Sect 8.1 Systems of Linear Equations
A system of linear equations is where all the equations in a system are linear ( variables
raised to the first power). Equations are typically written in standard form, Ax + By = C.
Three possible graphs.
One solution
Independent & consistent
NO solution
Independent & inconsistent
Ways to solve systems.
Graphing, Substitution, and Elimination
Infinite solutions
Dependent & consistent
Sect 8.1 Systems of Linear Equations
Ways to solve systems.
Graphing, Substitution, and Elimination
Solve for x & y.
3x  2 y  11
x y 3
Substitution Method.
1. Choose an equation and get x or y Step 1
by itself.
x y 3
y  x3
Step 2
3x  2 y  11
3 x  2 x  3  11
2. Substitute step 1 equation into the
second equation.
Step 3 3 x  2 x  6  11
5 x  6  11
3. Solve for the remaining variable.
4 Substitute this answer into the step 1 Step 4 y  x  3
equation.
y  1 3
5x  5
x 1
y4
(1,4) Is the intersection point
and solution.
Sect 8.1 Systems of Linear Equations
Solve for x & y.
3x  4 y  1
2 x  3 y  12
Elimination Method.
1. Choose variable to cancel out.
Look for opposite signs.
2. Add the equations together to cancel.
3. Solve for the remaining variable.
4 Substitute this answer into either
equation in the step 1 equations.
3x  4 y  13 
 9 x  12 y  3
+
2 x  3 y  124 
 8x  12 y  48
23  3 y  12
Step 4
Step 1
The y-terms are opposite signs.
Multiply the first equation by 3
and the second equation by 4.
6  3 y  12
3y  6
y2
Step 2
17 x  0 y  51
17 x 51 Step 3

17 17
x3
(3, 2) is the intersection
point and solution.
Sect 8.1 Systems of Linear Equations
Solve for x & y. Elimination
2( 3 x  2 y  4)
 6x  4 y  6
+
6x – 4y = 8
0 = 14
False statement…No Solution
Solve for x & y.
(8 x  2 y  4 )/2
+
 4x  y  2
Elimination
You can also
divide by 2.
4x – y = -2
0=0
True statement…Infinite Solutions!
How to write the answers. The
solutions is the graph of the line.
Convert one of the equations into
y = mx + b form.
-4x + y = 2
+4x
+4x
y = 4x + 2
Solution is ( x, 4x +2 )
Sect 8.1 Systems of Linear Equations Elimination Method with 3 by 3 systems.
Step 1. Choose a TERMINATOR equation!
Solve for x, y, & z.
Look for coefficients of 1!!
Equation #3.
3x  9 y  6 z  3
T
2x  y  z  2
x  yy  zz 22
x yz  2
+
2x  y  z  2
3x + 2y = 4
Step 2. Pair this equation together with the other two
equations. Also decide which variable to eliminate, it
must be the same variable for both pairings!
-6( x  y  z  2 )
3x  9 y  6 z  3
Cancel the z terms!
-6x – 6y – 6z = -12
-3x + 3y
= -9 Divide by 3.
-x + y
= -3
Step 3. Bring the two new equations together
Step 4. Back substitute.
as a 2 by 2 system and solve.
y = -1
3( -x + y
= -3 )
3x + 2y = 4
+
-3x +3y = -9
5y = -5
-x – 1 = -3
x=2
2 1  z  2
z 1
T
x, y, z   2,1,1
Solve for x, y, & z.
2 x  y  3z  0
4x  2 y  6z  0
T x yz 2
x yz 2
+
2 x  y  3z  0
3x
Step 3.
+
– 2z = 2
Step 1. Choose a
TERMINATOR
Step 2. Pair T with the
other two equations.
Cancel y’s.
2( x  y  z  2 )
4x  2 y  6z  0
+ 2x – 2y + 2z = 4
6x
– 4z = 4
Divide by -2.
-3x + 2z = -2
3x = 2 + 2z
0=0
2222zz
x
33
x, y, z   _____, _____, z 
2( 3x  y  4 z  9 )
x  2y  z  4
+ 6 x  2 y  8 z  18
7 x  7 z  14
x  z  2
x  z 2
Step 4.
Infinite solutions.
Let z be independent! 2  2 z
 yz 2
Solve for x.
3
2  2 z  3 y  3z  6
 3 y  4  5z
 4  5z
3
x  2y  z  4
z  2  2y  z  4
2 y  2 z  6
y z  3
x, y, z   _____, _____, z 
Solve for x, y, & z.
