1. Copyright © 2013, 2009, 2006 Pearson Education, Inc. 1 Section 3.1 Systems of Linear Equations in Two Variables Copyright © 2013, 2009, 2006 Pearson Education, Inc. 1

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3. 3 Systems of Equations Recall that an equation of the form Ax + By = C is a line when graphed. Two such equations are called a system of linear equations. A solution of a system of linear equations is an ordered pair that satisfies both equations in the system. For example, the ordered pair (2,1) satisfies the system 3x + 2y = 8 4x – 3y = 5.

4. Copyright © 2013, 2009, 2006 Pearson Education, Inc. 4 Systems of Equations Since two lines may intersect at exactly one point, may not intersect at all, or may intersect at every point; it follows that a system of linear equations will have exactly one solution, will have no solution, or will have infinitely many solutions.

5. Copyright © 2013, 2009, 2006 Pearson Education, Inc. 5 Systems of EquationsEXAMPLE SOLUTION Determine whether (3,2) is a solution of the system Because 3 is the x-coordinate and 2 is the y-coordinate of the point (3,2), we replace x with 3 and y with 2. Since the result is false, (3,2) is NOT a solution for the system. ? ? ? false

6. Copyright © 2013, 2009, 2006 Pearson Education, Inc. 6 Objective #1: Example

7. Copyright © 2013, 2009, 2006 Pearson Education, Inc. 7 Objective #1: Example

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9. 9 Systems of Equations Solve Systems of Two Linear Equations in Two Variables, x and y, by Graphing 1) Graph the first equation. 2) Graph the second equation on the same set of axes. 3) If the lines intersect at a point, determine the coordinates of this point of intersection. The ordered pair is the solution to the system. 4) Check the solution in both equations. NOTE: In order for this method to be useful, you must graph the lines very accurately.

10. Copyright © 2013, 2009, 2006 Pearson Education, Inc. 10 Systems of EquationsEXAMPLE SOLUTION Solve by graphing: 1) Graph the first equation. First rewrite the equation in slope-intercept form. Now graph the equation.

11. Copyright © 2013, 2009, 2006 Pearson Education, Inc. 11 Systems of Equations 2) Graph the second equation on the same set of axes. First rewrite the equation in slope-intercept form. CONTINUED Now graph the equation.

12. Copyright © 2013, 2009, 2006 Pearson Education, Inc. 12 Systems of Equations 3) Determine the coordinates of the intersection point. This ordered pair is the system’s solution. Using the graph below, it appears that the solution is the point (1,0). We won’t know for sure until after we check this potential solution in the next step. CONTINUED

13. Copyright © 2013, 2009, 2006 Pearson Education, Inc. 13 Systems of Equations 4) Check the solution in both equations. CONTINUED Because both equations are satisfied, (1,0) is the solution and {(1,0)} is the solution set. ? ?? ? true

14. Copyright © 2013, 2009, 2006 Pearson Education, Inc. 14 Objective #2: Example

15. Copyright © 2013, 2009, 2006 Pearson Education, Inc. 15 Objective #2: Example

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17. Copyright © 2013, 2009, 2006 Pearson Education, Inc. 17 Substitution Method Solving Linear Systems by Substitution 1) Solve either of the equations for one variable in terms of the other. (If one of the equations is already in this form, you can skip this step) 2) Substitute the expression found in step 1 into the other equation. This will result in an equation in one variable. 3) Solve the equation containing one variable. 4) Back-substitute the value found in step 3 into one of the original equations. Simplify and find the value of the remaining variable. 5) Check the proposed solution in both of the system’s given equations.

18. Copyright © 2013, 2009, 2006 Pearson Education, Inc. 18 Substitution MethodEXAMPLE SOLUTION Solve by the substitution method: 1) Solve either of the equations for one variable in terms of the other. We will isolate the variable y from the first equation. Solve for y by subtracting 4x from both sides

19. Copyright © 2013, 2009, 2006 Pearson Education, Inc. 19 Substitution MethodCONTINUED Distribute Add like terms Add 4 to both sides Divide both sides by 7 3) Solve the resulting equation containing one variable.

