1. Chapter 4
Systems of Equations
4.1
4.2
4.5
4.6
4.4
4.3

2. Solving Systems of Linear Equations in Two Variables
4.1
Solving Systems of Linear Equations in Two Variables
Objectives:
Determine whether an ordered pair is a solution of a system of two linear equations.
Solve a system by: graphing, substitution, & elimination

3. Systems of Linear Equations
A system of linear equations consists of two or more linear equations.
The solution of a system of two equations in two variables is an ordered pair (x, y) that makes both equations true.

4. Example 1a
Determine whether the ordered pair (3, 1) is a solution of the system. First equation Second equation
Since (3, 1) makes both equations true, it is a solution. The solution set is {(3, 1)}.

5. Example 1b
Determine whether the ordered pair (4, 2) is a solution of the system. First equation Second equation
Since (4, 2) does not make both equations true, it is not a solution of the system.

6. Example 2a Solve the system by graphing. Answer: (4, 2). 2x – y = 6
– y = -2x + 6
y = 2x – 6
x + 3y = 10
3y = -x + 10
y = -1/3 x + 10/3
I want to graph integral points so let’s try the x-intercept
x + 3(0)= 10
x = 10
(10,0)
(4, 2)
Answer: (4, 2).
Don’t forget to check the solution

7. Example 2b Solve the system by graphing.
–x + 3y = 6
3y = x + 6
y = 1/3 x + 2
3x – 9y = 9
– 9y = – 3x + 9
y = 1/3 x – 1
The graphs have the same slopes but different y-intercepts.
The lines are parallel. They never intersect so there is NO SOULTION

8. Example 2c Solve the system by graphing. x = 3y – 1 3y – 1 = x
The graphs are coinciding lines.
Meaning any ordered pair is a solution.
There is INFINITE SOLUTIONS.
These equation are said to be dependent equations.

9. One solution:
consistent system
independent equations
No solution:
inconsistent system
independent equations
Infinite number of solutions:
consistent system
dependent equations

10. Solving a System Algebraically
Graphing the equations of a system by hand is often a good method of finding approximate solutions of a system, but it is not a reliable method of finding exact solutions.
We turn instead to two algebraic methods of solving systems. We will use substitution method and then elimination method.

11. Example 3 Use the substitution method to solve the system.
Substitute 3x – 6 for y in the 2nd equation.
–4x + 2y = –8
–4x + 2(3x – 6) = –8
–4x + 6x – 12 = –8
2x – 12 = –8
2x = 4
x = 2
Now substitute 2 for x into the 1st equation
y = 3(2) – 6
y = 0
Answer:
(2,0)

12. Example 5
Use the elimination method to solve the system. Eliminate x by adding the two equations.
Substitute 2 for y into one of the equations
Answer:
(3,2)

13. Example 6 Use the elimination method to solve the system. Answer:
Substitute -1 for x into the 1st equation
Answer:
(–1,–1)

14. Example 7 Use the elimination method to solve the system.
The resulting equation 0 = 4, is false for all values of y or x.
Thus, the system has no solution.
This system is inconsistent, and the graphs of the equations are parallel lines.

15. Example 8 Use the elimination method to solve the system.
The resulting equation 0 = 0, is true for all possible values of y or x. The two equations are equivalent.
They have the same solution set and there are an infinite number of solutions.
Also known as consistent solutions and dependent equations.

16. For Honors class Solve the following system of equations 2
Then continue with elimination method

17. 4.1 summary
Objectives:
Determine whether an ordered pair is a solution of a system of two linear equations.
Solve a system by: graphing, substitution, & elimination
Key Vocabulary:
Break-even point, Consistent, Inconsistent, Dependent, Independent
Solving a System of Two Equations Using the Substitution Method
Solve one of the equations for one of its variables.
Substitute the expression for the variable found in Step 1 into the other equation.
Find the value of one variable by solving the equation from Step 2.
Find the value of the other variable by substituting the value found in Step 3 into the equation from Step 1.
Check the ordered pair solution in both original equations.
Solving a System of Two Linear Equations Using the Elimination Method
Rewrite each equation in standard form, Ax + By = C.
If necessary, multiply one or both equations by some nonzero number so that the coefficients of a variable are opposites of each other.
Add the equations.
Find the value of one variable by solving the equation from Step 3.
Find the value of the second variable by substituting the value found in Step 4 into either original equation.
Check the proposed ordered pair solution in both original equations.