2 x  y  z  2
x  2 y  z  9
T x  4y  z 1
x  4y  z 1
+
2 x  y  z  2
3 x  3 y  1
-3(
)
Step 1. Choose a
TERMINATOR
Step 2. Pair T with the
other two equations.
Cancel z’s.
x  4y  z 1
+
x  2 y  z  9
2 x  2 y  8 Divide by 2.
xx  yy  44
+
 3x  3 y  12
0  11
False statement…No Solution.
Sect 8.1 Systems of Linear Equations
2
Find the equation of the parabola y  ax  bx  c that passes through the points
(2, 4), (-1, 1), and (-2, 5).
We need to find a, b, and c in the equation. Three unknown variables means we
need to create three equations. Each point will generate an equation.
2,4 : 4  a22  b2  c  4a  2b  c  4
 1,1 : 1  a 12  b 1  c  a  b  c  1
 2,5 : 5  a 22  b 2  c  4a  2b  c  5
4a  2b  c  4
-1( a  b  c  1 ) T
4a  2b  c  5
Cancel the c’s.
 a  b  c  1
4a  2b  c  4
+
3a  3b  3
a  b 1
1.25  b  1
b  0.25
 a  b  c  1
1.25   0.25  c  1
4a  2b  c  5
y  1.25 x 2  0.25 x  0.5
c


0
.
5
3a  b  4
+
+
4a  5
5
a   1.25
4
Sect 8.2 Gauss-Jordan Method to solve Systems of Linear Equations
Convert a system into an augmented matrix. An augmented matrix is made up rows and
columns using only the coefficients on the variables and the constants. Answers for
3 ___
9 ___
6 ___
3
1 ___
0 ___
0 ___
2
___
___
3x  9 y  6 z  3
2 ___
1 ___
-1 ___
2
0 ___
1 ___
0 ___
-1
___
___
2x  y  z  2
1 ___
1 ___
1 ___
2
0 ___
0 ___
1 ___
1
___
___
x yz 2
Matrix row transformations.
Reduced row echelon form
1. Interchange any two rows.
2. Multiply or divide the elements of any row by a nonzero real number.
3. Replace any row in the matrix by the sum of the elements of that row and a multiple of
the elements of another row.
Steps for the Gauss-Jordan Method.
1. Obtain 1 as the first element in the 1st column.
2. Use the 1st row to transform the remaining entries in the 1st column to 0.
3. Repeat step 1 and 2 by obtaining the 1 as the 2nd element in the 2nd column and use the
2nd row to transform the remaining entries in the 2nd column to 0.
4. Repeat step 1 and 2 by obtaining the 1 as the 3rd element in the 3rd column and use the
3rd row to transform the remaining entries in the 3rd column to 0.
5. Repeat until the Coefficient matrix becomes the Identity Matrix.
Sect 8.2 Gauss-Jordan Method to solve Systems of Linear Equations
Solve for x and y.
3x  4 y  1
5 x  2 y  19
___
1 -___
10
-17
___
+85
___
5 -5 ___
2+50 19
___
-5R1 + R2  R2
R1
6-5
3
___
-8-2
-4
___
2-19
1
___
R2
5
___
2
___
19
___
2R1 – R2  R1
___
1 -___
10
-17
___
+20
___
1+0 -10
___+10 -17
___
___
1 ___
0
___
3
___
0
___
104
___
0
___
0
___
2
___
52
R2 / 52  R2
2 x  3 y  11
3 y  2 x  11
4 x  5 y  11
5 y  4 x  11
___
1
___
2
10R2 + R1  R1
R1
6-5 -4-4
-3 ___
2
___
-22-11
11
___
R2
5
___
11
___
4
___
___
1
( x, y ) = ( 3, 2 )
-2R1 – R2  R1
Y
X
___
1
-___
8
-33
___
___
1
-___
8
-33
___
___
1+0 -___
8 +8
-33
___+32
___
1
___
0
-___
1
___
5 -5 ___
4+40
11+165
___
___
0
176
___
___
0
___
4
___
0
___
1
___
4
-5R1 + R2  R2
___
44
R2 / 44 R2
___
1
8R2 + R1  R1
( x, y ) = ( 4, -1 )
=Y
= X
Sect 8.2 Gauss-Jordan Method to solve Systems of Linear Equations
Column 1, 1 first.
Column 1, 0’s 2nd . Column 2, 1 first.
