LESSON OBJECTIVES 8.1 Solving Systems of Equations by Graphing Identify systems of equations as dependent or independent. Solve systems of linear equations by graphing. 8.2 Solving Systems of Equations Algebraically Solve systems of linear equations by substitution. Solve systems of linear equations by elimination. 8.3 Systems of Linear Inequalities Solve systems of linear inequalities by graphing. 8.4 Linear Programming Find the vertices of feasible regions. Solve applications of linear systems. 8.5 Solve Systems of Equations in Three Variables Solve systems of linear equations in three variables. Mental Math Find the additive inverse of each expression. 1. 5 5 2. y y 3. 2a 2a 4. 4b 4b Skills Review Find the x- and y-intercepts of each equation. 1. y = 6x + 3 1 ( 2, 0 ), (0, 3) 2. 4x + 2y = 16 (4, 0), (0, 8) 3. 5x 3y = 15 (3, 0), (0, 5) 4. 7x 28y = 14 ( 2, 0), 0, 1 2 ( ) 344 Chapter 8 Systems of Equations and Inequalities
Look to Your Future Computer consultants use linear programming to minimize production cost and maximize revenue for businesses. The goal of linear programming is to help business owners make decisions that will not violate any constraints and achieve improvement in productivity. Even though business owners have a limited number of constraints, linear programming software has the capability to optimize results that contain an infinite number of variables and constraints. PLANNING THE CHAPTER Math Labs, pp. 377 379 Data Sheet (Lab Data Sheets) Math Applications, pp. 380 385 Chapter Review, pp. 386 387 Chapter Test, p. 388 Software Generated Assessment Standardized Test Practice, p. 389 Grid Response Form (CRB) Chapter Resource Book (CRB) Reteaching, pp. 33, 37, 43, 47, 53 Extra Practice, pp. 35, 39, 45, 49, 55 Enrichment, pp. 41, 51 Standardized Test Response Form, pp. 57, 58 Standardized Test Answers, p. 59 Classroom/Journal Topics What s Ahead? In this chapter, students will learn about solving systems of equations and inequalities. Equations and inequalities are often used to model real-life situations. Students should become familiar with solving and graphing systems of equations in two and three variables, and systems of inequalities in two variables. Students will also become familiar with linear programming. Chapter 8 Systems of Equations and Inequalities 345
LESSON PLANNING Vocabulary system of equations linear system solution independent system of equations dependent system of equations consistent system of equations inconsistent system of equations Extra Resources Reteaching 8.1 Extra Practice 8.1 Assignment In-class practice: 1 5 Homework: 6 35 1 4 1 4 Math Applications Exercises 2, 8, 9, 11 and 12 from pages 380 385 START UP Review the slope-intercept method of graphing. Have students graph equations such as y = 3x + 4 and 3x 2y = 6. Point out to students that they may need to solve the equation for y before graphing. R.E.A.C.T. Strategy Experiencing Have students make tables of values for y = 2x + 3 and y = 3x 7. Students should expand each table until a common point is found. Emphasize that an infinite number of points satisfies each of the individual equations given, but at most, only one point satisfies both equations. 346 Chapter 8 Systems of Equations and Inequalities
INSTRUCTION Have students explain, in their own words, why a system of linear equations can have 0, 1, or infinitely many solutions. Ask students to recall other types of equations they have graphed, such as quadratic equations or absolute value equations. Ask students to visualize how many solutions a system of two quadratic equations can have. (There may be 0, 1, 2, 3, 4, or infinitely many solutions.) Answer to Ongoing Assessment R.E.A.C.T. Strategy Cooperating Ask students to work in pairs. Have each student write one linear equation. Students should then graph both equations and find the solution to the system of equations. Many student pairs of equations will not result in integer solutions. Discuss why many of the solutions were not integers. Challenge students to write a system of equations that has integral solutions. Answer to Critical Thinking A set of perpendicular lines is an independent system because the graphs of the lines intersect at one point. 8.1 Solving Systems of Equations by Graphing 347
INSTRUCTION Remind students that: Parallel lines have the same slope, but different y intercepts. Coinciding lines have the same slope and the same y intercept. Intersecting lines have a different slope and may or may not have the same y-intercept. Reteaching 8.1 (CRB) Chalkboard Example Have students identify each system by name and find its solution. 1. y = 2x 6 3y + 18 = 6x The slope of both lines is 2. The y-intercept of both lines is 6. Therefore, the lines are coincident. 2. y = 3x + 8 y = 3 2 x + 8 The slope of the first line is 3. The slope of the second line is 3. The y-intercept of both lines is 8. 2 Therefore, the lines will intersect at (0, 8). 348 Chapter 8 Systems of Equations and Inequalities
INSTRUCTION Ask students to each write three separate systems of equations. The first system should be dependent, the second system should be inconsistent, and the third system should have a solution of (3, 1). Ask students to share their systems of equations as well as their strategies for writing each system. Extra Practice 8.1 (CRB) Common Student Misconceptions Students often confuse dependent and inconsistent systems once they begin solving systems of equations algebraically. Take time to reinforce the visual explanation as to why a dependent system of equations has an infinite number of solutions and why an inconsistent system of equations has no solution. Emphasize that students can determine how a system will look by analyzing the slopes and y intercepts of the equations. 8.1 Solving Systems of Equations by Graphing 349
WRAP UP To ensure mastery of objectives, students should be able to: Classify a system of equations by determining the number of solutions. Find the solution to an independent system of equations by graphing. Assignment In-class practice: 1 5 Homework: 6 35 Math Applications Exercises 2, 8, 9, 11, and 12 from pages 380 385 350 Chapter 8 Systems of Equations and Inequalities Think and Discuss Answers 1. The point where the graphs in a system of equations intersect is the solution of the system. 2. Intersecting lines have one solution; parallel lines have no solution; coinciding lines have infinitely many solutions. 3. The graph of an inconsistent system is parallel lines because the lines do not intersect, and therefore, there are no solutions. 4. Lines cannot intersect at exactly two points. 5. Graph each equation using a graphing calculator. Use the Intersect feature to find where the graphs intersect.
