1. Review

2. Ex: Check whether the ordered pairs are solutions of the system
Ex: Check whether the ordered pairs are solutions of the system. x-3y= -5 -2x+3y=10
(-5,0)
-5-3(0)= -5
-5 = -5
-2(-5)+3(0)=10
10=10
Solution
(1,4)
1-3(4)= -5
1-12= -5
-11 = -5
*doesn’t work in the 1st eqn, no need to check the 2nd.
Not a solution.

3. Solving a System Graphically
Graph each equation on the same coordinate plane. (USE GRAPH PAPER!!!)
If the lines intersect: The point (ordered pair) where the lines intersect is the solution.
If the lines do not intersect:
They are the same line – infinitely many solutions (they have every point in common).
They are parallel lines – no solution (they share no common points).

4. Ex: Solve the system graphically. 2x-2y= -8 2x+2y=4
(-1,3)

5. Ex: Solve the system graphically. 2x+4y=12 x+2y=6
1st eqn:
y = -½x + 3
2ND eqn:
What does this mean?
the 2 equations are for the same line!
Infinite many solutions

6. Ex: Solve graphically: x-y=5 2x-2y=9
1st eqn:
y = x – 5
2nd eqn:
y = x – 9/2
What do you notice about the lines?
They are parallel!
No solution!

7. Solving Systems of Equations using Substitution
Steps:
1. Solve one equation for one variable
(y= ; x= ; a=)
2. Substitute the expression from step one into the other equation. Then solve.
3. Substitute back into Step 1 and solve for the other variable.
4. Check the solution in both equations of the system.

8. 1) Solve the system using substitution
x + y = 5
y = 3 + x
Step 1: Solve an equation for one variable.
The second equation is
already solved for y!
Step 2: Substitute
x + y = 5 x + (3 + x) = 5
2x + 3 = 5
2x = 2
x = 1

9. 1) Solve the system using substitution
x + y = 5
y = 3 + x
y = 3 + x
Y = 3 + (1)
y = 4
Step 3: Plug back in to find the other variable.
(1, 4)
(1) + (4) = 5
(4) = 3 + (1)
Step 4: Check your solution.
The solution is (1, 4). What do you think the answer would be if you graphed the two equations?

10. Which answer checks correctly?
3x – y = 4
x = 4y - 17
(2, 2)
(5, 3)
(3, 5)
(3, -5)

11. 2) Solve the system using substitution
3y + x = 7
4x – 2y = 0
It is easiest to solve the
first equation for x.
3y + x = 7
-3y y
x = -3y + 7
Step 1: Solve an equation for one variable.
Step 2: Substitute
4x – 2y = 0
4(-3y + 7) – 2y = 0

12. 2) Solve the system using substitution
3y + x = 7
4x – 2y = 0
-12y + 28 – 2y = 0
-14y + 28 = 0
-14y = -28
y = 2
x = -3y + 7
x = -3(2) + 7
x =
x = 1
Step 3: Plug back in to find the other variable.

13. 2) Solve the system using substitution
3y + x = 7
4x – 2y = 0
Step 4: Check your solution.
(1, 2)
3(2) + (1) = 7
4(1) – 2(2) = 0
When is solving systems by substitution easier to do than graphing?
When only one of the equations has a variable already isolated (like in example #1).

14. 2x + 4y = 4 3x + 2y = 22 -4y + 4 -2y + 2 -2x + 4 -2y+ 22
If you solved the first equation for x, what would be substituted into the bottom equation.
2x + 4y = 4
3x + 2y = 22
-4y + 4
-2y + 2
-2x + 4
-2y+ 22

15. 3) Solve the system using substitution
x = 3 – y
x + y = 7
Step 1: Solve an equation for one variable.
The first equation is
already solved for x!
Step 2: Substitute
x + y = 7
(3 – y) + y = 7
3 = 7
The variables were eliminated!!
This is a special case.
Does 3 = 7? FALSE!
When the result is FALSE, the answer is NO SOLUTIONS.

16. 3) Solve the system using substitution
2x + y = 4
4x + 2y = 8
Step 1: Solve an equation for one variable.
The first equation is
easiest to solved for y!
y = -2x + 4
4x + 2y = 8
4x + 2(-2x + 4) = 8
Step 2: Substitute
4x – 4x + 8 = 8
8 = 8
This is also a special case.
Does 8 = 8? TRUE!
When the result is TRUE, the answer is INFINITELY MANY SOLUTIONS.

17. What does it mean if the result is “TRUE”?
The lines intersect
The lines are parallel
The lines are coinciding
The lines reciprocate
I can spell my name

18. Solving Systems of Equations using Elimination
Steps:
1. Place both equations in Standard Form
Ax + By = C.
2. Determine which variable to eliminate with Addition or Subtraction.
3. Solve for the variable left.
4. Go back and use the found variable in step 3 to find second variable.
5. Check the solution!!!!

