1. SOLVING EQUATIONS USING DETERMINANTS
Standards 2, 25
SOLVING EQUATIONS USING DETERMINANTS
SECOND ORDER DETERMINANT
CRAMER’S RULE
PROBLEM 1
PROBLEM 2
PROBLEM 3
PROBLEM 4
SOLVING SYSTEMS OF EQUATIONS
IN THREE VARIABLES
PROBLEM 5
PROBLEM 6
END SHOW
PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

2. ALGEBRA II STANDARDS THIS LESSON AIMS:
Students solve systems of linear equations and inequalities (in two or three variables) by substitution, with graphs, or with matrices.
Estándar 2:
Los estudiantes resuelven sistemas de ecuaciones lineares y desigualdades (en 2 o tres variables) por substitución, con gráficas o con matrices.
Standard 25:
Students use properties from number systems to justify steps in combining and simplifying functions.
Estándar 25:
Los estudiantes usan propiedades de sistemas numéricos para justificar pasos en combinar y simplificar funciones.
PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

3. SECOND ORDER DETERMINANT
Standards 2, 25
SECOND ORDER DETERMINANT a b c d
rows
columns
VALUE OF A SECOND ORDER DETERMINANT a b c d
= ad - cb
Find the value of the following determinants 2 -1 6 3 7
-14 -2 4
= (2)(3) –(6)(-1)
=6+6
=12
= (7)(4) –(-2)(-14)
=28-28 =0
PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

4. The solution to the system is (x,y) cx + dy = f
Standards 2, 25
CRAMER’S RULE
ax + by = e
The solution to the system
is (x,y)
cx + dy = f e b f d x= a b c d
where a b c d
= 0
and a e c f y= a b c d
PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

5. Solve the following system of equations using Cramer’s Rule:
Standards 2, 25
2x + y = 4
ax + by = e 1
5x + y = 7 1
cx + dy = f x= a b c d e f 4 1 7
(4)(1) –(7)(1)
4 - 7 = -3 x= = = =1
(2)(1) –(5)(1) 2 1 5 y= a b c d e f 2 4 5 7
(2)(7) –(5)(4) = -6 -3 y= = = =2
(2)(1) –(5)(1) 2 1 5
The system is consistent and independent, with a unique solution at (1,2)
PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

6. Solve the following system of equations using Cramer’s Rule:
Standards 2, 25
4x + 2y = 10
ax + by = e
5x + y = 17 1
cx + dy = f x= a b c d e f 10 2 17 1
(10)(1) –(17)(2) =
-24 -6 x=
= 4 = = 4 2 5 1
(4)(1) –(5)(2) y= a b c d e f 4 10 5 17
(4)(17) –(5)(10) = 18 -6 y= = =
= -3
(4)(1) –(5)(2) 4 2 5 1
The system is consistent and independent, with a unique solution at (4,-3)
PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

7. Solve the following system of equations using Cramer’s Rule:
Standards 2, 25
6x + 4y = 10
ax + by = e
3x + 2y = 5
cx + dy = f x= a b c d e f 10 4 5 2
(10)(2) –(5)(4) = x= = =
(6)(2) –(3)(4) 6 4 3 2
Since, the determinant from the denominator is zero, and division by zero is not defined: THIS SYSTEM DOES NOT HAVE A UNIQUE SOLUTION and Cramer’s Rule can’t be used.
PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

8. Solve the following system of equations Standards 2, 25
4.6x + 2.4y = 4.2
ax + by = e
8.2x - y = 28.6 1
cx + dy = f
This is a typical case when it is useful to use Cramer’s rule, because substitution or elimination are both difficult to apply. x= a b c d e f
4.2
2.4
28.6 -1
(4.2)(-1) –(28.6)(2.4) =
-72.84
-24.28 x= = =
= 3
(4.6)(-1) –(8.2)(2.4)
4.6
2.4
8.2 -1
4.6
4.2
8.2
28.6 y= a b c d e f
(4.6)(28.6) –(8.2)(4.2) =
97.12
-24.28 y= = =
= -4
(4.6)(-1) –(8.2)(2.4)
4.6
2.4
8.2 -1
The system is consistent and independent, with a unique solution at (3,-4)
PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

9. SYSTEMS OF EQUATIONS IN THREE VARIABLES
Standard 5
SYSTEMS OF EQUATIONS IN THREE VARIABLES
Eliminating one variable, this case z:
Solve: -2
3x + 2y -3z = -2
-6x - 4y + 6z = 4
12x +9y -6z = 12
3x + 2y -3z = -2 3
4x + 3y -2z = 4
6x + 5y = 16
4x + 3y -2z = 4
5x + 4y -6z = -5 -3
4x + 3y -2z = 4
-12x - 9y + 6z = -12
5x + 4y - 6z = -5
5x + 4y -6z = -5
-7x -5y = -17
Using the equations with two variables to eliminate other variable, this case y:
Using one of the two variable equations to substitute x and get the other variable, this case y:
Using one of the three variable equations to substitute x and y to get z:
6x + 5y = 16
-7x -5y = -17
3x + 2y -3z = -2
6x + 5y = 16
3( ) + 2( ) -3z = -2 1 2
- x = -1
6( )+ 5y = 16 1
3 + 4 – 3z = -2
6 + 5y = 16
7 – 3z = -2
x=1
5y = 10
-3z = -9
y = 2
z = 3
The system has unique solution at (1,2,3)
PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

10. SYSTEMS OF EQUATIONS IN THREE VARIABLES
Standard 5
SYSTEMS OF EQUATIONS IN THREE VARIABLES
Eliminating one variable, this case z:
Solve: -3
2x + 3y -2z = 5
-6x - 9y + 6z = -15 2
10x +6y -6z = 24
2x + 3y -2z = 5
5x + 3y -3z = 12
4x - 3y = 9
5x + 3y -3z = 12
4x + 2y -5z = 4 -5
5x + 3y -3z = 12
-25x - 15y + 15z = -60 3
12x + 6y -15z = 12
4x + 2y -5z = 4
-13x - 9y = -48
Using the equations with two variables to eliminate other variable, this case y:
Using one of the two variable equations to substitute x and get the other variable, this case y:
Using one of the three variable equations to substitute x and y to get z: -3
4x - 3y = 9
-13x -9y = -48
2x + 3y -2z = 5
4x - 3y = 9
2( ) + 3( ) -2z = 5 3 1
-12x + 9y = -27
4( )- 3y = 9 3
-13x - 9y = -48
6 + 3 – 2z = 5
12 - 3y = 9
-25x = -75
9 – 2z = 5
-3y = -3
x=3
-2z = -4
y = 1
z = 2
The system has unique solution at (3,1,2)
PRESENTATION CREATED BY SIMON PEREZ. All rights reserved