Practice and Problem Solving Additional Answers 12. independent; consistent; (3, 8) 13. dependent; consistent; infinitely many solutions 1 3 1 5 5 3 1 5 2 3 2 3 1 3 1 6 1 2 2 3 1 2 14. dependent; consistent; infinitely many solutions 16. inconsistent; no solution 17. independent; consistent; (2, 2) 15. independent; consistent; ( 5, 5) 8.1 Solving Systems of Equations by Graphing 351
Practice and Problem Solving Additional Answers 22. ( 1.5, 0.5) 23. (7.78, 0.89) 24. ( 4.58, 0.58) and (4.58, 8.58) 25. ( 2.30, 8.30) and (1.30, 4.70) 26. x + y = 5 and 8x + 6y = 34; 2 hours on laser tag and 3 hours on ice skating 1 2 5 4 27. 28. x + y = 7 and 550x + 800y = 4,600; 4 days at the amusement park and 3 days on the cruise 12 4 5b d 3 2 3 2 29. Answers to Math Applications Math Applications for this chapter are on pages 380 385. The notes and solutions shown below accompany the suggested applications to assign with this lesson. Due to limited space selected answers are given here. More detailed solutions to the Math Applications can be found on pages 380 and 383 384. 2. a. y = 8.5 11. a. 87 = a + b; 102 = 4a + b b. y = 0.35x + 5 b. 5 = a; b = 82 8. a. y = 34x c. y = 5x + 82 b. y = 30x + 60 d. 107 bpm after 5 minutes c. (15, 510) 12. a. y = 500x + 12,000 9. a. y = 85; y = 7.50x + 10 y = 475x + 15,000 (10, 85) b. (120, 72,000) 352 Chapter 8 Systems of Equations and Inequalities
LESSON PLANNING Vocabulary substitution method elimination method Extra Resources Reteaching 8.2 Extra Practice 8.2 Enrichment 8.2 Assignment In-class practice: 1 4 Homework: 5 34 Math Applications Exercises 1, 4, 6, and 13 from pages 380 385 START UP Have students solve the system of equations 3x + y = 13 and 6x y = 10 by graphing. Ask students to share their solutions. Because the system does not have an integral solution, students will have answers that vary slightly. Use this lack of precision of the solutions as motivation for learning the algebraic methods for solving systems of equations. Diversity in the Classroom Visual Learner Have students graph the equations x = 6 and y = 4. Ask students to determine the point of intersection. Emphasize that the point is a solution to the system of equations. (6, 4) 8.2 Solving Systems of Equations Algebraically 353
ACTIVE LEARNING Show students that decimals can be removed from the equations by multiplying the first equation by 100 and the second equation by 10. Explain that the solution will remain the same. Answers to Activity 2. 7.50x + 7.25y = 81 x + y = 11 4. 7.50(11 y) + 7.25y = 81 82.50 7.50y + 7.25y = 81 0.25y = 1.5 y = 6 INSTRUCTION Many students find the value of only one variable when solving a system of equations algebraically. Remind students that the solution is a point of intersection and therefore must have both coordinates for the solution point. R.E.A.C.T. Strategy Applying Landscape designers often use systems of equations to find the dimensions of a walkway around a garden or a pool. One equation may relate the length of the property to the width. Another equation may relate the length and width of the property to the perimeter or area of the garden or pool. Have students write and solve a system of equations to model one such situation. w = h + 6; 2h + 2w = 56; w = 17, h = 11 354 Chapter 8 Systems of Equations and Inequalities
Reteaching 8.2 (CRB) a b c c 5 1 4 -hour Enrichment 8.2 (CRB) Extra Practice 8.2 (CRB) Common Student Misconceptions Students should recognize that when both equations have one side equal to y, they can use a combination of symmetric and transitive properties to set the other sides of the equations equal to one another. For example, if the two original equations are y = 3x 4 and y = 6x + 8, the resulting equation is 3x 4 = 6x + 8, which is an equation in one variable. 8.2 Solving Systems of Equations Algebraically 355
Problem Solving Understand the Problem Find the number of ounces of each type of milk that are needed to get 12 ounces of 2% milk. Develop a Plan x + y = 12; 0.01x + 0.04y = 0.24 Carry Out the Plan y = x + 12; y = 0.25x + 6 y = x + 12 y = 0.25x + 6 1 11 1 5.75 2 10 2 5.5 3 9 3 5.25 4 8 4 5 5 7 5 4.75 6 6 6 4.5 7 5 7 4.25 8 4 8 4 You will need 8 ounces of low-fat milk and 4 ounces of whole milk. Check the Results Check the first equation. 8 + 4 = 12 12 = 12 Check the second equation. 0.01(8) + 0.04(4) = 0.02(12) 0.08 + 0.16 = 0.24 0.24 = 0.24 356 Chapter 8 Systems of Equations and Inequalities