19. 1) Solve the system using elimination.
2x + 2y = 6
3x – y = 5
Step 1: Put the equations in Standard Form.
They already are!
None of the coefficients are the same!
Find the least common multiple of each variable.
LCM = 6x, LCM = 2y
Which is easier to obtain?
2y (you only have to multiply the bottom equation by 2)
Step 2: Determine which variable to eliminate.

20. 1) Solve the system using elimination.
2x + 2y = 6
3x – y = 5
Multiply the bottom equation by 2
2x + 2y = 6
(2)(3x – y = 5)
8x = 16
x = 2
2x + 2y = 6
(+) 6x – 2y = 10
Step 3: Multiply the equations and solve.
2(2) + 2y = 6
4 + 2y = 6
2y = 2
y = 1
Step 4: Plug back in to find the other variable.

21. 1) Solve the system using elimination.
2x + 2y = 6
3x – y = 5
(2, 1)
2(2) + 2(1) = 6
3(2) - (1) = 5
Step 5: Check your solution.
Solving with multiplication adds one more step to the elimination process.

22. 2) Solve the system using elimination.
x + 4y = 7
4x – 3y = 9
Step 1: Put the equations in Standard Form.
They already are!
Find the least common multiple of each variable.
LCM = 4x, LCM = 12y
Which is easier to obtain?
4x (you only have to multiply the top equation by -4 to make them inverses)
Step 2: Determine which variable to eliminate.

23. 2) Solve the system using elimination.
x + 4y = 7
4x – 3y = 9
Multiply the top equation by -4
(-4)(x + 4y = 7)
4x – 3y = 9)
y = 1
-4x – 16y = -28
(+) 4x – 3y = 9
Step 3: Multiply the equations and solve.
-19y = -19
x + 4(1) = 7
x + 4 = 7
x = 3
Step 4: Plug back in to find the other variable.

24. 2) Solve the system using elimination.
x + 4y = 7
4x – 3y = 9
(3, 1)
(3) + 4(1) = 7
4(3) - 3(1) = 9
Step 5: Check your solution.

25. What is the first step when solving with elimination?
Add or subtract the equations.
Multiply the equations.
Plug numbers into the equation.
Solve for a variable.
Check your answer.
Determine which variable to eliminate.
Put the equations in standard form.

26. Which variable is easier to eliminate?
3x + y = 4
4x + 4y = 6 x y 6 4 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32

27. 3) Solve the system using elimination.
3x + 4y = -1
4x – 3y = 7
Step 1: Put the equations in Standard Form.
They already are!
Find the least common multiple of each variable.
LCM = 12x, LCM = 12y
Which is easier to obtain?
Either! I’ll pick y because the signs are already opposite.
Step 2: Determine which variable to eliminate.

28. 3) Solve the system using elimination.
3x + 4y = -1
4x – 3y = 7
Multiply both equations
(3)(3x + 4y = -1)
(4)(4x – 3y = 7)
x = 1
9x + 12y = -3
(+) 16x – 12y = 28
Step 3: Multiply the equations and solve.
25x = 25
3(1) + 4y = -1
3 + 4y = -1
4y = -4
y = -1
Step 4: Plug back in to find the other variable.

29. 3) Solve the system using elimination.
3x + 4y = -1
4x – 3y = 7
(1, -1)
3(1) + 4(-1) = -1
4(1) - 3(-1) = 7
Step 5: Check your solution.

30. What is the best number to multiply the top equation by to eliminate the x’s?
3x + y = 4
6x + 4y = 6 -4 -2 2 4 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32

31. Solve using elimination.
2x – 3y = 1
x + 2y = -3
(2, 1)
(1, -2)
(5, 3)
(-1, -1)

32. Find two numbers whose sum is 18 and whose difference 22.

33. Using Elimination to Solve a Word Problem:
Two angles are supplementary. The measure of one angle is 10 degrees more than three times the other. Find the measure of each angle.

34. Using Elimination to Solve a Word Problem:
Two angles are supplementary. The measure of one angle is 10 more than three times the other. Find the measure of each angle.
x = degree measure of angle #1
y = degree measure of angle #2
Therefore x + y = 180

35. Using Elimination to Solve a Word Problem:
Two angles are supplementary. The measure of one angle is 10 more than three times the other. Find the measure of each angle.
x + y = 180
x =10 + 3y

36. Using Elimination to Solve a Word Problem:
x + y = 180
x =10 + 3y
x = 180
x =
x = 137.5
(137.5, 42.5)
x + y = 180
-(x - 3y = 10)
4y =170
y = 42.5

37. Using Elimination to Solve a Word Problem:
The sum of two numbers is 70 and their difference is 24. Find the two numbers.

38. Using Elimination to Solve a Word problem:
The sum of two numbers is 70 and their difference is 24. Find the two numbers.
x = first number
y = second number
Therefore, x + y = 70

39. Using Elimination to Solve a Word Problem:
The sum of two numbers is 70 and their difference is 24. Find the two numbers.
x + y = 70
x – y = 24

40. Using Elimination to Solve a Word Problem:
x + y =70
x - y = 24
47 + y = 70
y = 70 – 47
y = 23
2x = 94
x = 47
(47, 23)

41. Now you Try to Solve These Problems Using Elimination.
Find two numbers whose sum is 18 and whose difference is 22.
The sum of two numbers is 128 and their difference is Find the numbers.