Solve for x, y, & z.
(3x  9 y  6 z  3 )/3
R1
2x  y  z  2
x yz 2
___
2
___
1
___
1
___
3
___
2
___
1
___
1 +0 ___
3 -3 ___
2- 3 ___
1+0
R2
___
2-2 ___
1-6 ___
-1-4
___
2-2
___
0
___
-5
___
-5
___
0
___
0
R3
___
1-1 ___
1-3 ___
1-2
___
2 -1
___
0
___
-2
___
-1
___
1
___
0 +0 ___
-2 +2 ___
-1+2 ___
1+0
Always reduce rows or
equations when possible.
___
1
___
3
- 2R1 + R2  R2
R2 / -5  R2
___
1
___
0
- 3R2 + R1  R1
- 1R1 + R3  R3
Column 2, 0’s 2nd .
___
1
2R2 + R3  R3
Column 3, 1 is done, 0’s 2nd .
___
1 +0 ___
0 + 0 ___
-1 +1 ___
1+ 1
___
1
___
0
___
0
___
2
___
___
___
___
___
___
___
___
___
0 +0 ___
1 +0 ___
1 -1 ___
0-1
___
0
___
1
___
0
___
-1
___
___
___
___
___
___
___
___
___
0
___
0
___
0
___
1
___
1
___
___
___
___
___
___
___
___
___
0
___
1
___
1
R3 + R1  R1
-1R3 + R2  R2
( x, y, z ) = ( 2, -1, 1 )
This process must be done by hand! This process is also programmed into your
calculator. Find the Matrix button, for most above x -1 button. There are 3
categories, Names, Math, and Edit. We want Edit 1st. Pick a Matrix and hit Enter,
select the dimensions of your Matrix, and enter the data values by rows. DOUBLE
CHECK THE ENTRIES, ONE MISTAKE AND THE ANSWER IS WRONG! 2nd
Quit for the home screen and go back to the Matrix window for Math. Scroll up to
select rref( and hit Enter. Back to Matrix window and stay in the Names window.
Select your matrix and hit Enter. Close the parenthesis and Enter.
3x  9 y  6 z  3
2x  y  z  2
x yz 2
Sect 8.2 Gauss-Jordan Method to solve Systems of Linear Equations
Solve for x, y, & z.
2x  y  4z  7
3 x  4 y  z  13
x yz 4
R1
x
___
4
___
1
___
1
___
1
___
4
___
1 +0 ___
1 -1 ___
1 +2 ___
4 -1
-3
___
3 -3 ___
4 -3 1___
___-12
13
___
0
___
1
___
-2
___
1
___
0
___
2-2
7___
-8
___
0
___
-1
___
2
___
-1
___
0 +0 ___
-1+ 1 ___
2 -2 ___
- 1 +1
___
1
R3
___
1
___
1-2
1___
4___
-2
___
1
___
-2
___
1
- 3R1 + R2  R2
-1R2 + R1  R1
- 2R1 + R3  R3
R2 + R3  R3
y
z
___
1
___
0
___
3
___
3
x  3z  3
x  3  3z
___
0
___
1
___
-2
___
1
y  2z  1
y  1 2z
___
0
___
0
___
0
___
0
Infinite Solutions. z is the independent variable.
( x, y, z ) = ( 3 – 3z, 1 + 2z, z )
Sect 8.2 Gauss-Jordan Method to solve Systems of Linear Equations
Solve for x, y, & z.
5 x  3 y  9 z  19
6___
5 8-___
3 2-___
9 26-___
19
___
1
___
5
___
-7
___
7
___
1
___
5
___
-7
___
7
3 x  4 y  z  13
___
3
___
4
___
1
___
13
___
3 -3 ___
4 - 15___
1 +21 ___
13-21
___
0
___
-11 ___
22
___
-8
3 x  2 y  5 z  12
___
3
___
2
___
5
___
12
___
3 -3 ___
2 -15 ___
5 +21 ___
12 -21
___
0
___
-13 ___
26
___
-9
2R2 – R1  R1
- 3R1 + R2  R2
- 3R1 + R3  R3
___
1
___
5
___
-7
___
7
___
1
___
5
___
-7
___
7
___
0
___
1
___
-2
___
8/11
___
0
___
1
___
-2
___
8/11
___
0
___
-1
___
2
-9/13
___
___
0
___
0
___
0 5/143
___
R2 + R3  R3
NO SOLUTION
R2 /-11  R2
R3 / 13  R3