Solving Systems of Equations
and Inequalities
Section
Section
Section
Section
Section
Section
3.1A-B
Two variable linear equations
3.1C Matrices Resolution of linear systems
3.1D
Three variable linear equations
3.1D Determinants & Inverses of Matrices
3.2A Solving Systems of Linear Inequalities
3.2B-D
Linear Programming
Section 3.1
Two Variable Linear Equations
 A set of two or more linear equations that each
contain two variables.
 These equations represent lines that will
intersect, overlap, or be parallel to each
other.
Section 3.1
Two Variable Linear Equations
 When these lines intersect or overlap each
other they are said to be consistent. This
means they have at least one point of
intersection. If there is only one point of
intersection, the lines are independent. If
there is more than one point of
intersection, the lines are dependent.
 When two lines do not intersect, they are
said to be inconsistent because the lines
are parallel.
Section 3.1
Two Variable Linear Equations
 Check your understanding:
 Given the following two linear equations,
determine the consistency & dependence of
the system. 2x + y = 5 & 4x + 2y = 8
a. Consistent, independent
b. Consistent, dependent
c. Inconsistent
Section 3.1
Two Variable Linear Equations
 Check your understanding:
 Given the following two linear equations,
determine the consistency & dependence of
the system. 2x + y = 5 & 4x + 2y = 8
a. Consistent, independent
b. Consistent, dependent
c. Inconsistent
Section 3.1
Two Variable Linear Equations
 Check your understanding:
 Given the following two linear equations,
determine the consistency & dependence of
the system. 3x + 2y = 5 & 6x + 2y = 8
a. Consistent, independent
b. Consistent, dependent
c. Inconsistent
Section 3.1
Two Variable Linear Equations
 Check your understanding:
 Given the following two linear equations,
determine the consistency & dependence of
the system. 3x + 2y = 5 & 6x + 2y = 8
a. Consistent, independent
b. Consistent, dependent
c. Inconsistent
Section 3.1 Graphing Linear
Equations
 Systems of two linear equations that
intersect at one point (consistent &
independent) can be graphed
separately by finding ordered pairs of
the solution set for each and then
locate the coordinates for the point of
intersection.
Section 3.1 Graphing Linear
Equations
In blue we have y = 3x - 2,
and in red we have
y = -x + 2.
y = 3x - 2
-1
-5
0
-2
1
1
2
4
y = -x + 2
-1 3
0
2
1
1
2
0
Section 3.1 Graphing Linear
Equations
In blue we have y = 2x + 1,
and in red we have
y = x + 3.
y = 2x + 1
-1
-1
0
1
1
3
2
5
y=x+3
-1 2
0
3
1
4
2
5
Section 3.1 Graphing Linear
Equations
In blue we have y = 2x + 1,
and in red we have
y = 2x + 3.
y = 2x + 1
-1
-1
0
1
1
3
2
5
y = 2x + 3
-1 1
0
3
1
5
2
7
Homework
 Section 3.1A pages 163-164 complete
problems 1 – 17.
Section 3.1
Two Variable Linear Equations
 Systems of two linear equations are usually
solved using either the elimination method
or the substitution method.
 The elimination method uses a process of
removing one of the two variables
simultaneously from both equations through
addition of equal but opposite coefficients.
In blue we have y = 3x - 2,
and in red we have y = -x + 2.
Section 3.1
Two Variable Linear Equations
In blue we have y = 3x - 2,
and in red we have y = -x + 2.
Blue
3x – y = 2
Red
x+y=2
4x
=4
With the variable y having equal but opposite values
we can add the two equations together and
eliminate the y variable. We get 4x = 4. This allows
