refers to the problem on page 142. Ask students why you want to eliminate one of the variables. [Eliminating one variable is a precondition for finding the value of the other variable.] Ask students how they can check their answers. [You could substitute the x- and y-values into both equations to see if they make true statements.] Now analyze the solution and compare methods for solving systems of equations. 10 What happens when you add opposites? Why do you want to get opposite coefficients for one of the variables? When you add opposites the sum is 0. If the coefficients are opposites, you can add and get 0 to eliminate that variable. 11 How did you get opposite coefficients for x in the solution on the previous page? Each term in 3y 5 20.5x 1 2 was multiplied by 2 to make the x coefficients opposites. 12 Why does the equation stay balanced when you add the values on each side of the equal sign? The values on each side of the equal sign are equal to each other, so you are adding the same values to both sides. Whenever you add the same value to both sides of an equation, it stays balanced. 13 Which equation in the system was used to find the value of x? 2x 2 2y 5 4 Can you use the other equation? Explain. Yes, it doesn t matter which of the two equations you use to find the value of x. You get the same value for x using both equations. 14 How is elimination like substitution? How is it different? Possible answer: With both methods you get one equation in one of the variables. With elimination, you Concept Extension Solve a system of three equations. Tell students you will determine, as a class, whether this system of three equations has a single solution: A 2x 1 y 5 4 B 3x 2 y 5 1 15 How can you check your answer? Substitute the values for the variables in the Try It create opposite coefficients, and add. With substitution, one variable is expressed in terms of the other and then substituted into one of the equations. other equation to be sure that you get true statements for both equations. Use what you just learned about elimination to solve this problem. Show your work on a separate sheet of paper. 16 2x 1 y 5 9 3x 2 y 5 16 (5, 21) 143 C x 2 2y 5 2 Tell students they will find a solution for two of the equations and then test it on the third one. Point out that equations A and B are ideal for using the elimination method to cancel out the y-values and find a value for x. Have the class perform the elimination method on equations A and B to solve for x. [x 5 1] Have the class solve equations A and B for x 5 1. What is the result? [y 5 2 for both equations] Ask, Expressed as an ordered pair, what is the solution to the system of equations A and B? [(1, 2)] TRY IT SOLUTION 16 Solution: (5, 21); Students may combine the two equations to eliminate y and solve 5x 5 25 to get x 5 5. Substituting in x 5 5, both 2(5) 1 y 5 9 and 3(5) 2 y 5 16 simplify to y 5 21. ERROR ALERT: Students who answered x 5 5 only found the x-value of the solution. Remind students that the solution to a system of equations in two variables contains both an x-value and a y-value. Have students substitute the x-value into either equation to find the y-value. Explain that for (1, 2) to be a complete solution, it must satisfy all three equations. As a class, test (1, 2) in equation C. [x 2 2y 5 2; 1 2 2(2) 5 2; 23 Þ 2] Ask, So, is (1, 2) a solution of the entire system? [no] Display a graph of the three-equation system for students. Ask, What would the graph have to look like to have a solution? [All three lines would have to cross at one point.] 159

Part 4: Guided Practice Part 4: Guided Practice Part 4: Guided Practice The student divided each term in the equation 2y 5 6x 2 2 by 2 to get an expression equal to y. Solve the problem using elimination. Can it help to write both equations in the same form? Discuss your solution methods. Do you prefer using substitution or elimination? Study the model below. Then solve problems 17 19. Solve this system of equations. 3y 5 x 1 1 2y 5 6x 2 2 Look at how you could use substitution to solve a system of equations. 2y 5 6x 2 2 2 y 5 3x 2 1 Student Model Since y 5 3x 2 1, I can substitute 3x 2 1 for y in the first equation. 3(3x 2 1) 5 x 1 1 9x 2 3 5 x 1 1 8x 5 4 x 5 1 2 3y 5 1 1; y 5 1 2 1 2 Solution: 1, 1 2 1 2 2 17 What ordered pair is a solution to y 5 x 1 5 and x 2 5y 5 29? Show your work. Possible answer: x 2 5y 5 29 2 5y 5 2x 2 9 Find y. Find x. y 5 x 1 5 y 5 x 1 5 25y 5 2x 2 9 1 5 x 1 5 24y 5 24 x 5 24 y 5 1 Solution: (24, 1) 18 Graph the equations. What is your estimate of the solution of this system of equations? y 5 3x 2 2 y 5 22x Show your work. 5 4 3 2 1 25 24 23 2221 0 21 1 2 3 4 5 22 23 24 25 Solution: Possible answer: (0.5, 21) 19 Which of these systems of equations has no solution? A y 5 x 4 1 2 B C y 5 4x 2 1 y 5 2x 3 2 3 y 5 2x 2 3 y 5 4x y 5 4x 2 5 D x 1 y 5 3 2y 5 22x 1 6 Sheila chose D as the correct answer. How did she get that answer? Possible answer: Sheila thought that if the equations in the system are the same, then there is no solution. Do the equations have the same or different slopes? Solve the problem algebraically and compare the solution to your estimate. What do you know about the lines in a system of equations with no solution? Graph the solution to verify your answer. 144 145 AT A GLANCE Students use what they ve learned to answer questions about systems of equations. STEP BY STEP Ask students to solve the problems individually and check their answers. When students have completed each problem, have them to discuss their solutions with a partner or in a group. SOLUTIONS Ex After using substitution to solve for x, the model then uses the x-value to solve for y. 1 2 17 Solution: (24, 1); Students could solve the problem by using the substitution or elimination method. (DOK 1) 18 Solution: around (0.4, 20.8); Students could solve the problem by seeing where the lines intersect on the graph. (DOK 1) 19 Solution: C; Sheila thought that if the equations are identical, the system has no solution; in fact, the system has an infinite number of solutions. Explain to students why the other two answer choices are not correct: A and B are not correct because both systems have exactly one solution. (DOK 3) 160