systems of equations. i. systems of linear equations four methods: 1. elimination by substitution 2....
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I. Systems of Linear Equations
Four Methods:
1. Elimination by Substitution
2. Elimination by Addition
3. Matrix Method
4. Cramer’s Rule
5. Geometric Method for a 2 by 2 system
Example (1)
1
,
2147
,),1()3(
)3(1024
,2)1(
lim:
1)2(25
,,
21474)25(23
,),2(
25),1(
lim:
:
)2(423
)1(52
:
y
getweEquationstheofanyinthatngSubstituti
xx
getweEquationtoEquationAdding
yx
byEquationgMultiplyin
additionbyinationEMethodSecond
y
getweEquationstheofanyinthatngSubstituti
xxxx
getweEquationinthatngSubstituti
xygetweEquationFrom
onsubstitutibyinationEdFirstMetho
Solution
yx
yx
systemfollowingtheSolve
Geometric method
Graph the lines represented by the equationsThe solution is the intersection of these lines
420-2-4
10
5
0
-5
x
y
x
y
)1,2(
Example (2)
solutionnohassystemThe
impossiblex
getweEquationtoEquationAdding
yx
byEquationgMultiplyin
additionbyinationEMethodSecond
solutionnohassystemThe
impossiblexx
getweEquationinthatngSubstituti
xygetweEquationFrom
onsubstitutibyinationEdFirstMetho
Solution
yx
yx
systemfollowingtheSolve
260
,),2()3(
)3(1024
,2)1(
lim:
4104)25(24
,),2(
25),1(
lim:
:
)2(424
)1(52
:
Geometric methodGraph the lines represented by the equations
These lines are parallel and do not intersect, and so no solution for the given system exists.
420-2-4
12.5
10
7.5
5
2.5
0
-2.5
-5
x
y
x
y
Example (3)
solutionsmanyekyhassystemThe
trueAlways
getweEquationtoEquationAdding
yx
byEquationgMultiplyin
additionbyinationEMethodSecond
solutionsmanyekyhassystemThe
trueAlwaysxx
getweEquationinthatngSubstituti
xygetweEquationFrom
onsubstitutibyinationEdFirstMetho
Solution
yx
yx
systemfollowingtheSolve
intinf
!00
,),2()3(
)3(1024
,2)1(
lim:
intinf
!101010)25(24
,),2(
25),1(
lim:
:
)2(1024
)1(52
:
Geometric method
Graph the lines represented by the two equations(they are equivalent equations) representing the same lines
420-2-4
12.5
10
7.5
5
2.5
0
-2.5x
y
x
y
3 by 3 Linear system
See the following examples:
1. Example (5) Page 144
2. Example (6) Page 145
3. Example (7) Page 146
Two Equations in Two Unknowns
00
0
221
111
222
121
2221
12110
22221
11211
0
:
yx
yx
yandx
Then
If
ca
caand
ac
ac
aa
aa
Let
Solution
cyaxa
cyaxa
systemthesolve
Example
221
421
21
21
0
4240215
82
2132431
38
2115635
32
:
135
832
:
00
0
0
yx
y
x
yandx
Then
If
haveWe
Solution
yx
yx
systemtheSolve
The case when Δ0 = 0The left side of the first equation is a k multiple of the left side of the second one, for
some real number k
The right side of the first equation is a k
multiple of the right side of the second one
→ There are finitely many solutions for the system
The right side of the first equation is not a
k multiple of the right side of the second
one.
→ There is no solution for the system
1664
832
)1(
yx
yx
Case
1564
832
)2(
yx
yx
Case
),0()
3
)0(28,0()2,1()
3
)1(28,1(:
)3
28,(
.3
28
)2(,2)1(
2
&2
0121264
32
1664
832
)1(
38
0
orExample
solutionaisr
rpairAny
xy
uknownstwowithequationonehaveweThus
EqgetwebyEqofsidebothgMultiplyin
firsttheofsiderigtthetimesisequationfirsttheofsidelrighttThe
firsttheofsideleftthetimesisequationfirsttheofsideleftThe
Solution
yx
yx
Case
systemtheforsolutionNo
impossibleiswhich
getweEqfromequationthisgSubtractin
yx
getwebyEqofsidebothgMultiplyin
firsttheofsiderigtthetimesnotisequationfirsttheofsiderightttheBut
firsttheofsideleftthetimesisequationfirsttheofsideleftThe
Solution
yx
yx
Case
10
),2(
1664
:,2)1(
2
2
0121264
32
1564
832
)2(
0
Three by Three Determinants
54156)1(4)5(3)3(2
)32(4)61(3)41(2
23
114
13
213
12
212
123
211
432
3231
222113
3331
232112
3332
232211
333231
232221
131211
Example
aa
aaa
aa
aaa
aa
aaa
aaa
aaa
aaa
A System of Three Linear Equations in Three Unknowns
000
0
33231
22221
11211
33331
23221
13111
33323
23222
13121
333231
232221
131211
0
3333231
2232221
1131211
,
0
:
zyx
z
y
x
zandyx
Then
If
caa
caa
caa
aca
aca
aca
aac
aac
aac
aaa
aaa
aaa
Let
Solution
czayaxa
czayaxa
czayaxa
systemtheSolve
Example
31
32
1
2,1
1
1
3)1(8)5(2)1(
712
611
821
2)5()1(8)1(
172
161
181
1)1()1(2)0(8
117
116
128
01)1()1(20
112
111
121
:,
72
6
82
:
0
0
zandyx
haveWe
Solution
zyx
zyx
zyx
systemtheSolve
x
z
y
x
Nonlinear SystemExample (1)
81)3(33
71)2(32
32
0)3)(2(
06
07)13(2
:),1(
13
,),2(
)2(013
)1(072
:
2
2
2
yx
oryx
xorx
xx
xx
xxx
getweEqinthatngSubstituti
xy
haveweEqFrom
Solution
yx
yxx
systemfollowingtheSolve
22
00
0)2(
02
042
04
:),1(
,),2(
:
)2(0
)1(04
:
)2(
2
2
22
22
xy
orxy
yy
yy
y
yyy
getweEqinthatngSubstituti
yx
haveweEqFrom
Solution
yx
yyx
systemfollowingtheSolve
Example
Geometric methodGraph the line and the quadratic function represented by the two
equations; y = - x2 + 2x + 7 and y = 3x + 1The points of their intersection are the solutions of the system
420-2-4
10
5
0
-5
-10
-15
x
y
x
y
)7,2(
)8,3(
Example (1)
22,
4)(
2)(
)(
secint
xiffexistsxthatnoticeweFirst
xxg
xxf
equationsofsystemfollowingtheSolve
functionsfollowingtheoftionertheFind
?7
)2,2())2(,2(secintint
42
7sin;2
}7,2{
2;0)7)(2(
2;0149
2;1682
2;)4(2
42
)()(
,intint
:
2
2
2
fromcomezerothedoesWhere
fistionerofpoThe
xx
equationthesatisfynotdoesxcex
x
xxx
xxx
xxxx
xxx
xx
xgxf
letwefunctionstwotheofctionserseofspothefindTo
Solution
Answer: Because, when we squared √(x+2), we introduced
the other function whose square is also equal x+2
Which function is this?
At which point does it intersect the line y = x+2