1. Chapter 9: Systems of Equations and Inequalities; Matrices

2. 9.1 Systems of Equations
A set of equations is called a system of equations.
The solutions must satisfy each equation in the system.
A linear equation in n unknowns has the form
where the variables are of first-degree.
If all equations in a system are linear, the system is a system of linear equations, or a linear system.

3. 9.1 Linear System in Two Variables
Three possible solutions to a linear system in two variables:
One solution: coordinates of a point,
No solutions: inconsistent case,
Infinitely many solutions: dependent case.

4. dependent & consistent
9.1
Characteristics of a system of two linear equations in two variables.
Graphs
# of solutions
Classification
Nonparallel lines
one
consistent
Identical lines
infinite
dependent & consistent
Parallel lines
No solutions
inconsistent

5. 9.1 Substitution Method Example Solve the system. (1) (2) Solution
Solve (2) for y.
Substitute y = x + 3 in (1).
Solve for x.
Substitute x = 1 in y = x + 3.
Solution set: {(1, 4)}

6. 9.1 Solving a Linear System in Two Variables Graphically
Example Solve the system graphically.
Solution Solve (1) and (2) for y.
(1)
(2)

7. Solve using the method of graphing.
x2 + y2 = 25
x2 + y = 19

8. Solving a Nonlinear System of Equations
Example Solve the system.
Solution Choose the simpler equation, (2), and
solve for y since x is squared in (1).
Substitute for y into (1) .
(1)
(2)

9. Solving a Nonlinear System of Equations
Substitute these values for x into (3).
The solution set is

10. 9.2 Elimination Method Example Solve the system. (1) (2)
Solution To eliminate x, multiply (1) by –2 and (2)
by 3 and add the resulting equations.
(1)
(2)
(3)
(4)

11. 9.2 Elimination Method
Substitute 2 for y in (1) or (2).
The solution set is {(3, 2)}.
Consistent System

12. Solving an Inconsistent System
Example Solve the system.
Solution Eliminate x by multiplying (1) by 2 and
adding the result to (2).
Solution set is .
(1)
(2)
Inconsistent System

13. Solving a System with Dependent Equations
Example Solve the system.
Solution Eliminate x by multiplying (1) by 2 and adding the
result to (2).
Consistent & Dependent System
Solution set is all R’s.
(1)
(2)

14. 9.3 Graphing
Systems of
Inequalities

15. Look at the two graphs. Determine the following:
A. The equation of each line.
B. How the graphs are similar.
C. How the graphs are different.
The equation of each line is y = x + 3.
The lines in each graph are the same and represent all of the solutions to the equation y = x + 3.
The graph on the right is shaded above the line and this means that all of these points are solutions as well.

16. Inequalities with Greater Than
Point: (-4, 5)
Pick a point from the shaded region and test that point in the equation y = x + 3.
This is incorrect. Five is greater than or equal to negative 1.
If a solid line is used, then the equation would be 5 ≥ -1.
If a dashed line is used, then the equation would be 5 > -1.
The area above the line is shaded.

17. Inequalities with Less Than
Point: (1, -3)
Pick a point from the shaded region and test that point in the equation y = -x + 4.
This is incorrect. Negative three is less than or equal to 3.
If a solid line is used, then the equation would be -3 ≤ 3.
If a dashed line is used, then the equation would be -3 < 3.
The area below the line is shaded.

18. Graphing Linear Inequalities
Write the inequality in slope-intercept form.
Use the slope and y-intercept to plot two points.
Draw in the line. Use a solid line for less than or equal to () or greater than or equal to (≥). Use a dashed line for less than (<) or greater than (>).
Pick a point above the line or below the line. Test that point in the inequality. If it makes the inequality true, then shade the region that contains that point. If the point does not make the inequality true, shade the region on the other side of the line.
Systems of inequalities – Follow steps 1-4 for each inequality. Find the region where the solutions to the two inequalities would overlap and this is the region that should be shaded.

19. Example Graph the following linear system of inequalities.
Use the slope and y-intercept to plot two points for the first inequality. x y
Draw in the line. For  use a solid line.
Pick a point and test it in the inequality. Shade the appropriate region.