3 4, 37 4 ( ) ( 19 4, 5 4 ) ( 2, 1 2) ( 3, 3 2) ( 7 6, 2 3) 3. The graphs of the lines in a system of equations may be found to intersect, but the exact point of intersection, or solution, may be difficult to determine by looking at the graph. 4. When the two equations in the system represent the same line there will be an infinite number of solutions; when the equations in the system represent parallel lines, there will be no solutions. WRAP UP To ensure mastery of objectives, students should be able to: Use the substitution method to solve a system of equations. Use the elimination method to solve a system of equations. Determine which algebraic method is most appropriate for a given system of equations. Assignment In-class practice: 1 4 Homework: 5 34 Math Applications Exercises 1, 4, 6, and 13 from pages 380 385 Think and Discuss Answers 1. It is easiest to use the elimination method when one of the coefficients is 1 or 1. 2. When both variables are eliminated and the two equations in the system are equal to the same value, there will be an infinite number of solutions; when both variables are eliminated and the equations in the system are not equal to the same value, there will be no solution. 8.2 Solving Systems of Equations Algebraically 357
Mixed Review Additional Answers 28. 8 6 4 2 x 8 6 4 2 2 2 4 6 8 4 6 8 y 2 3 29. 34. Answers to Math Applications Math Applications for this chapter are on pages 380 385. The notes and solutions shown below accompany the suggested applications to assign with this lesson. Due to limited space selected answers are given here. More detailed solutions to the Math Applications can be found on pages 380 382 and 385. 1. a. 8a + b = 410 15a + b = 550 a = 20; 250 = b b. y = 20x + 250 c. Yes, the hot air balloon will clear the tower by 50 feet 4. a. 300x + 275y = 868,750 b. x + y = 3,000 c. 25x = 43,750 x = 1,750; y = 1,250 6. a. 350x + 400y = 7,350 b. y = 7 13 x c. x = 13; y = 7 d. The pro shop manager should order 13 sets of steel clubs and 7 sets of graphite clubs for $7,350. 13. a. x + y = 500 b. 0.15x + 0.25y = 0.194(500) c. y = 220; x = 280 d. The chemist should use 280 liters of the 15% solution and 220 liters of the 25% solution. 358 Chapter 8 Systems of Equations and Inequalities
LESSON PLANNING Vocabulary system of inequalities Extra Resources Reteaching 8.3 Extra Practice 8.3 Assignment In-class practice: 1 4 Homework: 5 22 Math Applications Exercise 3 from pages 380 385 START UP Review the method of graphing linear inequalities. Have students graph inequalities such as y > 3x 2 and 4x 2y 12. Remind students that they may need to reverse the direction of the inequality as they put the inequality into slope-intercept form. Reteaching 8.3 (CRB) Chalkboard Examples Solve the inequalities x 8 20 and 3x + 7 < 16. Graph the solution sets for both inequalities on the same number line. Point out to students that the section of the number line that is shaded twice represents the solution set for both inequalities. 8.3 Systems of Linear Inequalities 359
INSTRUCTION Multiple shading can become cumbersome and confusing for students. Show students how to use arrows to indicate the directions of the shading. Instruct students to only shade the overlapping region. x y Answer to Ongoing Assessment 8 6 4 y 2 8 6 4 2 2 4 6 2 4 6 8 x 8 Extra Practice 8.3 (CRB) Enriching the Lesson Challenge students to write a system of two inequalities which has no solution and a system of two inequalities that has a solution that includes all points on a coordinate grid. Once students have successfully written the two systems of inequalities, ask students to write a word problem for each of the systems. Sample sytems: The system x > 4 and x 3 has no solution and the system y > 3x + 4 and y 3x + 4 has all points on the coordinate grid as solutions. 360 Chapter 8 Systems of Equations and Inequalities
INSTRUCTION When using the rule > is above the boundary line and < is below the boundary line to shade, make sure that students realize y must be on the left side of the inequality. Cultural Connection In the movie, Good Will Hunting, Professor Lambeau poses a difficult mathematics problem on a blackboard outside his classroom. The problem posed contained Parseval s Theorem for Fourier Analysis, along with part of its proof. The main character in the movie, Will Hunting, the school s janitor, mysteriously solves the challenging blackboard problem. WRAP UP To ensure mastery of objectives, students should be able to: Graph a system of linear inequalities and define the solution set as a shaded region on the coordinate plane. Assignment In-class practice: 1 4 Homework: 5 25 Think and Discuss Answers 1. Answers will vary. Sample: A system of inequalities uses the symbols >, <,, or and a system of equations uses the equal sign, =. 2. Sketch the line that corresponds to each inequality in the system. Shade in the appropriate side. The solution includes every point in the region where the shading of the graphs overlap. 3. Use a dashed line for inequalities containing the symbols < or >. 4. Switch the inequality symbol when multiplying or dividing by a negative number. Math Applications Exercise 3 from pages 380 385 8.3 Systems of Linear Inequalities 361
Practice and Problem Solving Additional Answers 8. 9. 10. 11. 1 5 3 10 1 x 3 y 5 10 12. 14. 90 y 15. 900 y 80 800 13. 70 60 50 40 30 20 10 1 2 3 4 5 6 7 8 9 x 700 600 500 400 300 200 100 x 50 100 150 200 250 300 350 400 450 362 Chapter 8 Systems of Equations and Inequalities
