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2. Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Systems of Equations CHAPTER 1Solving Systems of Linear Equations Graphically 2Solving Systems of Linear Equations by Substitution 3Solving Systems of Linear Equations by Elimination 4Solving Systems of Linear Equations in Three Variables 5Solving Systems of Linear Equations Using Matrices 6Solving Systems of Linear Equations Using Cramer’s Rule 7Solving Systems of Linear Inequalities 9

3. Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solving Systems of Linear Equations Graphically 1 1.Determine if an ordered pair is a solution for a system of equations. 2.Solve a system of linear equations graphically. 3.Classify systems of linear equations in two unknowns.

4. Slide 9- 4 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Objective 1 Determine if an ordered pair is a solution for a system of equations.

5. Slide 9- 5 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley System of equations: A group of two or more equations. Solution for a system of equations: An ordered set of numbers that makes all equations in a system true.

6. Slide 9- 6 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Checking a Solution to a System of Equations To verify or check a solution to a system of equations: 1. Replace each variable in each equation with its corresponding value. 2. Verify that each equation is true.

7. Slide 9- 7 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example Determine whether each ordered pair is a solution to the system of equations. a. (  3, 2)b) (3, 4) Solution a. (  3, 2) x + y = 7y = 3x – 2  3 + 2 = 7 2 = 3(  3) – 2  1 = 72 =  11 FalseFalse

8. Slide 9- 8 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley continued b. (3, 4) x + y = 7y = 3x – 2 3 + 4 = 7 4 = 3(2) – 2 7 = 74 = 4True Because (  3, 2) does not satisfy both equations, it is not a solution for the system. Because (3, 4) satisfies both equations, it is a solution to the system of equations.

9. Slide 9- 9 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Objective 2 Solve a system of linear equations graphically.

10. Slide 9- 10 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley A system of two linear equations in two variables can have one solution, no solution, or an infinite number of solutions. The graphs intersect at a single point. The equations have the same slope, the graphs are parallel. The graphs are identical. There are an infinite number of solutions.

11. Slide 9- 11 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solving Systems of Equations Graphically To solve a system of linear equations graphically: 1. Graph each equation. a. If the lines intersect at a single point, then the coordinates of that point form the solution. b. If the lines are parallel, then there is no solution. c. If the lines are identical, then there are an infinite number of solutions, which are the coordinates of all the points on that line. 2. Check your solution.

12. Slide 9- 12 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example Solve the system of equations graphically. Solution Graph each equation. The lines intersect at a single point, (  2, 4). We can check the point in each equation to verify and will leave that to you. y = 2 – x 2x + 4y = 12 (  2, 4)

13. Slide 9- 13 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example Solve the system of equations graphically. Solution Graph each equation. The lines appear to be parallel, which we can verify by comparing the slopes. The slopes are the same, so the lines are parallel.

14. Slide 9- 14 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Objective 3 Classify systems of linear equations in two unknowns.

15. Slide 9- 15 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Consistent system of equations: A system of equations that has at least one solution. Inconsistent system of equations: A system of equations that has no solution. Dependent linear equations in two unknowns: Equations with identical graphs. Independent linear equations in two unknowns: Equations that have different graphs.

16. Slide 9- 16 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Classifying Systems of Equations To classify a system of two linear equations in two unknowns, write the equations in slope-intercept form and compare the slopes and y-intercepts. Consistent system: The system has a single solution at the point of intersection. Independent equations: The graphs are different and intersect at one point. They have different slopes. Consistent system: The system has an infinite number of solutions. Dependent equations: The graphs are identical. They have the same slope and same y-intercept. Inconsistent system: The system has no solution. Independent equations: The graphs are different and are parallel. Though they have the same slopes, they have different y- intercepts.

17. Slide 9- 17 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example Returning to a previous example (1) is the system consistent or inconsistent, are the equations dependent or independent, and how many solutions does the system have? The graphs intersected at a single point. The system is consistent. The equations are independent (different graphs), and the system has one solution: (  2, 4).