20. Copyright © 2013, 2009, 2006 Pearson Education, Inc. 20 Substitution Method 4) Back-substitute the obtained value into one of the original equations. We back-substitute 1 for x into one of the original equations to find y. Let’s use the first equation. CONTINUED Replace x with 1 Multiply Subtract 4 from both sides Therefore, the potential solution is (1,0).

21. Copyright © 2013, 2009, 2006 Pearson Education, Inc. 21 Substitution Method 5) Check. Now we will show that (1,0) is a solution for both of the original equations. CONTINUED ? ?? ? true Because both equations are satisfied, (1,0) is the solution and {(1,0)} is the solution set.

22. Copyright © 2013, 2009, 2006 Pearson Education, Inc. 22 Objective #3: Example

23. Copyright © 2013, 2009, 2006 Pearson Education, Inc. 23 Objective #3: Example

24. Copyright © 2013, 2009, 2006 Pearson Education, Inc. 24 Objective #3: ExampleCONTINUED

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26. Copyright © 2013, 2009, 2006 Pearson Education, Inc. 26 Addition (Elimination) Method Solving Linear Systems by Addition 1) If necessary, rewrite both equations in the form Ax + By = C. 2) If necessary, multiply either equation or both equations by appropriate nonzero numbers so that the sum of the x-coefficients or the sum of the y- coefficients is 0. 3) Add the equations in step 2. The sum should be an equation in one variable. 4) Solve the equation in one variable (the result of step 3). 5) Back-substitute the value obtained in step 4 into either of the given equations and solve for the other variable. 6) Check the solution in both of the original equations. NOTE: As you now know, there is more than one method to solve a system of equations. The reason for learning more than one method is because sometimes one method will be preferable or easier to use over another method.

27. Copyright © 2013, 2009, 2006 Pearson Education, Inc. 27 Addition (Elimination) MethodEXAMPLE SOLUTION Solve by the addition method: 1) Rewrite both equations in the form Ax + By = C. We first arrange the system so that variable terms appear on the left and constants appear on the right. We obtain Add 2y to both sides4x + 2y = 4 Add 3x to both sides

28. Copyright © 2013, 2009, 2006 Pearson Education, Inc. 28 Addition (Elimination) Method Multiply by 2 4x + 2y = 4 CONTINUED No Change

29. Copyright © 2013, 2009, 2006 Pearson Education, Inc. 29 Addition (Elimination) Method 3) Add the equations. CONTINUED 4x + 2y = 4 Add: 10x + 0y = 10 4) Solve the equation in one variable. Now solve the equation 10x = 10. 10x = 10 x = 1

30. Copyright © 2013, 2009, 2006 Pearson Education, Inc. 30 Addition (Elimination) Method 5) Back-substitute and find the value of the other variable. Now we will use one of the original equations and replace x with 1 to determine y. Let’s use the second equation. CONTINUED y = 0 Replace x with 1 Multiply Add Therefore, the potential solution is (1,0).

31. Copyright © 2013, 2009, 2006 Pearson Education, Inc. 31 Addition (Elimination) Method 6) Check. Now check the potential solution (1,0) in both original equations. CONTINUED ? ? ? ? true 4 = 0 + 4 4 = 4 0 = 0 Because both equations are satisfied, (1,0) is the solution and {(1,0)} is the solution set.

32. Copyright © 2013, 2009, 2006 Pearson Education, Inc. 32 Objective #4: Example

33. Copyright © 2013, 2009, 2006 Pearson Education, Inc. 33 Objective #4: Example

34. Copyright © 2013, 2009, 2006 Pearson Education, Inc. 34 Objective #4: ExampleCONTINUED

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36. Copyright © 2013, 2009, 2006 Pearson Education, Inc. 36 Solving Systems of Equations Comparing Solution Methods MethodAdvantagesDisadvantages Graphing You can see the solutions.If the solutions do not involve integers or are too large to be seen on the graph, it’s impossible to tell exactly what the solutions are. Substitution Gives exact solutions. Easy to use if a variable is on one side by itself. Addition Gives exact solutions. Easy to use if a variable has a coefficient of 1 or -1. Solutions cannot be seen.

