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

2. 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.

3. 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

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

5. Linearity ax+by+c=0 y=mx+b
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

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

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

8. 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.

9. Try Again

10. 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?

11. Try Again

12. 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.

13. Yes, (1,6) Satisfies both equations!
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 = and -2x + y - 4 = 0,
-6(1)+(6)= and -2(1) +(6) - 4 = 0.
Yes, (1,6) Satisfies both equations!

14. Other Methods I'm ready for the test! Substitution Method
There are several other methods of solving systems of linear equations. Each is best used in different situations. Choose from the list:
Substitution Method
Elimination Method
Matrix Algebra
I'm ready for the test!

15. 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

16. Substitution Method y = 6x
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

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

18. 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.

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

20. 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.

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

22. 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.

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

24. Substitution Method We now have values for each variable:
x = 1/ 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.

25. Elimination Method
Elimination method is used when it appears easy to eliminate one variable from the system through transformation. Remember that linear transformations do not change the solutions of a system.

26. Elimination Method -6x + y = 0 -2x + y = 4
Examine our original system:
-2x + y = 4
-6x + y = 0
Notice that the y coefficients are 1, therefore we can multiply either equation by -1 and add the system, thus eliminating the y variable.

27. Elimination Method 6x – y = 0 + 6x - y = 0 4x = 4 -1(-6x + y = 0)
Let’s transform the second equation:
-1(-6x + y = 0)
6x – y = 0
Now we can add the equations together:
-2x + y = 4
+ 6x - y = 0
4x = 4

28. Elimination Method -2x + y = 4 -6x + y = 0
Solving the equation for x yields x=1.
Once we have the x value, we can plug it into either of our original equations and solve for y:
-2x + y = 4
-6x + y = 0

29. Elimination Method y = 6 Plugging x=1 into the second equation yields:
So our solution is (1,6), just as we obtained from looking at the graph!

30. Elimination Method x – 5y = -2 3x + 2y = 11
Now you try. Use the elimination method to solve the system:
x – 5y = -2
3x + 2y = 11
The first step is:
Add the equations together
Transform the first equation by multiplying it by -3
I’m not sure. I need to review the elimination method.

31. Elimination Method Adding the two equations together yields:
x – 5y = -2
+ 3x + 2y = 11
4x – 3y = 9
We have not eliminated any variables. We must transform one of the equations, and then add them together.

32. Elimination Method
Yes, notice that if we multiply the first equation by -3, we obtain additive inverses for the x coefficient. Our system is now:
-3x + 15y = 6
3x + 2y = 11
The next step is:
Add the two equations
Solve the second equation for y
I’m not sure. I need to review.

33. Elimination Method Solving the second equation for y yields:
y = .5(11 – 3x)
This does not help us solve the system. Notice that since we have additive inverses, we can eliminate the x variable by adding the equations.

34. Elimination Method -3x + 15y = 6 + 3x + 2y = 11
Yes, by adding the two equations we get:
-3x + 15y = 6
+ 3x + 2y = 11
17y = 17
Solving for y, we get y = 1. The next step is:
Plug y = 1 into either equation and solve for x
Plug x = 1 into either equation and solve for y
I’m not sure. I need to review.

35. Elimination Method
We found that y=1, not x=1. So we must plug y=1 into either equation.

36. Elimination Method
Yes, now we can plug y=1 into either of the original equations, or the transformed equation. Let’s choose the first original equation:
x – 5y = -2
x – 5(1) = -2
x = 3
So our solution is (3,1), which must be the only solution to a system of two lines.

37. Matrix Algebra -2x + y = 4 -6x + y = 0
We can use linear algebra to solve a system of linear equations. First, we take our x and y coefficients and place them in a matrix:
-2x + y = 4
-6x + y = 0

38. Matrix Algebra -2x + y = 4 -6x + y = 0
Next, we put the constant terms into a vector:
-2x + y = 4
-6x + y = 0
Then we augment the matrix and vector:

39. Matrix Algebra
Next, we use a calculator to put the matrix into row-reduced echelon form:
Now we can read off our values for x and y from the matrix. The first row reads x =1, the second reads y = 6.

40. 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 Again Take Test

41. Question 1 How many solutions exist for the system at the right? 1 21 2
Infinite

42. 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.

43. Question 1
Try Again

44. Question 2 2x + 3y = -12 Solve the system by substitution method:
The solution is:
(-3, -2)
(-2, 3)
(13, 4)
No Solution

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

46. Question 2
Try Again

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

48. Question 3
Try Again

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

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

51. Question 4
Try Again

52. 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?

53. 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.

54. 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?

55. Question 5 On Your Own The End
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.
Can you think of at least 3 other examples of systems of linear equations?
The End