Solve Systems of Linear Equations Using Elimination Method (40

Systems of Linear Equations
We need one piece of information to solve for one
unknown quantity.

unknown quantity. If 2 hamburgers cost 4 dollars and
x is the cost of a hamburger, then 2x = 4 or x = $2.

If there are two unknowns x and y, we need two
pieces of information to solve them.

Example A. Let x = cost of a hamburger,
y = cost of an order of fries. If two hamburgers and
an order of fries cost $7, and one hamburger and one
order of fries cost $5. What is the price of each?

We translate this information into two equations:

2x + y = 7

2x + y = 7
x + y = 5

2x + y = 7
x + y = 5{
A group of equations such as this is called a
system of equations.
Eq. 1
Eq. 2

We are to find values for x and for y that will satisfy all
equations in the system.

equations in the system. Such a set of values is
called a solution for the system.

called a solution for the system. To solve the system,
we note that the difference in the two orders is one
hamburger and the difference in price is $2.

hamburger and the difference in price is $2. So the
price of the hamburger must be $2.

price of the hamburger must be $2. In algebra, we
subtract the equations in the system:
2x + y = 7
x + y = 5)
Eq. 1
Eq. 2

2x + y = 7
x + y = 5)
x = 2
Eq. 1
Eq. 2

2x + y = 7
x + y = 5)
x = 2
Eq. 1
Eq. 2
Put x = 2 back into Eq. 2, we get y = 3.

2x + y = 7
x + y = 5)
x = 2
Eq. 1
Eq. 2
Hence a hamburger is $2 and an order of fries is $3.

2x + y = 7
x + y = 5)
x = 2
Eq. 1
Eq. 2
Hence a hamburger is $2 and an order of fries is $3.
This is called the elimination method - we eliminate
variables by adding or subtracting two equations.

Elimination method reduces the number of
variables in the system one at a time.

variables in the system one at a time. The method
changes a given system into a smaller system.

changes a given system into a smaller system. We
do this successively until the solution is clear.

Example B:
Let x = cost of a hamburger,
y = cost of an order of fries,
z = cost of a soda.
2 hamburgers, 3 fries and 3 soda cost $13.
1 hamburger, 2 fries and 2 soda cost $8.
3 hamburgers, 2 fries, 3 soda cost $13.
Find x, y, and z.

We translate the information into a system of three
equations with three unknowns.
Example B.
Let x = cost of a hamburger,
y = cost of an order of fries,
z = cost of a soda.
2 hamburgers, 3 fries and 3 soda cost $13.
1 hamburger, 2 fries and 2 soda cost $8.
3 hamburgers, 2 fries, 3 soda cost $13.
Find x, y, and z.

2x + 3y + 3z = 13
x + 2y + 2z = 8
3x + 2y + 3z = 13
{
E 1
E 2
E 3
2 burgers 3 fries, 3 sodas: $13
1 burger, 2 fries and 2 sodas: $8
3 burgers, 2 fries, 3 sodas: $13

2x + 3y + 3z = 13
x + 2y + 2z = 8
3x + 2y + 3z = 13
{
E 1
E 2
E 3
Select x to eliminate since there is 1x in E2.

2x + 3y + 3z = 13
x + 2y + 2z = 8
3x + 2y + 3z = 13
{
E 1
E 2
E 3
-2*E 2 + E1:

2x + 3y + 3z = 13
x + 2y + 2z = 8
3x + 2y + 3z = 13
{
E 1
E 2
E 3
-2*E 2 + E1:
-2x – 4y – 4z = -16
+) 2x + 3y + 3z = 13

2x + 3y + 3z = 13
x + 2y + 2z = 8
3x + 2y + 3z = 13
{
E 1
E 2
E 3
-2*E 2 + E1:
-2x – 4y – 4z = -16
+) 2x + 3y + 3z = 13
0 – y – z = - 3