x = 1, and by inserting x = 1 back into the equation
we can see that y = 1.
Section 3.1
Two Variable Linear Equations
 The substitution method uses a process of
rewriting one of the two equations by isolating
one of the variables and then substituting the
equation into the other equation.
In blue we have y = 3x - 2,
and in red we have y = -x + 2.
With both equations in the form of y
equals, we can substitute the 2nd
equation into the first and we get:
-x + 2 = 3x – 2. Solving for x we
get -4x = -4 and x = 1.
Practice Substitution Method
2x + y = 10  y = -2x + 10
3x – 4y = -10
 3x - 4(-2x + 10) = -10
3x + 8x – 40 = -10
11x – 40 = -10
11x = -10 + 40 = 30
x = 30/11 y = -2(30/11) + 10  y = -60/11 + 110/11 =
5x - y = 9  y = 5x - 9
2x + 4y = 42
x=
35/
22
 2x + 4(5x - 9) = -10
2x + 20x – 45 = -10
22x – 45 = -10
22x = -10 + 45 = 35
; y = 5(35/22) – 9 = 175/22 – 198/22 = -23/22
50/
11
Practice Elimination Method
2x + y = 10  4(2x + y = 10) 
3x – 4y = -10
2(30/11) + y = 10  y =
110/
11
-
60/
11
8x + 4y = 40
3x – 4y = -10
11x
= 30
x = 30/11
=
50/
11
;y=
50/
11
5x - y = 9  4(5x - y = 9) 
2x + 4y = 42
20x - 4y = 36
2x + 4y = 42
22x = 78
x = 39/11
5(39/11) - y = 9  y = 195/11 - 99/11 = 96/11 ; y =
96/
11
Section 3.1
Two Variable Linear Equations
 Check your understanding:
 Given the following two linear equations,
use the substitution method to start the
solution of the system.
5x + 3y = 22
a. Y = 10 – 6x
b. Y = 10 – 3x
c. Y = 10 + 3x
6x + 2y = 20
Section 3.1
Two Variable Linear Equations
 Check your understanding:
 Given the following two linear equations,
use the substitution method to start the
solution of the system.
5x + 3y = 22
a. Y = 10 – 6x
b. Y = 10 – 3x
c. Y = 10 + 3x
6x + 2y = 20
Homework
 Section 3.1B pages 173 – 174
problems 3 – 21 odd
Section 3.1 - Matrices
 Matrix – A rectangle array of terms (elements)
arranged in columns and rows. A matrix with
m rows and n columns is called an m x n
matrix, (read m by n matrix).
 Matrices are also used to determine solutions
for multiple variable linear equations. This
technique can be used as an alternative to
elimination or substitution methods.
Section 3.1 - Matrices
a11 a12 a13
a21 a22 a23
a31 a32 a33
3 x 3 Matrix
The first number indicates the row
(horizontal) and the second number
indicates the column number (vertical).
Equal Matrices – Two matrices are equal if and only they
have the same dimensions and are equal element by
element.
This expression states that
Y
X
=
2x – 6
2y
Y = 2x – 6 and x = 2y. Using the
substitution method, we see that
Y = 2(2y) – 6 and so y = 2, x = 4.
Section 3.1 - Matrices
Addition of Matrices – The sum of two m x n matrices is a
m x n matrix in which the elements are the sum of the
corresponding elements of the given matrices.
A = -2 0 1
0 5 -8
A+B
=
B
=
-6 7 -1
Solve for A + B.
4 -3 10
-2 + (-6)
0+4
0 + 7 1 + (-1)
5 + (-3)
-8 + 10
A+B
=
-8 7 0
4 2 2
Section 3.1 - Matrices
Subtraction of Matrices – The difference of two m x n
matrices is equal to the sum A + (-B) where (-B) is the
additive inverse of B.
A = -2 0 1
0 5 -8
A-B
=
B
=
-6 7 -1
Solve for A - B.
4 -3 10
-2 - (-6)
0-4
0-7
5 - (-3)
1 - (-1)
-8 - 10
A-B
=
4 -7 2
-4 8 -18
Section 3.1 - Matrices
Scalar Product – The product of a scalar k and an m x n
matrix A is an m x n matrix denoted by kA. Each element of
kA equals k times the corresponding element of A.
A = -2 0 1
0 5 -8
kA
=
k
=
5(-2)
5(0)
Solve for kA.
5
5(0 ) 5(1)
5(5) 5(-8)
kA
=
-10 0
5
0 25 -40
Section 3.1 Determinants and Inverses