18. Solving Systems of Linear Equations in Three Variables
4.2
Solving Systems of Linear Equations in Three Variables
Objective:
Solve a system of three equations in three variables

19. A linear equation in three variables contains three variables each variable raised to the power of 1.
Example: 3x – 2y + z = 5

20. Intersection in a Point

21. No Common Intersection
Three planes intersect at no point common to all three. This system has no solution. This system is inconsistent.

22. Intersection in a Line
Three planes intersect at all the points of a single line. The system has infinitely many solutions. The system is consistent.

23. Intersection in a Plane
Three planes coincide at all points on the plane. The system is consistent, and the equations are dependent.

24. Example 1 Solve the system. Plan to eliminate y 5x + z = 11 7x = 14
Sub 2 for x & 1 for z in eq. 1
x – y + 2z = 3
4x + y – z = 8
3x – y + z = 6
Sub 2 for x
5(2) + z = 11
10 + z = 11
z = 1
Answer:
(2,1,1)
Check on next slide

25. Ex 1 check
Solve the system. The solution is (2, 1, 1). To check, let x = 2, y = 1, and z = 1 in all three original equations. Equation (1) Equation (2) Equation (3)
All three statements are true, so the solution is (2, 1,1).

26. Example 2 (-1) –3x – 4y + 2z = –14 –2x – 1y – 2z = 9
Solve the system
(-1)
–3x – 4y + 2z = –14 –2x – 1y – 2z = 9
2x – 2y + 2z = -18
+ 2z
– 2z
(-1)
(2)
Planning to eliminate z
Add lines 1 & 2 and 2 & 3
-5x – 5y = –5
– 3y = – 9
-5x – 5(3) = –5
-5x – 15 = –5
-5x = 10
x = -2
3(-2) + 4(3) -2z = 14
– 2z = 14
6 – 2z = 14
-2z = 8
z = -4
y = 3
Solution: (-2, 3, -4)

27. Example 2 (differently)
Solve the system
3x + 4y – 2z = 14 2x + y + 2z = -9 2x – 2y + 2z = -18
(2)
Planning to eliminate z
Add lines 1 & 2 and 1 & 3
5x + 5y = 5
5x + 2y = -4
5x + 5(3) = 5 5x + 15 = x = x = -2
3(-2) + 4(3) -2z = – 2z = – 2z = z = z = -4
(-1)
5x + 5y = 5 -5x – 2y = 4
3y = 9
y = 3
Solution: (-2, 3, -4)

28. Example 3 Solve the system. Planning to eliminate x 2x – 4y + 8z = 2
(2)
Add lines 1 & 2 and 2 & 3
–10y + 10z = 24
–10y + 10z = 22
(-1)
–10y + 10z = 24
10y – 10z = –22
Since the statement is false, this system is inconsistent and has no solution.
0 = 2

29. Example 4
Solve the system. Planning on eliminating x -6x + 12y + 3z = -6 6x – 12y – 3z = 6 All three equations are identical, and therefore equations (1), (2), and (3) are all equivalent. There are infinitely many solutions of this system. The equations are dependent.
(3)
(6)

30. 4.2 summary
Objective:
Solve a system of three equations in three variables
Solving a System of Three Linear Equations by the Elimination Method 1. Write each equation in standard form Ax + By + Cz = D. 2. Choose a pair of equation and use the equations to eliminate a variable. 3. Choose any other pair of equations and eliminate the same variable as in Step Two equations in two variables should be obtained from Step 2 and Step 3. Use methods from Section 4.1 to solve this system for both variables. 5. To solve for the third variable, substitute values of the variables found in Step 4 into any of the original equations containing the third variable. 6. Check the ordered triple solution in all three original equations.