18. Slide 9- 18 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Which set of points is a solution to the system? a) (–1, 1) b) (–1, –1) c) (0, 2) d) (–3, 7)

19. Slide 9- 19 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Which set of points is a solution to the system? a) (–1, 1) b) (–1, –1) c) (0, 2) d) (–3, 7)

20. Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solving Systems of Linear Equations by Substitution 2 1.Solve systems of linear equations using substitution. 2.Solve applications involving two unknowns using a system of equations.

21. Slide 9- 21 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Objective 1 Solve systems of linear equations using substitution.

22. Slide 9- 22 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solving Systems of Two Equations Using Substitution To find the solution of a system of two linear equations using the substitution method: 1. Isolate one of the variables in one of the equations. 2. In the other equation, substitute the expression you found in step 1 for that variable. 3. Solve this new equation. (It will now have only one variable.) 4. Using one of the equations containing both variables, substitute the value you found in step 3 for that variable and solve for the value of the other variable. 5. Check the solution in the original equations.

23. Slide 9- 23 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example Solve the system of equations using substitution. Solution Step 1: Isolate a variable in one equation. The second equation is solved for x. Step 2: Substitute x = 1 – y for x in the first equation.

24. Slide 9- 24 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley continued Step 3: Solve for y. Step 4: Solve for x by substituting 4 for y in one of the original equations. The solution is (  3, 4).

25. Slide 9- 25 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example Solve the system of equations using substitution. Solution Step 1: Isolate a variable in one equation. Use either equation. 2x + y = 7 y = 7 – 2x Step 2: Substitute y = 7 – 2x for y in the second equation.

26. Slide 9- 26 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley continued Step 3: Solve for x. Step 4: Solve for y by substituting 2 for x in one of the original equations. The solution is (2, 3).

27. Slide 9- 27 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Inconsistent Systems of Equations The system has no solution because the graphs of the equations are parallel lines. You will get a false statement such as 3 = 4. Consistent Systems with Dependent Equations The system has an infinite number of solutions because the graphs of the equations are the same line. You will get a true statement such as 8 = 8.

28. Slide 9- 28 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example Solve the system of equations using substitution. Solution Substitute y = 4 – 3x into the second equation. True statement. The number of solutions is infinite.

29. Slide 9- 29 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Objective 2 Solve applications involving two unknowns using a system of equations.

30. Slide 9- 30 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solving Applications Using a System of Equations To solve a problem with two unknowns using a system of equations: 1. Select a variable for each of the unknowns. 2. Translate each relationship to an equation. 3. Solve the system.

31. Slide 9- 31 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example Terry is designing a table so that the length is twice the width. The perimeter is to be 216 cm. Find the length and width of the table. Understand We are given two relationships and are to find the length and width. Plan Select a variable for the length and another variable for the width, translate the relationships to a system of equations, and then solve the system.

32. Slide 9- 32 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley continued Execute Let l represent the length and w the width. Relationship 1:The length is twice the width. Translation: l = 2w Relationship 2:The perimeter is 216 cm. Translation: 2l + 2w = 216 System:

33. Slide 9- 33 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley continued Solve. Now find the value of l. l = 2w = 2(36) = 72 Answer The length should be 72 cm and the width 36 cm. Substitute 2w for l. Combine like terms. Divide both sides by 6 to isolate w.

34. Slide 9- 34 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solve the system. a) (10, 3) b) (1,  4) c) (10, 14) d) (3, 6)

35. Slide 9- 35 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solve the system. a) (10, 3) b) (1,  4) c) (10, 14) d) (3, 6)

36. Slide 9- 36 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solve the system. a) (2, 3) b) (  2, 6) c) (2,  6) d) (3, 2)

37. Slide 9- 37 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solve the system. a) (2, 3) b) (  2, 6) c) (2,  6) d) (3, 2)

38. Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solving Systems of Linear Equations by Elimination 3 1.Solve systems of linear equations using elimination. 2.Solve applications using elimination.

39. Slide 9- 39 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Objective 1 Solve systems of linear equations using elimination.