2x + 3y + 3z = 13
x + 2y + 2z = 8
3x + 2y + 3z = 13
{
E 1
E 2
E 3
-2*E 2 + E1:
-2x – 4y – 4z = -16
+) 2x + 3y + 3z = 13
0 – y – z = - 3
-3*E 2 + E3:

2x + 3y + 3z = 13
x + 2y + 2z = 8
3x + 2y + 3z = 13
{
E 1
E 2
E 3
-2*E 2 + E1:
-2x – 4y – 4z = -16
+) 2x + 3y + 3z = 13
0 – y – z = - 3
-3*E 2 + E3:
-3x – 6y – 6z = -24
+) 3x + 2y + 3z = 13

2x + 3y + 3z = 13
x + 2y + 2z = 8
3x + 2y + 3z = 13
{
E 1
E 2
E 3
-2*E 2 + E1:
-2x – 4y – 4z = -16
+) 2x + 3y + 3z = 13
0 – y – z = - 3
-3*E 2 + E3:
-3x – 6y – 6z = -24
+) 3x + 2y + 3z = 13
0 – 4y – 3z = -11

2x + 3y + 3z = 13
x + 2y + 2z = 8
3x + 2y + 3z = 13
{
E 1
E 2
E 3
-2*E 2 + E1:
-2x – 4y – 4z = -16
+) 2x + 3y + 3z = 13
0 – y – z = - 3
-3*E 2 + E3:
-3x – 6y – 6z = -24
+) 3x + 2y + 3z = 13
0 – 4y – 3z = -11
Group these into a system of two equations with
two unknowns.

2x + 3y + 3z = 13
x + 2y + 2z = 8
3x + 2y + 3z = 13
{
E 1
E 2
E 3
-2*E 2 + E1:
-2x – 4y – 4z = -16
+) 2x + 3y + 3z = 13
0 – y – z = - 3
-3*E 2 + E3:
-3x – 6y – 6z = -24
+) 3x + 2y + 3z = 13
0 – 4y – 3z = -11
Group these into a system of two equations with
two unknowns.
Hence we reduced the system to a simpler one.

– y – z = - 3
– 4y – 3z = -11{

– y – z = - 3
– 4y – 3z = -11{ 4y + 3z = 11
{ y + z = 3 Eq.4
Eq. 5
*(-1)

– y – z = - 3
– 4y – 3z = -11{ 4y + 3z = 11
{ y + z = 3 Eq.4
Eq. 5
*(-1)
Let's eliminate z.

– y – z = - 3
– 4y – 3z = -11{ 4y + 3z = 11
{ y + z = 3 Eq.4
Eq. 5
-3*E 4 + E5:
*(-1)
Let's eliminate z.

– y – z = - 3
– 4y – 3z = -11{ 4y + 3z = 11
{ y + z = 3 Eq.4
Eq. 5
-3*E 4 + E5:
+) 4y + 3z = 11
-3y – 3z = -9
*(-1)
Let's eliminate z.

– y – z = - 3
– 4y – 3z = -11{ 4y + 3z = 11
{ y + z = 3 Eq.4
Eq. 5
-3*E 4 + E5:
+) 4y + 3z = 11
-3y – 3z = -9
y + 0 = 2
y = 2
*(-1)
Let's eliminate z.

– y – z = - 3
– 4y – 3z = -11{ 4y + 3z = 11
{ y + z = 3 Eq.4
Eq. 5
-3*E 4 + E5:
+) 4y + 3z = 11
-3y – 3z = -9
y + 0 = 2
y = 2
To get z, set 2 for y in E4:
*(-1)
Let's eliminate z.

– y – z = - 3
– 4y – 3z = -11{ 4y + 3z = 11
{ y + z = 3 Eq.4
Eq. 5
-3*E 4 + E5:
+) 4y + 3z = 11
-3y – 3z = -9
y + 0 = 2
y = 2
To get z, set 2 for y in E4: 2 + z = 3  z = 1
*(-1)
Let's eliminate z.