31. Systems of Linear Inequalities
4.5
Systems of Linear Inequalities
Objectives:
Graph a system of linear inequalities

32. Two or more equations make a system of linear equations, two or more linear inequalities make a system of linear inequalities. A solution of a system of linear inequalities is an ordered pair that satisfies each inequality in the system.
Graphing the Solutions of a System of Linear Inequalities
Graph each inequality in the system on the same set of axes.
The solutions of the system are the points common to the graphs of all the inequalities in the system.

33. Example 1 Graph the solution of the system:
Use a solid line since the inequality is greater than or equal to.
Use a dashed line since the inequality is less than.
Solution Region

34. Example 2 There is no overlap. No intersection.y 1 2 3 4 1 2 3 4
There is no overlap. No intersection.
So this system has no solution.
y = 2x + 4
y = 2x – 2

35. Example 3
Graph the solution of the system:

36. 4.5 summary Objectives: Graph a system of linear inequalities
Reminder:
When multiplying or dividing by a negative number remember to flip the inequality symbol.

37. Linear Programming 4.6 Objective:
Use an objective function and a system of linear inequalities to find the values for x and y where the maximum and minimum occurs.

38. Linear programming is a method for solving problems in which a particular quantity that must be maximized or minimized is limited by other factors.
These other limiting factors are called constraints.
In this section, the constraints are given as a system of linear inequalities. The graph of these constraints is called the feasible region.
Optimization is the process of finding the maximum or minimum value of these quantities.
An objective function is an algebraic expression in two or more variables describing a quantity that must be maximized or minimized.

39. Example 1
Find the maximum value and the minimum value of the objective function z = 2x + y subject to the following constraints:
Maximum value is 7 at (3,1) x y
(3,1)
2(3) + (1) = 7
(0,2.5)
2(0) + (2.5) =
2.5
x + 2y < 5
X-int (5,0)
Slope = –1/2
(0,0)
2(0) + (0) =
(2,0)
2(2) + (0) = 4
x – y < 2
X-int (2,0)
Y-int (0,–2)
Minimum value is 0 at (0,0)

40. Example 2
Find the maximum value and the minimum value of the objective function z = 8x + 5y subject to the following constraints:
Constraints:
0 < x < 10
0 < y < 5
3x + 2y > 6
8(0) + 5(5) = 25
8(10) + 5(5) = 105
(0,5)
(10,5)
(0,3)
8(0) + 5(3) = 15
(2,0)
(10,0)
8(2) + 5(0) = 16
Minimum value is (0,3)
Maximum value is (10,5)
8(10) + 5(0) = 80
Copyright © 2008 Pearson Addison-Wesley. All rights reserved.

41. Some Examples of real world problems (we wont solve them in here)
Suppose a student earns $10 per hour for tutoring and $8 per hour working at a newspaper office. The student decides that he should work no more than 15 hours per week so as to save time for his studies.
He must spend at least 3 hours per week tutoring since the tutoring center requires that.
He can not spend more than 8 hours per week tutoring since that is all the tutoring center will allow.
What is the maximum amount the student may earn?

42. Some Examples of real world problems (we wont solve them in here)
On June 24, 1948 the former Soviet Union blocked all land and water routes through East Germany to Berlin. A gigantic airlift was organized using American and British planes to bring food, clothing, and other supplies to the more than 2 million people in West Berlin. The cargo capacity was 30,000 cubic feet for an American plane and 20,000 cubic feet for a British plane. To break the Soviet blockade, the Western Allies had to maximize cargo capacity, but were subject to the following restrictions:
No more than 44 planes could be used.
The larger American planes required 16 personnel per flight, double that of the requirement for the British planes. The total number of personnel available could not exceed 512.
The cost of an American flight was $9,000 and the cost of a British flight was $5,000. Total weekly costs could not exceed $300,000.
Find the number of American planes and the number of British planes that were used to maximize cargo capacity.

43. 4.6 summary Solving a Linear Programming Problem
Let z = ax + by be an objective function that depends on x and y. Furthermore, z is subject to a number of constraints on x and y. If a maximum or minimum value of z exists, it can be determined as follows:
1. Graph the system of inequalities representing the constraints.
2. Find the value of the objective function at each corner, or vertex, of the graphed region. The maximum and minimum of the objective function occur at one or more of the corner points.

44. Solving Systems of Equations by Matrices
4.4
Solving Systems of Equations by Matrices
Objective:
Use matrices to solve a system of two and three equations

45. In this section we introduce solving a system of equations by a matrix.
A matrix is a rectangular array of numbers.
The numbers aligned horizontally are in the same row.
The numbers aligned vertically are in the same column.
The matrix has 2 rows and 3 columns
It is called a 2  3 matrix.
Each number in a matrix is called an element.

46. A matrix can be created from the coefficients of a system of equations.
System of Equations Corresponding Matrix
2x + 3y = – 4
x – 5y = 8

47. Example 1
Use matrices to solve the system. The corresponding matrix is: 2nd matrix → edit A Size: 2 x 3 and input the matrix 2nd matrix → math ↑ rref 2nd matrix A enter
The ordered pair solution is (5, 2).
Don’t forget to check your answers.

48. Example 2
Use matrices to solve the system. The corresponding matrix is: The equation 0 = 1 is false for all y or x values. The system is inconsistent and has no solution.
1 -3 1
rref =

49. Example 3
Use matrices to solve the system. The corresponding matrix is: The equation 0 = 0 is true for all y or x values, the lines are the same. Choose either line as the representing equation for the solution. There are infinite solutions.
2/3
1 -1/3
rref =

50. This same methodology can be extended to a linear system of three equations with 3 variables.
Now the matrix will have 3 rows and 4 columns.
We will still want ones along the diagonal from upper left to lower right and zeros everywhere else.

51. Example 3
Use matrices to solve the system. The corresponding matrix is:
The ordered triple is (0, 5, 4)
You should check to see that it satisfies all three equations in the original system.

52. 4.4 summary
Objective:
Use matrices to solve a system of two and three equations
Calculator steps:
2nd matrix → edit A
Input the size and the elements of the matrix
2nd matrix → math ↑ rref 2nd matrix A enter
Reminder:
You want one’s on the main diagonal from the upper left to the lower right and zeros everywhere else

53. Systems of Linear Equations and Problem Solving
4.3
Systems of Linear Equations and Problem Solving
Objectives:
Solve problems:
that can be modeled by a system of two linear equations
with cost and revenue functions
that can be modeled by a system of three linear equations
Hint: College Algebra solves these types of problems with a basic calculator only… but ACT/SAT allows a graphing calculator … please be able to do both

54. Example 1
Hilton University Drama Club sold 311 tickets for a play. Student tickets cost $0.50 each; non-student tickets cost $1.50 each. If total receipts were $ How many of each ticket were sold? Let s = student tickets Let n = non-student tickets
Use matrices to solve the system of equations
There were 81 student tickets and 230 non-student tickets sold for the play.

55. Example 2
One number is 4 more than twice the second number. Their total is 25. Find the numbers. x = first number y = second number x – 2y = 4 x + y = 25
Use matrices to solve the system of equations
The ordered pair solution of the system is (18, 7).

56. Example 3
Terry Watkins can row a boat 10.6 kilometers in 1 hour downstream and 6.8 kilometers upstream in 1 hour. Find how fast he can row in still water, and find the speed of the current.
Rate
Time (hr)
Distance
Downstream
r + c 1
10.6
Upstream
r  c
6.8
r + c = 10.6
r – c = 6.8
Use matrices to solve the system of equations
Terry’s rate in still water is 8.7 km/hr and the rate of the water current is 1.9 km/hr

57. Example 4
The Candy Barrel shop manager mixes candy worth $2 per pound with trail mix worth $1.50 per pound. Find how many pounds of each she should use to make 50 pounds of a party mix worth $1.80 per pound.
x = the amount of candy y = the amount of trail mix
1.80(50)
Use matrices to solve the system of equations
The store manager needs to mix 30 pounds of candy with 20 pounds of trail mix to create 50 pounds of party mix worth $1.80 per pound.

58. Example 5
The measure of the largest angle of a triangle is 90 more than the measure of the smallest angle, and the measure of the remaining angle is 30 more than the measure of the smallest angle. Find the measure of each angle.
x = smallest angle y = middle angle z = largest angle
x + y + z = 180 (sum of the angles = 180)
z = 90 + x (largest angle is 90 more than smallest)
y = 30 + x (remaining angle is 30 more than smallest angle)
Use substitution
x + (30 + x) + (90 + x) = 180
3x = 180
3x = 60
x = 20
The measure of the smallest angle is 20, the measure of the largest angle is 110 and the measure of the remaining angle is 50.

59. Example 6
Given the cost function C(x) and the revenue function R(x), find the number of units x that must be sold to break even
C(x) = 95x R(x) = 105x
To break even is when the cost function = revenue function
95x = 105x
1100 = 10x
110 = x
The company will break even when they sale 110 units.

60. Example 7
A store sells tents, sleeping bags, and camp stools. A customer buys a tent, 2 sleeping bags, and 5 camp stools for $138. The price of a tent is 9 times the price of a camp stool. The cost of a sleeping bag is $13more than the cost of a camp stool. Find the cost of each item.
Let each represent the cost for one:
T = tent B = sleeping bags C = camp stools
T + 2B + 5C = 138
T = 9C
B = C + 13
Use Substitution to solve
(9C) + 2(C + 13) + 5C = 138
16C + 26 = 138
16C = 112
C = 7
The tent cost $63, the sleeping bag $20,and the camp stool $7.

61. Example 8 ax2 + bx + c = y 25a – 5b + c = 7 a – b + c = –2
Find the equation of the parabola y = ax2+ bx + c that passes through the points (–5, 7), (–1, –2), and (3, 5).
The three points must satisfy the equation. Substitute each ordered pair into the equation to obtain a system of equations.
ax2 + bx + c = y
(–5, 7)
25a – 5b + c = 7
(–1, –2)
a – b + c = –2
9a + 3b + c = 5
(3, 5)
Use matrices to solve
Copyright © 2008 Pearson Addison-Wesley. All rights reserved.

62. Add on to notes:
An aircraft flew 6 hours with the wind. The return trip took 7 hours against the wind. If the speed of the plane in still air is 552 miles per hour more than the speed of the wind, find the wind speed and the speed of the plane in still air.
R x T = D
dist with wind = dist against wind
(P + W)6 = (P – W)7
6P + 6W = 7P – 7W
-7P+ 7W P + 7W
–P + 13W = 0
P = W + 552
-W -W
P – W = 552
–P + 13W = 0
P – W = 552
12W = 552
W = 46
P = = 598
Plane is traveling 598 mph and the wind is blowing 46 mph.

63. Add on to notes:
Find how many quarts of 5% butterfat milk and 2% butterfat milk should be mixed to yield 60 quarts of 4% butterfat milk?
X + Y = 60
0.05X Y = 0.04(60)
X qts
5% fat
Y qts
2% fat
(100)
(-2)
X + Y = 60
5X + 2Y = 4(60)
(X + Y) qts 4%
-2X + -2Y = -120
5X + 2Y = 240
3X = 120
40 quarts of 5% butterfat milk and 20 quarts of 2% butterfat milk are required to make 60 quarts of 4% butterfat milk
x = 40
y = 20

64. Add on to notes:
Find how many quarts of 5% butterfat milk and 2% butterfat milk should be mixed to yield 60 quarts of 4% butterfat milk?
5% fat
2 parts 4%
X qts
1 part
Y qts
2% fat
3 parts
60 qts
x = 40
y = 20
40 quarts of 5% butterfat milk and 20 quarts of 2% butterfat milk are required to make 60 quarts of 4% butterfat milk

65. Add on to notes:
Two motorcyclists start at the same point and travel in opposite directions. One travels 2 mph faster than the other. In 6 hours they are 192 miles apart. How fast is each traveling?
RxT=D
R1 xT1
R2 xT2
(R) x 6
(R + 2) x 6
6R + 6R + 12 = 192
12R = 180
R = 15
R + 2 = 17
The motorcycles were traveling 15 and 17 mph.

66. 4.3 summary Objectives: Solve problems:
that can be modeled by a system of two linear equations
with cost and revenue functions
that can be modeled by a system of three linear equations
Reminders:
Line up the variables
If a variable is missing then place a zero in the matrix as a place holder
D = RxT