Post on 22-May-2015
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Chapter 6: Systems of Linear Equations
Two Variable System
DefinitionsSystem – a conjunction of two or more equations
0111 cybxa0222 cybxa
GENERAL FORM:
Solution – an ordered pair of values that will satisfy all equations in the system
SOLUTION OF TWO VARIABLE SYSTEM
Graphical
Method
Assigning 2 points for each line
Converting each line into
y=mx + b
Computing the intercepts
of each line
SOLUTION OF TWO VARIABLE SYSTEM
Algebraic
Method
Cramer’s Rule
Elimination Substit
ution
Eliminating one variable and solving the other
Solving for the determinants
Isolating one variable
GRAPHICAL METHOD
Only intersecting equations have a unique or finite solution. They are called consistent equations. However, there are also systems that has infinite solutions or no solutions at all. Inconsistent equations – if two lines are parallel, then the system has no solution
-4 -3 -2 -1 1 2 3 4 5 x
-4
-3
-2
-1
1
2
3
4
5
y
O
Dependent equations – if two lines coincide, then the system has infinite solutions
-5 -4 -3 -2 -1 1 2 3 4 x
-5
-4
-3
-2
-1
1
2
3
4
y
O
Illustration
132
46
yx
yx
3
2,0 yx
For equation (1)
3
2,0
0,4 yx 0,4
For equation (2)
(1)
(2)
3
1,0 yx
3
1,0
0,2
1 yx
0,
2
1
-5 -4 -3 -2 -1 1 2 3 4 x
-5
-4
-3
-2
-1
1
2
3
4
y
O
3
1,0
0,
2
1
-5 -4 -3 -2 -1 1 2 3 4 x
-5
-4
-3
-2
-1
1
2
3
4
y
O
3
1,0
0,
2
1
-5 -4 -3 -2 -1 1 2 3 4 x
-5
-4
-3
-2
-1
1
2
3
4
y
O
0,4
3
2,0
3
1,0
0,
2
1
-5 -4 -3 -2 -1 1 2 3 4 x
-5
-4
-3
-2
-1
1
2
3
4
y
O
0,4
3
2,0
3
1,0
0,
2
1
-5 -4 -3 -2 -1 1 2 3 4 x
-5
-4
-3
-2
-1
1
2
3
4
y
O
0,4
3
2,0
3
1,0
0,
2
1
1,2
Algebraic Solution
Elimination Method (Gaussian Elimination)
1. Find equations equivalent to the two given equations such that the coefficients of one variable are additive inverse or the same
2. By addition or subtraction of the two equations, we eliminate one variable
3. Derive a linear equation in one variable and solve that variable4. Substitute the value of the solved variable in the other equation
and solve for the second variable
Illustration:
432
14
yx
yx Multiply (1) by 3(1)
(2)
432
3312
yx
yx Add (1) and (2)(1)
(2)
7014 x
2
1
14
7x
12
14
y Substitute the value of x to (1)
1y
1,2
1Thus, is the solution to the system
Substitution Method
432
14
yx
yxConsider the same system:
xy 41We solve for y in (1)
(1)
(2)
(3)
Then we substitute (3) to (2) 4)41(32 xx
2
1
14
7
714
41232
x
x
xx
2
141y Substitute the value of x to (3)
1y
1,2
1Thus, is the solution to the system
Cramer’s Rule
111 cybxa 222 cybxa
The determinant of the coefficients of the system is
22
11
ba
baD 1221 baba
Replace the coefficients of x in D by the constant terms to get
22
11)(bc
bcxD 1221 bcbc
Similarly, replace the coefficients of y in D by the constants
22
11)(ca
cayD 1221 caca
If D ≠ 0 then the solution (x, y) of the system is computed as
D
xDx
)(
D
yDy
)(
432
14
yx
yxUsing the same system:
32
14
D 14212
34
11)(
xD 743
2
1
14
7)(
D
xDx
42
14)( yD 14216
114
14)(
D
yDy
ExerciseUse Elimination, Substitution or Cramer’s Rule to solve for the solution of the system
723
1185
yx
yx
Using Cramer’s Rule:
23
85
D 342410
27
811)(
xD 345622
134
34)(
D
xDx
73
115)(
yD 683335
234
68)(
D
yDy
723
1185
yx
yx
Three Variable System
3333
2222
1111
dzcybxa
dzcybxa
dzcybxa
To determine the solution to a system of 3 variables, we can use either Elimination or Cramer’s Rule
Elimination: 3 Variable System
Equation 1
Equation 2
Equation 3
VAR 1
VAR 1
Equation 4
Equation 5
VAR 2
Value of VAR
3
Apply Back Substitution
2 VAR SYSTEM
Example3 Variable System
8233
122
632
zyx
zyx
zyx
8233
122
632
zyx
zyx
zyx
2424
632
zyx
zyx
(1)
(2)
(3)
4703 zx
8233
3636
zyx
zyx
5803 zx
583
473
zx
zx
583
473
zx
zx
1z 1
5)1(83
583
x
x
zx
1
6)1(32)1(
632
y
y
zyx 1,1,1
8233
122
632
zyx
zyx
zyx
233
212
321
D
3
2
1
3
1
2
18122 3869
238
211
326
)(
xD
8
1
6
3
1
2
93212 343624
13
3)(
D
xDx
(1)
(2)
(3)
283
212
361
)(
yD
3
2
1
8
1
6
48362 324169
13
3)(
D
yDy
1
63)1(2)1(
632
z
z
zyx 1,1,1
(1)