– y – z = - 3
– 4y – 3z = -11{ 4y + 3z = 11
{ y + z = 3 Eq.4
Eq. 5
-3*E 4 + E5:
+) 4y + 3z = 11
-3y – 3z = -9
y + 0 = 2
y = 2
For x, set 2 for y , set 1 for z in E2:
*(-1)
Let's eliminate z.

– y – z = - 3
– 4y – 3z = -11{ 4y + 3z = 11
{ y + z = 3 Eq.4
Eq. 5
-3*E 4 + E5:
+) 4y + 3z = 11
-3y – 3z = -9
y + 0 = 2
y = 2
x + 2*2 + 2*1 = 8 
*(-1)
Let's eliminate z.

– y – z = - 3
– 4y – 3z = -11{ 4y + 3z = 11
{ y + z = 3 Eq.4
Eq. 5
-3*E 4 + E5:
+) 4y + 3z = 11
-3y – 3z = -9
y + 0 = 2
y = 2
x + 2*2 + 2*1 = 8  x + 6 = 8  x = 2
*(-1)
Let's eliminate z.

– y – z = - 3
– 4y – 3z = -11{ 4y + 3z = 11
{ y + z = 3 Eq.4
Eq. 5
-3*E 4 + E5:
+) 4y + 3z = 11
-3y – 3z = -9
y + 0 = 2
y = 2
x + 2*2 + 2*1 = 8  x + 6 = 8  x = 2
*(-1)
Hence the hamburger costs $2, and an order of fries
costs $2 and the drink costs $1.
Let's eliminate z.

A matrix is a rectangular table of numbers.
Matrix Notation

For example
5 2 -3
4 -1 0
-1 2 -1
6 -2 3
11 9 -4
Matrix Notation
are matrices.

For example
5 2 -3
4 -1 0
-1 2 -1
6 -2 3
11 9 -4
Matrix Notation
are matrices.
Systems of linear equations can be put into matrices
and solved using matrix notation.

For example
5 2 -3
4 -1 0
-1 2 -1
6 -2 3
11 9 -4
Matrix Notation
are matrices.
x + 4y = -7
2x – 3y = 8{
For example, the system
1 4 -7
2 -3 8
may be written as the matrix:
x y #

For example
5 2 -3
4 -1 0
-1 2 -1
6 -2 3
11 9 -4
Matrix Notation
are matrices.
x + 4y = -7
2x – 3y = 8{
1 4 -7
2 -3 8
x y #
This is called the augmented matrix for the system.

For example
5 2 -3
4 -1 0
-1 2 -1
6 -2 3
11 9 -4
Matrix Notation
are matrices.
x + 4y = -7
2x – 3y = 8{
1 4 -7
2 -3 8
x y #
This is called the augmented matrix for the system.
Each row of the matrix corresponds to an equation,
each column corresponds to a variable and the last
column corresponds to numbers.

Matrix Notation
An augmented matrix can be easily converted back to
a system .

Matrix Notation
a system .
For example, the augmented matrix
2 -3 8
1 4 -7
is the system x + 4y = -7{2x – 3y = 8

Operations of the equations correspond to operations
of rows in the matrices.
Matrix Notation
a system .
2 -3 8
1 4 -7

of rows in the matrices. Three such operations are
important. These are the elementary row operations:
Matrix Notation
a system .
2 -3 8
1 4 -7

Matrix Notation
I. Switch two rows.
a system .
2 -3 8
1 4 -7

Matrix Notation
I. Switch two rows. Switching row i with row j is
notated as Ri Rj.
a system .
2 -3 8
1 4 -7

Matrix Notation
1 4 -7
2 -3 8
For example,
R1 R2
notated as Ri Rj.
a system .
2 -3 8
1 4 -7

Matrix Notation
1 4 -7
2 -3 8
For example,
R1 R2 2 -3 8
1 4 -7
notated as Ri Rj.
a system .
2 -3 8
1 4 -7

Matrix Notation
II. Multiply a row by a constant k = 0.

Matrix Notation
II. Multiply a row by a constant k = 0. Multiply row i by
k is notated as k*Ri.

Matrix Notation
1 4 -7
2 -3 8
-3*R2
For example,

Matrix Notation
1 4 -7
2 -3 8
-3*R2 1 4 -7
-6 9 -24
For example,

III. Add the multiple of a row to another row.
Matrix Notation
1 4 -7
2 -3 8
-3*R2 1 4 -7
-6 9 -24
For example,

III. Add the multiple of a row to another row. k times
row i added to row j is notated as “k*Ri add  Rj”.
Matrix Notation
1 4 -7
2 -3 8
-3*R2 1 4 -7
-6 9 -24
For example,

1 4 -7
2 -3 8
For example,
-2*R1 add  R2
Matrix Notation
1 4 -7
2 -3 8
-3*R2 1 4 -7
-6 9 -24
For example,

1 4 -7
2 -3 8
For example,
-2*R1 add  R2
-2 -8 14
write a copy of -2*R1
Matrix Notation
1 4 -7
2 -3 8
-3*R2 1 4 -7
-6 9 -24
For example,

1 4 -7
2 -3 8
For example,
-2*R1 add  R2 1 4 -7
0 -11 22
-2 -8 14
Matrix Notation
1 4 -7
2 -3 8
-3*R2 1 4 -7
-6 9 -24
For example,

1 4 -7
2 -3 8
For example,
-2*R1 add  R2 1 4 -7
0 -11 22
-2 -8 14
Fact: Performing elementary row operations on a
matrix does not change the solution of the system.
Matrix Notation
1 4 -7
2 -3 8
-3*R2 1 4 -7
-6 9 -24
For example,

Matrix Notation
The elimination method may be carried out with
matrix notation.

Elimination Method in Matrix Notation:
Matrix Notation
matrix notation.

Matrix Notation
matrix notation.
1. Apply row operations to transform the matrix

* * * *
* * * *
* * * *
Row operations
Matrix Notation
matrix notation.
1. Apply row operations to transform the matrix

* * * *
* * * *
* * * *
Row operations
Matrix Notation
matrix notation.
1. Apply row operations to transform the matrix to the
upper diagonal form where all the entries below the
main diagonal (the lower left triangular region) are 0 .

* * * *
* * * *
* * * *
Row operations * * * *
* * *
* *
0
0 0
Matrix Notation
matrix notation.

* * * *
* * * *
* * * *
* * *
* *
0
0 0
2. Starting from the bottom row, get the answer
for one of the variables.
Matrix Notation
matrix notation.

* * * *
* * * *
* * * *
* * *
* *
0
0 0
for one of the variables. Then go up one row, use
the solution already obtained to get another answer of
another variable.
Matrix Notation
matrix notation.

* * * *
* * * *
* * * *
* * *
* *
0
0 0
for one of the variables. Then go up one row, use
the solution already obtained to get another answer of
another variable. Repeat the process, working from
the bottom row to the top row to extract all solutions.
Matrix Notation
matrix notation.

x + 4y = -7
2x – 3y = 8{ Eq. 1
Eq. 2
Example C: Solve using matrix notation:
Matrix Notation

1 4 -7
2 -3 8
x + 4y = -7
2x – 3y = 8{ Eq. 1
Eq. 2
Put the system into a matrix:
Matrix Notation

1 4 -7
2 -3 8
-2*R1 add  R2
x + 4y = -7
2x – 3y = 8{ Eq. 1
Eq. 2
Matrix Notation

1 4 -7
2 -3 8
-2*R1 add  R2
-2 -8 14
x + 4y = -7
2x – 3y = 8{ Eq. 1
Eq. 2
Matrix Notation

1 4 -7
2 -3 8
-2*R1 add  R2 1 4 -7
0 -11 22
-2 -8 14
x + 4y = -7
2x – 3y = 8{ Eq. 1
Eq. 2
Matrix Notation

1 4 -7
2 -3 8
-2*R1 add  R2 1 4 -7
0 -11 22
-2 -8 14
x + 4y = -7
2x – 3y = 8{ Eq. 1
Eq. 2
Starting from the bottom row: -11y = 22  y = -2
Matrix Notation

1 4 -7
2 -3 8
-2*R1 add  R2 1 4 -7
0 -11 22
-2 -8 14
x + 4y = -7
2x – 3y = 8{ Eq. 1
Eq. 2
Go up one row to R1 and set y = -2:
Matrix Notation

1 4 -7
2 -3 8
-2*R1 add  R2 1 4 -7
0 -11 22
-2 -8 14
x + 4y = -7
2x – 3y = 8{ Eq. 1
Eq. 2
x + 4y = -7
x + 4(-2) = -7
Matrix Notation

1 4 -7
2 -3 8
-2*R1 add  R2 1 4 -7
0 -11 22
-2 -8 14
x + 4y = -7
2x – 3y = 8{ Eq. 1
Eq. 2
x + 4y = -7
x + 4(-2) = -7  x = 1
Matrix Notation
Hence the solution is {x = 1, y = -2}
or as the ordered pair (1, -2).

Example D: Solve using matrix notation:
2x + 3y + 3z = 13
x + 2y + 2z = 8
3x + 2y + 3z = 13
{
E 1
E 2
E 3
Matrix Notation

2x + 3y + 3z = 13
x + 2y + 2z = 8
3x + 2y + 3z = 13
{
E 1
E 2
E 3
2 3 3 13
1 2 2 8
3 2 3 13
Matrix Notation

2x + 3y + 3z = 13
x + 2y + 2z = 8
3x + 2y + 3z = 13
{
E 1
E 2
E 3
2 3 3 13 R1 R2
1 2 2 8
3 2 3 13
2 3 3 13
1 2 2 8
3 2 3 13
Matrix Notation

2x + 3y + 3z = 13
x + 2y + 2z = 8
3x + 2y + 3z = 13
{
E 1
E 2
E 3
2 3 3 13 R1 R2
-2 -4 -4 -16Put the system into a matrix:
1 2 2 8
3 2 3 13
2 3 3 13
1 2 2 8
3 2 3 13
-2*R1 add R2
Matrix Notation

2x + 3y + 3z = 13
x + 2y + 2z = 8
3x + 2y + 3z = 13
{
E 1
E 2
E 3
2 3 3 13 R1 R2
1 2 2 8
3 2 3 13
2 3 3 13
1 2 2 8
3 2 3 13
-2*R1 add R2
0 -1 -1 -3
1 2 2 8
3 2 3 13
Matrix Notation

2x + 3y + 3z = 13
x + 2y + 2z = 8
3x + 2y + 3z = 13
{
E 1
E 2
E 3
2 3 3 13 R1 R2
1 2 2 8
3 2 3 13
2 3 3 13
1 2 2 8
3 2 3 13
-2*R1 add R2
0 -1 -1 -3
1 2 2 8
3 2 3 13
-3* R1 add R3
-3 -6 -6 -24
Matrix Notation

2x + 3y + 3z = 13
x + 2y + 2z = 8
3x + 2y + 3z = 13
{
E 1
E 2
E 3
2 3 3 13 R1 R2
1 2 2 8
3 2 3 13
2 3 3 13
1 2 2 8
3 2 3 13
-2*R1 add R2
0 -1 -1 -3
1 2 2 8
3 2 3 13
-3* R1 add R3
-3 -6 -6 -24
0 -1 -1 -3
1 2 2 8
0 -4 -3 -11
Matrix Notation

2x + 3y + 3z = 13
x + 2y + 2z = 8
3x + 2y + 3z = 13
{
E 1
E 2
E 3
2 3 3 13 R1 R2
1 2 2 8
3 2 3 13
2 3 3 13
1 2 2 8
3 2 3 13
-2*R1 add R2
0 -1 -1 -3
1 2 2 8
3 2 3 13
-3* R1 add R3
-3 -6 -6 -24
0 -1 -1 -3
1 2 2 8
0 -4 -3 -11
-4*R2 add R3
0 4 4 12
Matrix Notation

2x + 3y + 3z = 13
x + 2y + 2z = 8
3x + 2y + 3z = 13
{
E 1
E 2
E 3
2 3 3 13 R1 R2
1 2 2 8
3 2 3 13
2 3 3 13
1 2 2 8
3 2 3 13
-2*R1 add R2
0 -1 -1 -3
1 2 2 8
3 2 3 13
-3* R1 add R3
-3 -6 -6 -24
0 -1 -1 -3
1 2 2 8
0 -4 -3 -11
-4*R2 add R3
0 4 4 12
0 -1 -1 -3
1 2 2 8
0 0 1 1
Matrix Notation

0 -1 -1 -3
1 2 2 8
0 0 1 1
Matrix Notation
Extract the solution starting
from the bottom row.

0 -1 -1 -3
1 2 2 8
0 0 1 1
From R3, we get z = 1.
Matrix Notation

0 -1 -1 -3
1 2 2 8
0 0 1 1
Go up one row to R2, we get
-y – z = -3
-y – (1) = -3
Matrix Notation

0 -1 -1 -3
1 2 2 8
0 0 1 1
-y – z = -3
-y – (1) = -3
3 – 1 = y
2 = y
Matrix Notation

0 -1 -1 -3
1 2 2 8
0 0 1 1
-y – z = -3
-y – (1) = -3
3 – 1 = y
2 = y
Up to R1, we get x + 2y + 2z = 8
x + 2(2) + 2(1) = 8
Matrix Notation

0 -1 -1 -3
1 2 2 8
0 0 1 1
-y – z = -3
-y – (1) = -3
3 – 1 = y
2 = y
x + 2(2) + 2(1) = 8
x + 6 = 8
x = 2
Matrix Notation

0 -1 -1 -3
1 2 2 8
0 0 1 1
-y – z = -3
-y – (1) = -3
3 – 1 = y
2 = y
x + 2(2) + 2(1) = 8
x + 6 = 8
x = 2
So the solution is (2, 2, 1).
Matrix Notation

A soda costs $1, a hot dog costs $3 and an ice cream costs
$2. Find the cost of the following orders.
1) 2 sodas, 2 hotdogs and 1 ice cream. ______
2) 3 sodas, 3 hotdogs and 2 ice creams. ______
3) 4 sodas, 6 hotdogs and no ice creams. ______
4) 8 sodas, 6 hotdogs and 7 ice creams. ______
Now let a soda cost $x, a hot dog cost $y and an ice cream
cost $z. Find an algebraic expression for the cost of the
above orders.
5) _______________ 6) _____________
7) _____________ 8) _____________
9) If order #1 is $15, order #2 is $24 and order #3 is $32,
write the system of equations and solve.
10) If order #1 is $17, order #3 is $36 and order #4 is $67,
write the system of equations and solve.

1) $10 3) $22 5) 2x +2y + z 7) 4x + 6y
9) 2x + 2y + z = $15
3x + 3y + 2z = $24
4x + 6y = $32.
x = 2, y = 4, z = 3
11) x + 2y = 7
4y = 8
x = 3, y = 2
13) x + 2y + 4z = 3
y + z = 4
z = 6
x = -17, y = -2, z =6
13) 2 1 1
0 -2.5 7.5
17) -1 2 2 1
1 1 1 2
2 2 1 1
(Answers to the odd problems)

Solve Systems of Linear Equations Using Elimination Method (40

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