42. Matrix Operations

43. What is a Matrix?
MATRIX: A rectangular arrangement of numbers in rows and columns.
The ORDER of a matrix is the number of the rows and columns.
The ENTRIES are the numbers in the matrix.
This order of this matrix is a 2 x 3.
columns
rows

44. What is the order? (or square matrix) 3 x 3
(Also called a column matrix)
1 x 4
3 x 5
(or square matrix)
2 x 2
4 x 1
(Also called a row matrix)

45. Adding Two Matrices
To add two matrices, they must have the same order. To add, you simply add corresponding entries.

46. = 7 7 4 5 = 7 5 7

47. Subtracting Two Matrices
To subtract two matrices, they must have the same order. You simply subtract corresponding entries.

48. 2-0
-4-1
3-8 -5 2 -5
8-3
0-(-1)
-7-1 5 -8 1 = =
1-(-4)
5-2
0-7 5 3 -7

49. Multiplying a Matrix by a Scalar
In matrix algebra, a real number is often called a SCALAR. To multiply a matrix by a scalar, you multiply each entry in the matrix by that scalar.

50. -3 3 -2 6 -5
-2(-3)
-2(3) 6 -6
-12
-2(6)
-2(-5) 10

51. Multiplying Matrices

52. In order to multiply matrices...
A • B = AB
m x n
n x p
m x p
Ex 1.
Can you multiply? What will the dimensions be? A B AB
2 x 3
3 x 4
2 x 4 AB A B
5 x 3
5 x 2
Not possible

53. How to multiply... ac ad =
2 x 1
1 x 2
2 x 2

54. How to multiply... ac ad = bc bd
2 x 1
1 x 2
2 x 2

55. Ex. 1

56. Ex. 2
Find AB
-16
+ 1
-12 +2
-15
-10 =
- 2
- 4 -2
- 4

57. Ex. 3
Find BA

58. How to multiply... ag
+bi
+ck ah
+bj
+cm =
2 x 3
3 x 2
2 x 2

59. 2 x 3 3 x 2 2 x 2 = How to multiply... ag +bi +ck ah +bj +cm dg +ei
+fk dh
+ej
+fm
2 x 3
3 x 2
2 x 2

60. Ex. 4 -1(4) +5(6) -1(-3) +5(8) -4 + 30 3 + 40 5(4) +2(6) 5(-3) +2(8)20
+12
-15
+ 16 =
+ -24
+ -32
0(4)
+-4(6)
0(-3)
+-4(8) 26 43 32 1
-24
-32

61. If we are multiplying matrices, we multiply each row of the first matrix by each column in the second matrix!!
multiply each row of the first matrix by each column in the second matrix!! + + 1 + + 2 + + 1 + +

62. 2 X 3 3 X 3 2(3)+(-3)(8)+5(-3) 2(9)+(-3)(0)+ 5(1) 2(1)+(-3)(-4)+ 5(5)
-1(3)+(6)(8)+8(-3)
-1(9)+(6)(0)+ 8(1)
-1(1)+(6)(-4)+ 8(5)

63. Ex. 5

64. Examples: 2(3) + -1(5) 2(-9) + -1(7) 2(2) + -1(-6) 3(3) + 4(5)
3(-9) + 4(7)
3(2) + 4(-6)

65. *They don’t match so can’t be multiplied together.*
Dimensions: 2 x x 2
*They don’t match so can’t be multiplied together.*

66. 2 x 2 2 x 2 *Answer should be a 2 x 2 0(4) + (-1)(-2) 0(-3) + (-1)(5)
1(4) + 0(-2)
1(-3) +0(5)

67. Solving Systems of Equations with Matrices

68. A system of equations may be represented as a matrix equation
A system of equations may be represented as a matrix equation. For example, the system of equations
may be represented by the matrix equation

69. Write the matrix equation that represents the system:

70. Write the matrix equation that represents the system:

71. A matrix equation is in the form AX = B, where A is the coefficient matrix, X is the variable matrix, and B is the constant matrix.

72. Solving AX=B Real Numbers ax=b (1/a)(ax) = (1/a)b (๋1/a)(a)x = b/a
Note: 1/a must exist to solve ax = b

73. Solving AX=B Real Numbers ax=b (1/a)(ax) = (1/a)b (๋1/a)(a)x = b/a
Matrices
AX=B
A-1(AX)=A-1B
(A-1A)X=A-1B
IX=A-1B
X=A-1B
Note: A-1 must exist to solve AX=B

74. Solve the system of equations using matrices.
x = -7
y = 15

75. Solve the system of equations using matrices.
x = 1
y = -1

76. Ex. 2 Solve using matrices.
AX = B
X = A-1B
x = -7
y = -4 A B
(-7, -4)

77. Ex. 3 Solve using matrices
y = 2
(5/7, 2)