40. Slide 9- 40 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example Solve the system of equations. Solution We can add the equations. Notice that y is eliminated, so we can easily solve for the value of x. Divide both sides by 4 to isolate x. Now that we have the value of x, we can find y by substituting 4 for x in one of the original equations. x + y = 9 4 + y = 9 y = 5 The solution is (4, 5).

41. Slide 9- 41 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example—Multiplying Solve the system of equations. Solution Because no variables are eliminated if we add, we rewrite one of the equations so that it has a term that is the additive inverse of one of the terms in the other equation. Multiply the first equation by 4. Solve for y. x + y = 8 3 + y = 8 y = 5 The solution is (3, 5).

42. Slide 9- 42 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example—Multiplying Solve the system of equations. Solution Choose a variable to eliminate, y, then multiply both equations by numbers that make the y terms additive inverses. Multiply the first equation by 2. Multiply the second equation by  5. Multiply by 2. Multiply by  5. Add the equations to eliminate y.

43. Slide 9- 43 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley continued Substitute x = 1 into either of the original equations. 4x – 5y = 19 4(1) – 5y = 19 4 – 5y = 19 –5y = 15 y =  3 The solution is (1, –3).

44. Slide 9- 44 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example—Fractions or Decimals Solve the system of equations. Solution To clear the decimals in Equation 1, multiply by 100. To clear the fractions in Equation 2, multiply by 5. Multiply by 100. Multiply by 5. Multiply equation 2 by  1 then combine the equations.

45. Slide 9- 45 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley continued Substitute to find y. The solution is (  1,  3).

46. Slide 9- 46 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solving Systems of Equations Using Elimination To solve a system of two linear equations using the elimination method: 1. Write the equations in standard form (Ax + By = C). 2. Use the multiplication principle to clear fractions or decimals. 3. Multiply one or both equations by a number (or numbers) so that they have a pair of terms that are additive inverses. 4. Add the equations. The result should be an equation in terms of one variable. 5. Solve the equation from step 4 for the value of that variable. 6. Using an equation containing both variables, substitute the value you found in step 5 for the corresponding variable and solve for the value of the other variable. 7. Check your solution in the original equations.

47. Slide 9- 47 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Inconsistent Systems and Dependent Equations When both variables have been eliminated and the resulting equation is false, such as 0 = 5, there is no solution. The system is inconsistent. When both variables have been eliminated and the resulting equation is true, such as 0 = 0, the equations are dependent, so there are an infinite number of solutions.

48. Slide 9- 48 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Objective 2 Solve applications using elimination.

49. Slide 9- 49 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example The length of Richard’s garden is 4 meters greater than 3 times the width. The perimeter of his garden is 72 meters. What are the dimensions of the garden? Understand We are to find the dimensions of the garden. The formula for the perimeter is P = 2l + 2w. Plan Translate the relationships to a system of equations, and then solve the system. Execute Let l = the length Let w = the width

50. Slide 9- 50 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley continued Relationship 1 The length is 4 m more than 3 times the width Translation l = 3w + 4. Relationship 2 The perimeter of the garden is 72 m Translation 2l + 2w = 72. Solve the system: rewrite Multiply by  2.

51. Slide 9- 51 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley continued Now we can find l using l = 3w + 4 l = 3(8) + 4 l = 28 Answer The garden has a length of 28 meters and a width of 8 meters. Check 2l + 2w = 72 2(28) + 2(8) = 72 56 + 16 = 72 True

52. Slide 9- 52 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solve the system. a) (2, 3) b) (7, 0) c) (–2,  3) d) (5, 5)

53. Slide 9- 53 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solve the system. a) (2, 3) b) (7, 0) c) (–2,  3) d) (5, 5)

54. Slide 9- 54 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solve the system. a) (0, 10) b) (1, 5) c) (–2,  3) d) no solution

55. Slide 9- 55 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solve the system. a) (0, 10) b) (1, 5) c) (–2,  3) d) no solution

56. Slide 9- 56 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solve the system. a) (1, 1) b) (1, 5) c) (–1, 1) d) no solution

57. Slide 9- 57 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solve the system. a) (1, 1) b) (1, 5) c) (–1, 1) d) no solution