37. Copyright © 2013, 2009, 2006 Pearson Education, Inc. 37 Objective #5: Example

38. Copyright © 2013, 2009, 2006 Pearson Education, Inc. 38 Objective #5: Example

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40. Copyright © 2013, 2009, 2006 Pearson Education, Inc. 40 Solving Systems of Equations The Number of Solutions to a System of Two Linear Equations Number of SolutionsWhat This Means Graphically Graphical Examples Exactly one ordered-pair solution The two lines intersect at one point. No solutionThe two lines are parallel. Infinitely Many Solutions The two lines are identical.

41. Copyright © 2013, 2009, 2006 Pearson Education, Inc. 41 Solving Systems of Equations NOTE: It is extremely helpful to understand these relationships as well as any other relationship between an equation and it’s graph. NOTE: To determine that a system has exactly one solution, solve the system using one of the methods. A single solution will occur as in the previous examples. NOTE: To determine that a system has no solution, solve the system using one of the methods. Eventually, you’ll get an obviously false statement, like 3 = 4.

42. Copyright © 2013, 2009, 2006 Pearson Education, Inc. 42 Solving Systems of EquationsEXAMPLE (a) How many tickets can be sold and supplied for $50 per ticket? (b) Find the ticket price at which supply and demand are equal. At this price, how many tickets will be supplied and sold?

43. Copyright © 2013, 2009, 2006 Pearson Education, Inc. 43 Solving Systems of EquationsSOLUTION CONTINUED The number of tickets that can be sold for $50 per ticket is found using the demand model: (a) How many tickets can be sold and supplied for $50 per ticket? The number of tickets that can be supplied for $50 per ticket is found using the supply model: N = 5(50) + 6000 = 250 + 6000 = 6250 tickets supplied.

44. Copyright © 2013, 2009, 2006 Pearson Education, Inc. 44 Solving Systems of EquationsCONTINUED N = 5p + 6000 1) Solve either of the equations for one variable in terms of the other. The system of equations already has N isolated in both equations. (b) Find the ticket price at which supply and demand are equal. At this price, how many tickets will be supplied and sold?

45. Copyright © 2013, 2009, 2006 Pearson Education, Inc. 45 Substitution MethodCONTINUED Add 25p to both sides 3) Solve the resulting equation containing one variable. Divide both sides by 30 Subtract 6000 from both sides 7800 = 30p + 6000 1800 = 30p 60 = p

46. Copyright © 2013, 2009, 2006 Pearson Education, Inc. 46 Substitution Method Replace p with 60 CONTINUED Multiply Add 4) Back-substitute the obtained value into one of the original equations. Back-substitute 60 for p into one of the original equations to find N. Use the first equation. N = 6300

47. Copyright © 2013, 2009, 2006 Pearson Education, Inc. 47 Substitution MethodCONTINUED Therefore, the solution is (60,6300). Therefore, supply and demand will be equal when the ticket price is $60 and 6300 tickets are sold. 5) Check. Now show that (60,6300) is a solution for both of the original equations. ?? ?? true N = 5p + 6000 6300 = 5(60) + 6000 6300 = 300 + 6000 6300 = 6300

48. Copyright © 2013, 2009, 2006 Pearson Education, Inc. 48 Objective #6: Example

49. Copyright © 2013, 2009, 2006 Pearson Education, Inc. 49 Objective #6: Example

50. Copyright © 2013, 2009, 2006 Pearson Education, Inc. 50 Objective #6: Example

51. Copyright © 2013, 2009, 2006 Pearson Education, Inc. 51 Objective #6: Example