Practice and Problem Solving Additional Answers 16. 17. Sample: y 3, y 2, x 4, x 3 19. 20. 3 3 8 2 2 " 9 4 3 4 8 7 + 7 " 14 + 14 2 + 4 2 2 2 4+ 3 " 2 2 + 6 1 + 2 2 16 5 4 2 2 3 28 " + + 3 21 24 2 3( 7) Mixed Review Additional Answers 18. a. Answers to Math Applications Math Applications for this chapter are on pages 380 385. The notes and solutions shown below accompany the suggested applications to assign with this lesson. More detailed solutions to the Math Applications can be found on page 380. 3. a. x + y 20 b. 200x + 250y 4,000 c. Answers will vary. Sample answer: (4, 16), (5, 15), (8, 12). 8.3 Systems of Linear Inequalities 363
LESSON PLANNING Vocabulary linear programming objective function constraints feasible region Extra Resources Reteaching 8.4 Extra Practice 8.4 Enrichment 8.4 Assignment In-class practice: 1 4 Homework: 5 21 Math Applications Exercise 7 from pages 380 385 START UP Semester-long college courses are devoted to the topic of linear programming. These courses are taken by students seeking degrees in mathematics, engineering, and other math related fields. 364 Chapter 2 Systems of Equations and Inequalities R.E.A.C.T. Strategy Cooperating Have students work in pairs to write a system of inequalities that would result in a polygonal region that is triangular in shape with an area of 24 square units. Tell students that the triangular shape must span all four quadrants of the coordinate plane. Sample answer: x 4 y 2 y 7 10 x 3 3
7 4 3 4 25 1 2 7 2 ACTIVE LEARNING Ask students the following question. If Noah needs to make a profit of at least $16,000 in order to pay for overhead costs, how he can increase overall profit given the constraints of the problem? (Noah can charge more for each camera.) INSTRUCTION Discuss with students the concept of discrete variables versus continuous variables. Answers to Activity 4. 6. P = 60(0) + 70(0) = 0 P = 60(0) + 70(215) = 15,050 P = 60(160) + 70(90) = 15,900 P = 60(250) + 70(0) = 15,000 Common Student Misconceptions 7. Noah can maximize his profit by selling 160 digital cameras that have 32MB of storage and 90 digital cameras that have 64 MB of storage capacity. Students often forget to evaluate an objective function when solving linear programming problems. This leads students to give the highest vertex as the answer. Remind students they must use the objective function to evaluate P at each vertex point in order to determine the maximum profit. 2.4 Linear Programming 365
INSTRUCTION Example 2 Students may need to use the algebraic process of substitution to determine the exact points where 5x + 2y 400 will intersect the lines y 20 and y 70. The graphs may not yield precise points of intersection. Students may not be familiar with the word feasible. Help students become familiar with this word by using the synonyms workable, doable, and viable. Reteaching 8.4 (CRB) R.E.A.C.T. Strategy Transferring Have students use the Internet to research real world applications of linear programming. Have students compile a list of five such applications. Students should briefly summarize each application and should provide an example which clearly illustrates the optimization that is being calculated in the application. 366 Chapter 2 Systems of Equations and Inequalities
Answer to Ongoing Assessment Answer to Critical Thinking There are only two points of intersection. Even though the vertex (0, 7) has a value of 105, there are other points in the feasible region that give greater values. WRAP UP To ensure mastery of objectives, students should be able to: Find the vertices of a feasible region by graphing a system of inequalities. Solve optimization problems by evaluating objective functions with the vertices of a feasible region. Think and Discuss Answers 1. The objective function is used to determine the maximum or minimum. 2. After graphing a system of linear inequalities, the feasible region is the overlapping region. 3. A minimizing cost problem might not have a maximum because the cost could infinitely increase. 4. The value of the objective function is checked at each vertex of the feasible region. Assignment In-class practice: 1 4 Homework: 5 21 Math Applications Exercise 7 from pages 380 385 2.4 Linear Programming 367
Extra Practice 8.4 (CRB) Enrichment 8.4 (CRB) Practice and Problem Solving Additional Answers 5. C is minimized at 125 when x = 25 and y = 50; C is not maximized because the feasible region is unbounded. 6. C is minimized at 26 when x = 6 and y = 4; C is maximized at 66 when x = 11 and y = 11. 7. C is minimized at 6 when x = 6 and y = 0; C is maximized at 2 when x = 2 and y = 4. 8. C is minimized at 6 when x = 6 and y = 0; C is maximized at 13 when x = 2 and y = 5. 9. C is minimized at 14 when x = 7 and y = 0; C is maximized at 1 when x = 2 and y = 5. 10. C is minimized at 3 when x = 0 and y = 3; C is maximized at 9 when x = 3 and y = 0. 11. C is minimized at 0 when x = 0 and y = 0; C is maximized at 42 when x = 7 and y = 0. 12. C is minimized at 4 when x = 4 and y = 1; C is not maximized because the feasible region is unbounded. 368 Chapter 2 Systems of Equations and Inequalities
Mixed Review Additional Answers 21. a. y 0, x 0 y 3 and x y 65; Answers to Math Applications Math Applications for this chapter are on pages 380 385. The notes and solutions shown below accompany the suggested applications to assign with this lesson. 7. a. 12x + 10y 240 20x + 25y 500 b. P = 22x + 20y c. d. Substitute the points (0, 0), (0, 20), (10, 12), (20, 0) into the profit function. P = 22(0) + 20(0) P = 0 P = 22(0) + 20(20) P = 400 P = 22(10) + 20(12) P = 460 P = 22(20) + 20(0) P = 440 The manager should order 10 cases of ice cream and 12 cases of frozen custard for a cost of $500 and a profit of $460. (0, 0), (0, 20), (10, 12), (20, 0) 2.4 Linear Programming 369
LESSON PLANNING Vocabulary ordered triplet triangular form Extra Resources Reteaching 8.5 Extra Practice 8.5 Assignment In-class practice: 1 4 Homework: 5 28 Math Applications Exercises 5, 10, and 14 from pages 380 385 START UP Give students the system of equations 2x + y 3z = 9 and x 2y + 4z = 12. Ask students to algebraically solve the system of equations. Obviously, students will not be able to find three numeric solutions for the variables. Ask students to explain the reason they are not able to find numeric values for (x, y, z). (They are unable to eliminate two variables at one time.) R.E.A.C.T. Strategy Experiencing Help students visualize ordered triples by using the classroom to represent 3-dimensional space. Use the floor to represent the xy-plane and one of the vertical edges to represent the z-axis. Hold a ball in your hand and ask students to direct the ball to specified ordered triples. Then place the ball and ask students to specify the ordered triple. Students will enjoy telling you to go downstairs, next door, or outside for the ordered triples (8, 10, 12), (8, 10, 12) and ( 8, 10, 12). 370 Chapter 8 Systems of Equations and Inequalities
INSTRUCTION Instruct students to check the ordered triple in all three equations in order to verify they have obtained the correct solution to the system of equations. Show students how to solve this system of equations could also be solved using the elimination method. Diversity in the Classroom Auditory Learners Ask students to explain to a partner the steps to use to solve this system of four equations with four unknown variables. 2a b + 4c + d = 11 a + b + c + d = 3 2a + b + c d = 5 3a 2b + c + d = 17 If time permits, have students determine the solution. 8.5 Solve Systems of Equations in Three Variables 371
INSTRUCTION Use three pieces of colored tag board or cardstock to represent three planes given by three equations in three variables. Cut slits into the cardboard so you can slide the pieces of cardboard together. Use the cardboard planes to help students visualize solution scenarios such as one point, one line, one plane, or no solution. Answer to Ongoing Assessment a + b + c = 86 b + c = 54 c = 25 The other two numbers are 29 and 32. INSTRUCTION Example 3 Make sure that students understand that this system could be solved by eliminating either y or z in order to create a system in two variables. R.E.A.C.T. Strategy Applying Give students the task of determining the number of calories contained in one gram of fat, one gram of protein, and one gram of carbohydrate. Provide students with nutrition labels from three separate food packages as a basis for writing three equations. For example: a frozen single serving meal contains 7 grams of fat, 50 grams of carbohydrate, 20 grams of protein, and 343 calories. The equation representing this label would be 7f + 50c + 20p = 343. 372 Chapter 8 Systems of Equations and Inequalities
INSTRUCTION Encourage students to label and show all work as they solve a system of three equations. Mistakes are common, and it will save time when students are able to trace backwards in a solution process in order to find the point of error. Reteaching 8.5 (CRB) R.E.A.C.T. Strategy Experiencing Show students how to use the graphing calculator to solve the system of equations in Example 3. Have students enter the coefficients of the three equations into Matrix A. Next, have students enter the constants into Matrix B. Finally, tell students that they are going to use inverse operations to solve for the three variables. Tell students that this method will be studied in more detail later in the book. 3 5 2 100. 60 A= 3 8 3 B = 129. 66 6 3 7 200. 54 8.5 Solve Systems of Equations in Three Variables 373
WRAP UP To ensure mastery of objectives, students should be able to: Solve a system of three linear equations using the elimination and substitution methods. Solve a system of three linear equations using the elimination method. Assignment In-class practice: 1 4 Homework: 5 28 Math Applications Exercises 5, 10, and 14 from pages 380 385 Think and Discuss Answers 1. To verify a solution in three variables, substitute the value of each variable in each equation in the system. 2. 0, 1, or infinite; when three planes do not have a common point there are no solutions; when three planes intersect at a common point there is one solution; when three planes intersect at all the points in a common line there are infinite solutions. 3. To solve a system of equations in three variables by elimination, obtain a system of two equations in two variables by using elimination. Then solve the system of equations in two variables. 4. Solve a system in triangular from by substitution when there is a coefficient of 1 in each equation for its first non zero term. 374 Chapter 8 Systems of Equations and Inequalities
Extra Practice 8.5 (CRB) 1 2 lb 1 4 lb 1 lb 2 3 4 lb 1 lb 2 1 3 lb 1 lb 2 1 3 8.5 Solve Systems of Equations in Three Variables 375
Practice and Problem Solving Additional Answers 21. The team made 9 foul shots, 13 two point shots, and 4 three point shots. Answers to Mixed Review 23. C is minimized at 28 when x = 4 and y = 4; C is maximized at 29 when x = 5 and y = 3. 24. C is not minimized because the feasible region is unbounded; C is maximized at 36 when x = 4 and y = 6. Answers to Math Applications Math Applications for this chapter are on pages 380 385. The notes and solutions shown below accompany the suggested applications to assign with this lesson. Due to limited space selected answers are given here. More detailed solutions to the Math Applications can be found on pages 381 and 380 385. 5. a. c = 20,000 b = 12,000 a = 10,000 b. 1,050,000(0.95) = 10,000(35) + 12,000(25) + c(20) 347,500 = 20c 17,375 = c 10. a. 100a + 200b + 100c = 400(0.1) 150a + 100b + 150c = 400(0.125) 100a + 300b + 150c = 550(0.1) b. c = 0.2 b = 0.05 a = 0.1 c. Solution A is 0.1 100 = 10% saline, solution B is 0.05 100 = 5% saline, and solution C is 0.2 100 = 20% saline. 14. a. z = 120 y = 80 x = 150 b. 150(8) + 80(8) + 120(8) = 1,200 + 640 + 960 = 2,800 376 Chapter 8 Systems of Equations and Inequalities
MATH LAB FOLLOW-UP As a class discuss how the problem changes if one student drew the shape and a different student cut out the shapes. Based on the solutions found in each pair of students, within each pair, have the students decide who should be drawing and who should be cutting to maximize the work. They can make a prediction about how many will be completed and then test their predictions. Decide as a class which two students (They can be from different pairs.) would be able to draw and cut out the most shapes. Choose two sets of students. Have the class make a prediction about how many will be completed for each pair. Then test the predictions. Activity 1 PREPARE Arrange students into pairs. Each pair of students needs a compass, a ruler, a pair of scissors, plain paper, and a stop watch. Have each pair of students set the compass to draw a circle with a 2 inch diameter. Explain that this activity is a simulation of workplace skills and therefore they cannot take shortcuts, such as drawing one circle and then cutting through three pieces of paper to get three circles from only drawing one circle. TEACH Discuss that each pair of students will be able to write two equations in two variables for each student. Be sure that students connect the situation with one that can be solved using a system of equations. Roles for pairs of students (both students have to play each role) 1. timer 2. worker Point out to students that the times given in Step 2 and 4 are in minutes, but their times from each task are in seconds. Be certain that they realize the time in minutes needs to be written as seconds. Math Labs 377
Math Lab Solutions and Notes 1. Times will vary. Sample answer: drawing: 34 seconds per square, 14 seconds per circle; cutting: 9 seconds per square; 16 seconds per circle 5. Sample answer: You should round down to 7 squares and 3 circles. drawing: 7 34 = 238; 3 16 = 48; 238 + 48 = 286. 286 is less than 300; cutting: 7 9 = 63; 3 16 = 48; 63 + 48 = 111. 111 is less than 120 6. Sample answer: 300 286 = 14, which is enough time to draw another circle. But 120 111 = 9, which is not enough time to cut out the circle. Activity 2 PREPARE Remind students of the boxed battleship game they may have played. You might want to bring the battleship game into class or ask one of your students to bring one in. Review how to graph linear inequalities. Be sure students understand they will only be working in the first quadrant and that each unit on the grid represents one square mile. TEACH Each student works from a coordinate grid to which only they have access. Remind students they are simulating a game where one ship is trying to enter an ocean region patrolled by enemy battleships. They should choose their courses carefully. FOLLOW-UP Have students graph their battleship territories on poster board. Use one color to represent the single ship opponent and a different color to represent their enemy. Display the poster boards around the classroom and invite students to present their battles to the class. Have the class examine the different territories and determine which single ship would have been able to travel in the ocean for the longest distance without entering enemy patrolled waters. 378 Chapter 8 Systems of Equations and Inequalities