A determinant is a square array of numbers (written within a
pair of vertical lines) which represents a certain sum of
products.


Calculating a 2 × 2 Determinant
In general, we find the value of a 2 × 2 determinant with elements a,
b, c, d as follows:
We multiply the diagonals (top left × bottom right first), (bottom left x top
right) then subtract the first product minus the second.
det
a b
c d
=
a b
c d
=
ad - cb
Section 3.1 Determinants and Inverses
det
a b
c d
=
a b
c d
=
ad - cb
Ex:  3
4

  (3x 2)  (4 x1)  6  4  2
1 2
 2 6 

  (2 x 2)  (1x6)  4  (6)  2
 1 2 
4 6

  (4 x  2)  (1x6)  8  (6)  2
 1 2 
3.1 Determinants & Inverses
Solve for the inverse by crisscrossing the first and fourth
terms and reversing the sign of the second and third
term. Multiply this with the reciprocal of the determinant
for the matrix ( 1 ).
det
Examples of inverses for a 2 x 2 matrix.
Determinant = {(4  -2) – (-16)} = -8-(-6) = -2
(6)   2 6 
 4 6   2


 


4  1 4
 1 2   (1)
 2 6 
3
 2 2   1

1  2 6 

   1
=


4  
2 
 1
2  1 4 

  2

 2 2 
This is the
inverse of the
given matrix.
4 6



1

2


3.1 Determinants & Inverses

For 2 x 2 matrices, the product of a determinant and the
inverse of a matrix can be used to solve systems of two linear
equations.
 3 1
det 
  (3x 4)  ((2) x(1))  12  ( 2)  14
 2 4 
1
 4
 
1  4 1  14
14
inverse 




3 
14  2 3   2


 14 14 
Determinants & Inverses
If we have a two variable system, we
can use the determinant and
inverse of a matrix to find the
solution to the system.
 3 2   x
 5




 
 

 1 4   y
  5 
det(3  4)  (1  2)  12  2  10
 4 2 
1  4 2   10 10 
inverse 


10  1 3   1 3 


 10 10 
 4 2 
 20 10
 10 10   5
 10  10



  ( x, y )  

1

3

5

 
 5  15




 10 10 
 10 10

 10 10
, )  (1,1)
(
10
10



3 x  2 y  5
x  4 y  5
We solved
for the value
of x and y
using
matrices
and found
that x = -1
and y = 1.
Determinants & Inverses
If we have a two variable
system, we can use the
determinant and inverse
of a matrix to find the
solution to the system.
 2 3   x
 3 




5

7
y
9

 
 

2x  3y  3
5x  7 y  9
det(2  7)  (5  3)  14  15  1
1  7 3   7 3 
inverse 



5
2

5
2
1
 

 7 3   3 
 21  27


  ( x, y )  
 5 2   9 
 15  18

  (6,3)

We solved
for the
value of x
and y using
matrices
and found
that x = -1
and y = 1.
Homework
 Section 3.1D pages 190-191
problems 2 - 6
Cramer’s Rule
Cramer’s Rule begins with the solving
of the determinant for the system
followed by the determinants for each
of the variables within the system.
The determinant for each of the
variables is calculated by first
substituting the solution column
values for the variable column values
and then worked on as a 2 x 2
matrix.
Cramer’s Rule (continued)
Cramer’s Rule (conclusion)
Section 3.1 Three variable linear systems
 Three linear equation systems are solved using
the elimination and substitution methods.
 The technique requires the isolation of one of
the variables amongst the three equations.
 This gets repeated for a 2nd variable and the
remaining variable is then determined.
 Afterwards, the other variables are determined.
Section 3.1 Three variable linear systems
 Sample
X + 2y + 2z = 10
2x – y + 2z = 6
X – 3y + 2z = 1
Solve for x, y,& z.
Isolate the z value first.
Combine line 1 & 2 and then
combine line 3 & 2.
X + 2y + 2z = 10  X + 2y + 2z = 10
2x – y + 2z = 6
-2x +y – 2z = -6
-x + 3y = 4
X – 3y + 2z = 1 
2x – y + 2z = 6 
X – 3y + 2z = 1
-2x +y – 2z = -6
-x - 2y = -5
Section 3.1 Three variable linear systems
 Sample
X + 2y + 2z = 10
2x – y + 2z = 6
X – 3y + 2z = 1
Solve for x, y,& z.
-x + 3y = 5  -x + 3y = 5
-x -2y = -5  x + 2y = 5
5y = 10
Y = 2, x = 1
Isolate the x value next.
Combine line 1 & 2 answer and
the line 3 & 2 answer.
X + 2y + 2z = 11  X + 2y + 2z = 11
2x – y + 2z = 6 
-2x +y – 2z = -6
-x + 3y
=5
X – 3y + 2z = 1 
2x – y + 2z = 6 
X – 3y + 2z = 1
-2x +y – 2z = -6
-x - 2y
= -5
Solve for z by reinserting y & x
values. 1 + 2(2) + 2z = 10
Z = 2.5