2. Systems of Linear Equations A Tutorial Designed for Mr. Wilson’s Algebra Classes. Click on my alma mater whenever it appears to continue!

3. Objective This interactive presentation is used as a supplement to our unit on solving systems of linear equations in two variables. After completing this unit, the student will be able to employ several methods to solve a system of linear equations, and apply their use in real- world examples.

4. Michigan Standards Our unit on linear systems addresses the state standards of: Mathematics, Strand V: Standard 1, Benchmarks 1,4 Standard 2, Benchmarks 3,4

5. Review Before we begin, let’s review some things about linearity and linear equations…

6. Linearity A Linear Equation in 2 variables is of the form: ax+by+c=0 Notice that this can be arranged into normal slope-intercept form: y=mx+b

7. What is a System? (Click on your answer) A system of linear equations is: a. A set of parabolas A set of parabolas A set of parabolas b. A set of two or more lines A set of two or more lines A set of two or more lines c. A stereo component A stereo component A stereo component

8. What are Solutions? A solution to a system of equations is any point where the graphs of the lines intersect: TrueFalse TrueFalse

9. YES! A system of linear equations is a set of two or more lines, grouped by the word “and”, such as y=x+1 and y=x-1.

10. Try Again

11. TRUE! A solution to a system of equations is any point that satisfies all of the equations. Graphically, this is any point of intersection of the system. Since we are dealing with systems of two linear equations, there can be at most one solution. Can you think of why this is?

12. Try Again

13. Sample System Examine our system at the right. There is one solution to our system at (1,6), because this is a point of intersection.

14. Solution to Sample We must always verify a proposed solution algebraically. We propose (1,6) as a solution, so now we plug it in to both equations to see if it works: y = 6x and y = 2x + 4, -6x +y = 0 and -2x + y - 4 = 0, -6(1)+(6)= 0 and -2(1) +(6) - 4 = 0. Yes, (1,6) Satisfies both equations!

15. Other Methods There are several other methods of solving systems of linear equations. Each is best used in different situations. Choose from the list:  Substitution Method Substitution Method Substitution Method  Elimination Method Elimination Method Elimination Method  Matrix Algebra Matrix Algebra Matrix Algebra

16. Substitution Method Substitution method is used when it appears easy to solve for one variable in terms of the other. The goal is to reduce the system to two equations of one unknown each. Let’s examine our example from earlier: -2x + y = 4 -6x + y = 0

17. Substitution Method Notice that we can arrange the second equation to slope-intercept form: y = 6x Now we have y solved in terms of x. The second step is to plug our value for y back into the first equation: -2x + 6x = 4 -2x + 6x = 4

18. Substitution Method Now we solve the equation for its only unknown, x: -2x + 6x = 4 -2x + 6x = 4 4x = 4 4x = 4 x = 1 x = 1 Notice that our x value matches our earlier value (1, 6).

19. Substitution Method Now we plug our x value back into the first equation: y = 6x y = 6(1) y = 6 Again, notice that our y value is the same as what we obtained earlier. Thus, we have found the same point, (1,6), as a solution to our system.

20. Substitution Method Now you try. Use the substitution method to solve this system: 5x + 6y = 10 2y = 3 Your first step is to: a. Solve the first equation for x Solve the first equation for x Solve the first equation for x b. Solve the second equation for y Solve the second equation for y Solve the second equation for y c. I forgot, I need to review the substitution method I forgot, I need to review the substitution method I forgot, I need to review the substitution method

21. Substitution Method You cannot solve the first equation for x, because you have no value for the y variable. Take another look at the problem, you’ll notice that it is easier to solve the second equation for y in terms of x.

22. Substitution Method Yes, solving the second equation for y yields: y = 3/2 The next step is: a. Plug y = 3/2 into 5x + 6y = 10 Plug y = 3/2 into 5x + 6y = 10 Plug y = 3/2 into 5x + 6y = 10 b. Plug y = 3/2 into 2y = 3 Plug y = 3/2 into 2y = 3 Plug y = 3/2 into 2y = 3 c. I’m not sure, I need to review I’m not sure, I need to review I’m not sure, I need to review

23. Substitution Method We already solved 2y = 3 to obtain y = 3/2, so we cannot plug this value back into the same equation. Go back and examine your choices again.

24. Substitution Method Yes, we now plug our y value into the first equation: 5x + 6(3/2) = 10 5x + 6(3/2) = 10 Simplifying, we get: 5x + 9 = 10 5x + 9 = 10 5x = 1 5x = 1 x = 1/5 x = 1/5

25. Substitution Method We now have values for each variable: x = 1/5 and y = 3/2 So the solution to our system is the point: (1/5, 3/2) We know that this can be the only solution because two lines can intersect in at most one point.

26. Test This review test will help you study for our unit test on systems of linear equations. Once you begin this practice test, you will not be allowed to go back and review. Review AgainTake Test Review AgainTake Test

27. Question 1 How many solutions exist for the system at the right? a.00 b.11 c.22 d.InfiniteInfinite

28. YES! The lines are parallel, so they never intersect. A solution to a system of equations is a point where the lines intersect, thus we have no solution to our system.

29. Question 1

30. Question 2 Solve the system by substitution method: 5x – y = -13 2x + 3y = -12 The solution is: a. (-3, -2) (-3, -2) (-3, -2) b. (-2, 3) (-2, 3) (-2, 3) c. (13, 4) (13, 4) (13, 4) d. No Solution No Solution No Solution

31. YES! The solution is (-3, -2). You can verify this by plugging it into the system: 5(-3) – (-2) = -13 5(-3) – (-2) = -13 2(-3) + 3(-2) = -12 2(-3) + 3(-2) = -12

32. Question 2

33. Question 3 Solve the system using elimination method: 2x + 5y = 7 3x + y = -9 The solution is: a. (12, -4) (12, -4) (12, -4) b. (-4, 12) (-4, 12) (-4, 12) c. (4, -21) (4, -21) (4, -21) d. No Solution No Solution No Solution

34. Question 3

35. YES! The solution is (4, -21). You can verify this by plugging it into the system: 2(4) + 5(-21) = 7 2(4) + 5(-21) = 7 3(4) + (-21) = -9

36. Question 4 Use any method to solve the system: -2x + 3y = 10 -2x + 3y = -10 The solution is: a. (2, 3) (2, 3) (2, 3) b. (2, -3) (2, -3) (2, -3) c. (3, -2) (3, -2) (3, -2) d. No Solution No Solution No Solution

37. Question 4

38. YES! Notice from our system that both equations have the same slope, but different y- intercepts. Thus, they can never intersect. Can you think of how they could instead have infinite solutions?

39. Question 5 On Your Own Question 5 is a story problem that demonstrates a real-world use of linear systems. Study it carefully, solve the question on your own, and think about how powerful linear systems can be.

40. Question 5 On Your Own Your company currently uses widgets and gadgets to produce your best selling product, the Ultimate. Looking over your books you see that in May you bought 200 widgets and 400 gadgets for $500, and in June you bought 250 widgets and 250 gadgets for the same cost, $500. How much does one widget cost? One gadget? If a new supplier offered to sell you widgets for 75% cost of what you currently pay, but gadgets would cost 10% more than what you currently pay, should you switch to this new supplier or stay with your current supplier?

41. Question 5 On Your Own Solutions to question 5 will be discussed in class. Solutions to question 5 will be discussed in class. Now that you have completed this presentation, you have at least one use for linear systems theory. Now that you have completed this presentation, you have at least one use for linear systems theory. Can you think of at least 3 other examples of systems of linear equations? Can you think of at least 3 other examples of systems of linear equations?