Math Lab Solutions and Notes 3. If the single ship enters at the origin, listed below are sample equations. y = 2x y = 4x y = 15.5x If the single ship enters at along the y-axis, the equation can be written using the slope-intercept form; listed below are sample equations. y = 2x + 8 y = 4x + 12 y = 15.5x + 2 If the single ship enters at along the x-axis, the equation after either a slope is determine or the y-intercept is determine; listed below are sample equations. y = x + 8 y = 1 3 x + 4 y = 5x + 20 5. Sample systems of equations are listed: y 8 x 12 y x 6. Answers will vary. Sample answer: y 4 x 18 y 4x + 10 y 14 y 8x y = 10x + 18 Sample region used for graph: x 15, y 5 3 x + 10, y x + 10 Math Labs 379
Math Applications Solutions and Notes 1. a. 8a + b = 410 15a + b = 550 Use the elimination method. Multiply the first equation by 1, and then add the second equation. 7a = 140 a = 20 Substitute 20 for a and solve for b. 550 = 15(20) + b 550 = 300 + b 250 = b b. y = 20x + 250 c. y = 20x + 250 y = 20(20) + 250 y = 400 + 250 y = 650 Yes, the hot air balloon will clear the tower by 50 feet. 2. a. y = 8.5 b. y = 0.35x + 5 c. Find the point of intersection (10, 8.5). If Kim writes 10 checks each month, the cost will be $8.50 with both online banking service and the current bank. d. If Kim writes fewer than 10 checks each month, it will be less expensive to use her current bank. If Kim pays more than 10 bills each month, the online banking service will be less expensive. 3. a. x + y 20 b. 200x + 250y 4,000 c. Answers will vary. Sample answer: (4, 16), (5, 15), (8, 12). The store could have 4 tractor A s and 16 tractor B s, 5 tractor A s and 15 tractor B s, or 8 tractor A s and 12 tractor B s in stock each month. 380 Chapter 8 Systems of Equations and Inequalities
Substitute 12,000 for b and 20,000 for c in the third equation. Then solve for a. a + 12,000 20,000 = 2,000 a 8,000 = 2,000 a = 10,000 b. 1,050,000(0.95) = 10,000(35) + 12,000(25) + c(20) 347,500 = 20c 17,375 = c Math Applications Solutions and Notes 4. a. 300x + 275y = 868,750 b. x + y = 3,000 c. Multiply the second equation by 275. Then subtract the second equation from the first equation. 25x = 43,750 x = 1,750 Substitute 1,750 for x. Then solve for y. 1,750 + y = 3,000 y = 1,250 The farmer should plant 1,750 acres of corm and 1,250 acres of soybeans to earn a total revenue of $868,750. 5. a. 35a + 25b + 20c = 1,050,000 a + b + c = 42,000 a + b c = 2,000 Subtract the third equation from the second equation. Then solve for c. 2c = 40,000 c = 20,000 Substitute 20,000 for c in the second equation. Then solve for a. a = 22,000 b Substitute 22,000 b for a and 20,000 for c in the first equation. Then solve for b. 35(22,000 b) + 25b + 20(20,000) = 1,050,000 770,000 35b + 25b + 400,000 = 1,050,000 10b = 120,000 b = 12,000 Math Applications 381
Math Applications Solutions and Notes 6. a. 350x + 400y = 7,350 b. y = 7 13 x c. Substitute 7 13 x for y in the first equation. Then solve for x. 350x + 400 ( 7 13)x = 7,350 565.4x = 7,350 x = 13 Substitute 13 for x in the second equation. Then solve for y. 7 y = ( 13)13 y = 7 d. 7 = 7 ( 13) = (13) 7 = 7 350(13) + 400(7) = 7,350 4,550 + 2,800 = 7,350 7,350 = 7,350 The pro shop manager should order 13 sets of steel clubs and 7 sets of graphite clubs for $7,350. 7. a. 12x + 10y 240 20x + 25y 500 b. P = 22x + 20y c. 7 13 (0, 0), (0, 20), (10, 12), (20, 0) d. Substitute the points (0, 0), (0, 20), (10, 12), (20, 0) into the profit function. P = 22(0) + 20(0) P = 0 P = 22(0) + 20(20) P = 400 P = 22(10) + 20(12) P = 460 P = 22(20) + 20(0) P = 440 The manager should order 10 cases of ice cream and 12 cases of frozen custard for a cost of $500 and a profit of $460. 382 Chapter 8 Systems of Equations and Inequalities
10. a. Add the amount of solution A, B, and C to find the total amount of in each mixture: 100 + 200 + 100 = 400. Then multiply the total amount in each mixture by the saline percent for each mixture. 100a + 200b + 100c = 400(0.1) 150a + 100b + 150c = 400(0.125) 100a + 300b + 150c = 550(0.1) b. Subtract the third equation from the first equation. Then solve for b. 100b 50c = 15 100b = 50c 15 b = 0.5c + 0.15 Subtract the third equation from the second equation. Substitute 0.5c + 0.15 for b. Then solve for a. Math Applications Solutions and Notes 8. a. y = 34x b. y = 30x + 60 c. The point of intersection is (15, 510). d. If a customer uses the service for 14 or fewer months, it would be less expensive to rent a modem. If a customer uses the service for 15 months or more, it would be less expensive to purchase a modem. 9. a. First, write the equations and then graph the equations. y = 85 y = 7.50x + 10 The point of intersection is (10, 85) b. If Mr. Williams uses the power washer for 10 or more hours, it would be less expensive to purchase it. Math Applications 383
Math Applications Solutions and Notes 10. b. (cont d.) 50a 200( 0.5c + 0.15) = 5 50a + 100c 30 = 5 50a = 100c + 25 a = 2c + 0.5 Substitute 0.5c + 0.15 for b and 2c 0.5 for a in the first equation. Then solve for c. 100( 2c + 0.5) + 200( 0.5c + 0.15) + 100c = 40 200c + 50 100c + 30 + 100c = 40 200c = 40 c = 0.2 b = 0.5(0.2) + 0.15 = 0.05 a = 2(0.2) + 0.5 = 0.1 c. Multiply each variable by 100 to change from a decimal to a percent. Solution A is 0.1 100 = 10% saline, solution B is 0.05 100 = 5% saline, and solution C is 0.2 100 = 20% saline. 11. a. 87 = a + b 102 = 4a + b b. Substitute 87 a for b. Then solve for a. 102 = 4a + (87 a) 102 = 3a + 87 15 = 3a 5 = a b = 87 a = 87 5 = 82 c. y = 5x + 82 d. y = 5(0) + 82 82 bpm before jogging y = 5(5) + 82 y = 25 + 87 y = 107 107 bpm after 5 minutes 12. a. y = 500x + 12,000 y = 475x + 15,000 b. 72,000 66,000 60,000 54,000 48,000 42,000 36,000 30,000 24,000 18,000 12,000 6,000 (120, 72,000) x 20 40 60 80 100 120 140 160 180 The point of intersection is the solution (120, 72,000). c. If the company uses the machine for more than 120 weeks, them Machine B would be less expensive. 384 Chapter 8 Systems of Equations and Inequalities
Math Applications Solutions and Notes 13. a. x + y = 500 b. 0.15x + 0.25y = 0.194(500) c. Substitute 500 y for x. Then solve for y. 0.15(500 y) + 0.25y = 97 75 0.15y + 0.25y = 97 0.10y = 22 y = 220 Substitute 220 for y. Then solve for x. x + 220 = 500 x = 280 d. 280 + 220 = 500 500 = 500 0.15(280) + 0.25(220) = 0.194(500) 42 + 55 = 97 97 = 97 The chemist should use 280 liters of the 15% solution and 220 liters of the 25% solution. 14. a. 4x + 2y + 2z = 1,000 3x + 3y + 2z = 930 2x + 3y + 3z = 900 Subtract the third equation from the second equation. Then solve for x. x z = 30 x = 30 + z Substitute 30 + z for x in the third equation. Then solve for y. 2(30 + z) + 3y + 3z = 900 60 + 2z + 3y + 3z = 900 3y = 5z + 840 y = 5 3 z + 280 Substitute 30 + z for x and 5 z + 280 for y in the first 3 equation. Then solve for z. y = 5 (120) + 280 = 80 3 x = 30 + 120 = 150 Machine X produces boards at a rate of 150 per hour, machine Y produces boards at a rate of 80 per hour, and machine Z produces boards at a rate of 120 per hour. b. 150(8) + 80(8) + 120(8) = 1,200 + 640 + 960 = 2,800 Math Applications 385
Vocabulary Review Cramer s Rule (3-5) consistent system of equations (8.1) constraints (8.4) dependent system of equations (8.1) elimination method (8.2) feasible region (8.4) inconsistent system of equations (8.1) independent system of equations (8.1) linear programming (8.4) linear system (8.1) objective function (8.4) ordered triplet (8.5) solution (8.1) substitution method (8.2) system of equations (8.1) system of inequalities (8.3) triangular form (8.5) Chapter Review Additional Answers Lesson 8.1 1. inconsistent; no solution 2. 3. y = 2.25x + 1.50 and y = 4.50; after 3 refills independent; consistent; (2, 0) y 8 6 4 2 5 4 3 2 1 2 1 2 3 4 5 4 6 8 x 386 Chapter 8 Systems of Equations and Inequalities
Chapter Review Additional Answers Lessons 8-3 and 8-4 9. C is minimized at 8 when x = 0 and y = 8; C is maximized at 8 when x = 4 and y = 0. 10. C is minimized at 16 when x = 0 and y = 4; C is maximized at 42 when x = 6 and y = 0. Chapter Review 387
Chapter Test Additional Answers 3. 4. 5. C is minimized at 39 when x = 9 and y = 3; C is maximized at 52 when x = 4 and y = 10. 6. C is minimized at 26 when x = 2 and y = 2; C is maximized at 98 when x = 10 and y = 6. 9. x + y = 450 and 0.25x + 0.5y = 178.5; 185 ft of blue lace streamer and 265 ft of white sheer streamer; 13. a. y 2x and y 33 x; 10. 25 child size pair of gloves and 0 adult size pair of gloves 11. 6 twenty dollar bills, 6 ten dollar bills, and 9 five dollar bills 388 Chapter 8 Systems of Equations and Inequalities
2x 3y = 0 x + 2y = 7 x 6 x 0 y 8 y 0 4x + 6y = 2 4x + 9y = 1 Standardized Test Practice Additional Answers Open Ended Response 6. The system of equations is independent; ( 2, 1) Standardized Test Response Form (CRB) Extended Response 7. a. 3f + 2p + n = 3.05 5f + 3p = 3.00 4f + 3p + 2n = 4.95 b. Solve the second equation for p: p = 1 5 3 f Substitute the expression equivalent to p into the other two equations: 3f + 2 ( 1 5 ) 3 f + n = 3.05 4f + 3 ( 1 5 f 3 ) + 2n = 4.95 Simplify each equation: 1 3 f + n = 1.05 f + 2n = 1.95 Multiply the first equation by 2: 2 3 f 2n = 2.1 f + 2n = 1.95 Add the equations: 1 3 f = 0.15 Solve for f: f = 0.45 Use substitution to solve for p and n: p = 0.25; n = 1.20 c. 4(0.45) + 6(0.25) + 2(1.20) = $5.70 Chapter Assessments 389