20. Example Graph the following linear system of inequalities. y
The region above the line should be shaded.
Now do the same for the second inequality.

21. Example Graph the following linear system of inequalities.
Use the slope and y-intercept to plot two points for the second inequality. x y
Draw in the line. For < use a dashed line.
Pick a point and test it in the inequality. Shade the appropriate region.

22. Example Graph the following linear system of inequalities. y
The region below the line should be shaded.

23. Example Graph the following linear system of inequalities.
The solution to this system of inequalities is the region where the solutions to each inequality overlap. This is the region above or to the left of the green line and below or to the left of the blue line.
Shade in that region. x y

24. You Try It
Graph the following linear systems of inequalities.

25. Problem 1 x y
Use the slope and y-intercept to plot two points for the first inequality.
Draw in the line.
Shade in the appropriate region.

26. Problem 1 x y
Use the slope and y-intercept to plot two points for the second inequality.
Draw in the line.
Shade in the appropriate region.

27. Problem 1 x y
The final solution is the region where the two shaded areas overlap (purple region).

28. Sect. 9.4 Linear Programming
Goal Find Maximum and Minimum values of a function
Goal Solve Real World Problems with Linear Programming

29. When graphing a System of Linear Inequalities
* Each linear inequality is called a Constraint
* The intersection of the graphs is called the Feasible Region
* When the graphs of the constraints is a polygonal region, we say the region is Bounded.

30. Sometimes it is necessary to find the Maximum or Minimum values that a linear function has for the points in a feasible region.
The Maximum or Minimum value of a related function Always occurs at one of the Vertices of the Feasible Region.

31. Example 1 y  4 y  - x + 6 y  y  6x + 4
Graph the system of inequalities. Name the coordinates of the vertices of the feasible region. Find the Maximum and Minimum values of the function f(x, y) = 3x + 2y for this polygonal region.
y  4
y  - x + 6
y 
y  6x + 4

32. y  4
y  - x + 6
y 
y  6x + 4
The polygon formed is a quadrilateral with vertices at (0, 4), (2, 4), (5, 1), and (- 1, - 2).

33. Use a table to find the maximum and minimum values of the function.
(x, y)
3x + 2y
f(x, y)
(0, 4)
3(0) +2(4) 8
(2, 4)
3(2) +2(4) 14
(5, 1)
3(5) + 2(1) 17
(- 1, - 2)
3(- 1) + 2(- 2)
- 7
The maximum value is 17 at (5, 1). The Minimum value is – 7 at (- 1, - 2).

34. Bounded Region Example 1 Example 2 y  - 3 x  - 2 y 
Graph the system of inequalities. Name the coordinates of the vertices of the feasible region. Find the Maximum and Minimum values of the function f(x, y) = 2x - 5y for this polygonal region.
y  - 3
x  - 2
y 

35. The polygon formed is a triangle with vertices at (- 2, 12), (- 2, - 3), and (4, - 3)
(x, y)
2x – 5y
F(x, y)
(- 2, 12)
2(-2) – 5(12)
- 64
(- 2, - 3)
2(- 2) – 5(- 3) 11
(4, - 3)
2(4) – 5(- 3) 23
The Maximum Value is 23 at (4, - 3). The Minimum Value is – 64 at (- 2, 12).

36. In this case, the region is said to be Unbounded.
Sometimes a system of Inequalities forms a region that is not a closed polygon.
In this case, the region is said to be Unbounded.

37. Example 3 Unbounded Region y + 3x  4 y  - 3x – 4 y  8 + x
Graph the system of inequalities. Name the coordinates of the vertices of the feasible region. Find the Maximum and Minimum values of the function f(x, y) = 4y – 3x for this region.
y + 3x  4
y  - 3x – 4
y  8 + x

38. There are only two points of intersection (- 1, 7) and (- 3, 5)
There are only two points of intersection (- 1, 7) and (- 3, 5). This is an Unbounded Region.
(x, y)
4y – 3x
F(x, y)
(- 1, 7)
4(7) – 3(- 1) 31
(- 3, 5)
4(5) – 3(- 3) 29
The Maximum Value is 31 at (- 1, 7). There is no minimum value since there are other points in the solution that produce lesser values. Since the region is Unbounded, f(x, y) has no minimum value.

