2. Copyright © 2012 Pearson Education, Inc. Publishing as Addison Wesley CHAPTER 6: Systems of Equations and Matrices 6.1 Systems of Equations in Two Variables 6.2 Systems of Equations in Three Variables 6.3 Matrices and Systems of Equations 6.4 Matrix Operations 6.5 Inverses of Matrices 6.6 Determinants and Cramer’s Rule 6.7 Systems of Inequalities and Linear Programming 6.8 Partial Fractions

3. Copyright © 2012 Pearson Education, Inc. Publishing as Addison Wesley 6.1 Systems of Equations in Two Variables  Solve a system of two linear equations in two variables by graphing.  Solve a system of two linear equations in two variables using the substitution and the elimination methods.  Use systems of two linear equations to solve applied problems.

4. Slide 6.1 - 4 Copyright © 2012 Pearson Education, Inc. Publishing as Addison Wesley Systems of Equations A system of equations is composed of two or more equations considered simultaneously. Example: 5x  y = 5 4x  y = 3 This is a system of two linear equations in two variables. The solution set of this system consists of all ordered pairs that make both equations true. The ordered pair (2, 5) is a solution of this system.

5. Slide 6.1 - 5 Copyright © 2012 Pearson Education, Inc. Publishing as Addison Wesley Solving Systems of Equations Graphically When we graph a system of linear equations, each point at which the graphs intersect is a solution of both equations and therefore a solution of the system of equations.

6. Slide 6.1 - 6 Copyright © 2012 Pearson Education, Inc. Publishing as Addison Wesley Solving Systems of Equations Graphically Let’s solve the previous system graphically. 5x  y = 5 4x  y = 3 Solution: We see that the graph intersects at the single point (2, 5), so this is the solution of the system of equations.

7. Slide 6.1 - 7 Copyright © 2012 Pearson Education, Inc. Publishing as Addison Wesley Systems of Equations If a system of equations has at least one solution, it is consistent. If the system has no solutions, it is inconsistent. If a system of two linear equations in two variables has an infinite number of solutions, the equations are dependent. Otherwise, they are independent.

8. Slide 6.1 - 8 Copyright © 2012 Pearson Education, Inc. Publishing as Addison Wesley Illustration of Graphs Graphs of linear equations may be related to each other in one of three ways.

9. Slide 6.1 - 9 Copyright © 2012 Pearson Education, Inc. Publishing as Addison Wesley Substitution Method The substitution method is a technique that gives accurate results when solving systems of equations. It is most often used when a variable is alone on one side of an equation or when it is easy to solve for a variable. One equation is used to express one variable in terms of the other, then it is substituted in the other equation.

10. Slide 6.1 - 10 Copyright © 2012 Pearson Education, Inc. Publishing as Addison Wesley Example Use substitution to solve the system 5x  y = 5, 4x  y = 3. Solution Solve the first equation for y: y = 5x  5 Then we substitute 5x  5 for y in the second equation to give an equation in one variable. 4x  (5x  5) = 3 4x  5x + 5 = 3 x = 2

11. Slide 6.1 - 11 Copyright © 2012 Pearson Education, Inc. Publishing as Addison Wesley Solution continued Now we use back-substitution and substitute 2 for x in either original equation. 4x  y = 3 4(2)  y = 3 8  y = 3 y = 5 We find the solution to the system of equations to be (2, 5), once again.

12. Slide 6.1 - 12 Copyright © 2012 Pearson Education, Inc. Publishing as Addison Wesley Elimination Method Using the elimination method, we eliminate one variable by adding the two equations. If the coefficients of a variable are opposites, that variable can be eliminated by simply adding the original equations. If the coefficients are not opposites, it is necessary to multiply one or both equations by suitable constants, before we add.

13. Slide 6.1 - 13 Copyright © 2012 Pearson Education, Inc. Publishing as Addison Wesley Example Solve the system using the elimination method. 6x + 2y = 4 10x + 7y =  8 Solution If we multiply the first equation by 5 and the second equation by  3, we will be able to eliminate the x variable. 30x + 10y = 20 Substituting: 6x + 2y = 4  30x  21y = 24 6x + 2(  4) = 4  11y = 44 6x  8 = 4 y =  4 6x = 12 The solution is (2,  4). x = 2

14. Slide 6.1 - 14 Copyright © 2012 Pearson Education, Inc. Publishing as Addison Wesley Another Example Solve the system. x  3y =  9 (1) 2x  6y = 3 (2) Solution:  2x + 6y = 18 Mult. (1) by  2 2x  6y = 3 0 = 21 There are no values of x and y in which 0 = 21. So this system has no solution. The graphs of the equations are of parallel lines.

15. Slide 6.1 - 15 Copyright © 2012 Pearson Education, Inc. Publishing as Addison Wesley Another Example Solve the system. 9x + 6y = 48 (1) 3x + 2y = 16 (2) Solution: 9x + 6y = 48  9x  6y =  48 Mult. (2) by  3 0 = 0 When we obtain the equation 0 = 0, we know the equations are dependent. There are infinitely many solutions. The graphs of the equations are identical.

16. Slide 6.1 - 16 Copyright © 2012 Pearson Education, Inc. Publishing as Addison Wesley Application--Example Ethan and Ian are twins. They have decided to save all of the money they earn, at their part-time jobs, to buy a car to share at college. One week, Ethan worked 8 hours and Ian worked 14 hours. Together they saved $256. The next week, Ethan worked 12 hours and Ian worked 16 hours and they earned $324. How much does each twin make per hour?

17. Slide 6.1 - 17 Copyright © 2012 Pearson Education, Inc. Publishing as Addison Wesley Solution Letting E represent Ethan and I represent Ian, the following system can be obtained. 8E + 14I = 256 First week 12E + 16I = 324 Second week Mult by 12 96E + 168I = 3072 Mult by  8  96E  128I =  2592 40I = 480 I = 12

18. Slide 6.1 - 18 Copyright © 2012 Pearson Education, Inc. Publishing as Addison Wesley Solution Solve for E. 8E + 14(12) = 256 8E = 88 E = 11 Ian makes $12 per hour while Ethan makes $11 per hour.

19. Slide 6.1 - 19 Copyright © 2012 Pearson Education, Inc. Publishing as Addison Wesley Example At Max’s Munchies, caramel corn worth $2.50 per pound is mixed with honey roasted mixed nuts worth $7.50 per pound in order to get 20 lb of a mixture worth $4.50 per pound. How many of each snack is used? Solution Make a guess. 16 lb of caramel corn 4 lb mixed nuts $2.50(16) = $40 $7.50(4) = $30 Total value = $70 $4.50(20) = $90 so our guess is incorrect

20. Slide 6.1 - 20 Copyright © 2012 Pearson Education, Inc. Publishing as Addison Wesley Example--continued At Max’s Munchies, caramel corn worth $2.50 per pound is mixed with honey roasted mixed nuts worth $7.50 per pound in order to get 20 lb of a mixture worth $4.50 per pound. How many of each snack is used? 2. Translate x + y = 20 2.50x + 7.50y = 90 Carmel cornNutsMixture Price per pound $2.50$7.50$4.50 Number of pounds xy20 Value of Mixture 2.50x7.50y4.50(20) = 90

21. Slide 6.1 - 21 Copyright © 2012 Pearson Education, Inc. Publishing as Addison Wesley Example--continued At Max’s Munchies, caramel corn worth $2.50 per pound is mixed with honey roasted mixed nuts worth $7.50 per pound in order to get 20 lb of a mixture worth $4.50 per pound. How many of each snack is used? 3. Carry out x + y = 20 2.50x + 7.50y = 90 Clear the decimals in the second equation by multiplying by 10. 25x + 75y = 900 Use elimination, multiply equation (1) by ‒ 25 and add to equation (2).

22. Slide 6.1 - 22 Copyright © 2012 Pearson Education, Inc. Publishing as Addison Wesley Example--continued At Max’s Munchies, caramel corn worth $2.50 per pound is mixed with honey roasted mixed nuts worth $7.50 per pound in order to get 20 lb of a mixture worth $4.50 per pound. How many of each snack is used? 3. Carry out 4. Check If 12 lb of caramel corn and 8 lb of nuts are used, the mixture weighs 12 + 8, or 20 lb. The value of the mixture is $2.50(12) + $7.50(8). $30 + $60 = $90 5. State The mixture should consist of 12 lb of caramel corn and 8 lb or honey roasted mixed nuts.

23. Slide 6.1 - 23 Copyright © 2012 Pearson Education, Inc. Publishing as Addison Wesley Example An airplane flies the 3000-mi distance from Los Angeles to New York, with a tailwind, in 5 hr. The return trip, against the wind, takes 6 hr. Find the speed of the airplane and the speed of the wind. Solution Make a drawing. p + w = traveling with tailwind p – w = traveling with headwind

24. Slide 6.1 - 24 Copyright © 2012 Pearson Education, Inc. Publishing as Addison Wesley Example--continued An airplane flies the 3000-mi distance from Los Angeles to New York, with a tailwind, in 5 hr. The return trip, against the wind, takes 6 hr. Find the speed of the airplane and the speed of the wind. 2. Translate 3. Carry out DistanceRateTime With Tailwind 3000p + w5 With headwind 3000p – w6

25. Slide 6.1 - 25 Copyright © 2012 Pearson Education, Inc. Publishing as Addison Wesley Example--continued An airplane flies the 3000-mi distance from Los Angeles to New York, with a tailwind, in 5 hr. The return trip, against the wind, takes 6 hr. Find the speed of the airplane and the speed of the wind. 3. Carry out p + w = 600 550 + w = 600 w = 50 4. Check If p = 550 and w = 50, then the speed of the plane with the tailwind is 550 + 50 or 600 mph, and the speed with the headwind is 550 – 50 or 500 mph. At 600 mph, the travel time for 3000 miles is 5 hr. At 500 mph, the travel time for 3000 miles is 6 hr.

26. Slide 6.1 - 26 Copyright © 2012 Pearson Education, Inc. Publishing as Addison Wesley Example--continued An airplane flies the 3000-mi distance from Los Angeles to New York, with a tailwind, in 5 hr. The return trip, against the wind, takes 6 hr. Find the speed of the airplane and the speed of the wind. 5. State The speed of the plane is 550 mph, and the speed of the wind is 50 mph.