39. Linear Programming Procedures.
Step 1: Define the Variables
Step 2: Write a system of Inequalities
Step 3: Graph the System of Inequalities
Step 4: Find the coordinates of the vertices of the feasible region.
Step 5: Write a function to be maximized or minimized.
Step 6: Substitute the coordinates of the vertices into the function.
Step 7: Select the greatest or least result. Answer the problem.

40. Example 4
Ingrid is planning to start a home-based business. She will be baking decorated cakes and specialty pies. She estimates that a decorated cake will take 75 minutes to prepare and a specialty pie will take 30 minutes to prepare. She plans to work no more than 40 hours per week and does not want to make more than 60 pies in any one week. If she plans to charge $34 for a cake and $16 for a pie, find a combination of cakes and pies that will maximize her income for a week.

41. Step 1: Define the Variables
C = number of cakes
P = number of pies
Step 2: Write a system of Inequalities
Since number of baked items can’t be negative, c and p must be nonnegative
c  0
p  0
A cake takes 75 minutes and a pie 30 minutes. There are 40 hours per week available.
75c + 30p  hours = 2400 min.
She does not want to make more than 60 pies each week
p  60

42. Step 3: Graph the system of Inequalities
cakes
Pies

43. Step 4: Find the Coordinates of the vertices of the feasible region.
The vertices of the feasible region are (0, 0), (0, 32), (60, 8), and (60, 0).
Step 5: Write a function to be maximized or minimized.
The function that describes the income is:
f(p, c) = 16p + 34c

44. (p, c) 16p + 34c f(p, c) (0, 0) 16(0) + 34(0) (0, 32) 16(0) + 34(32)
Step 6: Substitute the coordinates of the vertices into function.
(p, c)
16p + 34c
f(p, c)
(0, 0)
16(0) + 34(0)
(0, 32)
16(0) + 34(32)
1088
(60, 8)
16(60) + 34(8)
1232
(60, 0)
16(60) + 34(0)
960
Step 7: Select the Greatest or Least result. Answer the Problem.
The maximum value of the function is 1232 at (60, 8). This means that the maximum income is $1232 when Ingrid makes 60 pies and 8 cakes per week.

45. Section 9.5 Solving Systems using Matrices
What is a Matrix? [ ]
Augmented Matrices
Solving Systems of 3 Equations
Inconsistent & Dependent Systems

46. Concept: A Matrix
Any rectangular array of numbers arranged in rows and columns, written within brackets
Examples:

47. Matrix Row Transformations
Streamlined use of echelon methods

48. 9.5 Solution of Linear Systems by Row Transformations
This is called an augmented matrix where each member of the array is called an element or entry.
The rows of an augmented matrix can be treated just like the equations of a linear system.

49. Concept: Augmented Matrices
Are used to solve systems of linear equations:
Arrange equations in simplified standard form
Put all coefficients into a 2x3 or 3x4 augmented matrix

50. Reduced Row Echelon Method with the Graphing Calculator
Example Solve the system.
Solution The augmented matrix of the system is
shown below.

51. 9.5 Reduced Row Echelon Method with the Graphing Calculator
Using the rref command we obtain the row reduced
echelon form.

52. 9.5 Solving a System with No Solutions
Example Show that the following system is
inconsistent.
Solution The augmented matrix of the system is

53. 9.5 Solving a System with No Solutions
The final row indicates that the system is
inconsistent and has solution set .

54. 9.5 Solving a System with Dependent Equations
Example Show that the system has dependent
equations. Express the general solution using an
arbitrary variable.
Solution The augmented matrix is

55. 9.5 Solving a System with Dependent Equations
The final row of 0s indicates that the system has
dependent equations. The first two rows represent
the system

56. 9.5 Solving a System with Dependent Equations
Solving for y we get
Substitute this result into the expression to find x.
Solution set written with z arbitrary: