Download system of equations

Systems of Equations
and Inequalities
•Systems of Linear Equations:
Substitution and Elimination
Matrices
Determinants
•Systems of Non-linear Equations
•Systems of Inequalities
A system of equations is a collection of two or
more equations, each containing one or more
variables.
AAsolution
solutionofofaasystem
systemofofequations
equationsconsists
consistsofof
values
valuesfor
forthe
thevariables
variablesthat
thatreduce
reduceeach
eachequation
equation
ofofthe
thesystem
systemtotoaatrue
truestatement.
statement.
To
Tosolve
solveaasystem
systemofofequations
equationsmeans
meanstotofind
findall
all
solutions
solutionsofofthe
thesystem.
system.
When
Whenaasystem
systemofofequations
equationshas
hasatatleast
leastone
one
solution,
solution,ititisissaid
saidtotobe
beconsistent;
consistent;otherwise
otherwiseititisis
called
calledinconsistent.
inconsistent.
An equation in n variables is said to be
linear if it is equivalent to an equation of
the form
a1x1  a2 x2 an xn  b
where x1 , x2 , , xn are n distinct variables,
a1 , a2 , , an , b are constants, and at least
one of the a’s is not zero.
If each equation in a system of equations
is linear, then we have a system of linear
equations.
If the graph of the lines in a system of two linear
equations in two variables intersect, then the system
of equations has one solution, given by the point of
intersection. The system is consistent and the
equations are independent.
y
Solution
x
If the graph of the lines in a system of two linear
equations in two variables are parallel, then the
system of equations has no solution, because the
lines never intersect. The system is inconsistent.
y
x
If the graph of the lines in a system of two linear
equations in two variables are coincident, then the
system of equations has infinitely many solutions,
represented by the totality of points on the line. The
system is consistent and dependent.
y
x
Two Algebraic Methods for Solving
a System
1. Method of substitution
2. Method of elimination
 2 x  4 y  22
Solve: 
 x  3 y  15
(1)
(2)
STEP 1: Solve for x in (2)
 x  3 y  15
 x  3 y  15
x  3 y  15
STEP 2: Substitute for x in (1)
2 x  4 y  22
 


x  3 y  15
2 ( 3 y  15)  4 y  22
STEP 3: Solve for y
2( 3 y  15)  4 y  22
6 y  30  4 y  22
2 y  8
y  4
STEP 4: Substitute y = -4 into (2)
 x  3 y  15
 x  3( 4 )  15
 x  12  15
 x  3
x3
Solution: (3, -4)
STEP 5: Verify solution
(1)
 2 x  4 y  22

 x  3 y  15 (2)
(1): 2 ( 3)  4 ( 4 )  6  16  22
(2):  3  3( 4 )  3  12  15
2 x  4 y  22
 4 y  2 x  22
 2 x  22 1
11
y
 x
4
2
2
 x  3 y  15
3 y  x  15
x  15 1
y
 x 5
3
3
Rules for Obtaining an Equivalent
System of Equations
1. Interchange any two equations of the
system.
2. Multiply (or divide) each side of an
equation by the same nonzero constant.
3. Replace any equation in the system by
the sum (or difference) of that equation
and any other equation in the system.
 3x  2 y  7
Solve: 
 9 x  6 y  12
(1)
(2)
Multiply (1) by 3
 9 x  6 y  21 (1)

 9 x  6 y  12 (2)
Replace (2) by the sum of (1) and (2)
9 x  6 y  21 (1)

0 x  0 y  33 (2)
Equation (2) has no solution. System is
inconsistent.
5x  y  1
y  5 x  1
 15 x  3 y  3



y  5 x  1
 15 x  3  5 x  1  3
 15x  15x  3  3
 3  3
 x , y  x is any real number, y  5x  1
(1)
2 x  y  z  8

Solve:  2 x  3 y  z  4 (2)
4 x  2 y  3z  3 (3)

Replace (2) by the sum of (1) and (2)
 2x  y  z  8

2 y  2z  4

4 x  2 y  3z  3

Multiply (1) by -2
(1)
(2)
(3)
 4 x  2 y  2 z  16 (1)

2 y  2z  4
(2)

 4 x  2 y  3z  3 (3)

Replace (3) by the sum of (1) and (3)
 4 x  2 y  2 z  16

2 y  2z  4


4 y  5z  19

(1)
(2)
(3)
Multiply (2) by -2
 4 x  2 y  2 z  16

 4 y  4 z  8


4 y  5z  19

(1)
(2)
(3)
Replace (3) by the sum of (2) and (3)
 4 x  2 y  2 z  16

 4 y  4 z  8


 9 z  27

(1)
(2)
(3)
Multiply (3) by -1/9
 4 x  2 y  2 z  16

 4 y  4 z  8


z3

(1)
(2)
(3)
Back-substitute; let z = 3 in (2) and solve for y
 4 y  4 ( 3)  8
 4y  4
y  1
Back-substitute; let y = -1 and z = 3 in (1) and solve
for x.
 4 x  2 ( 1)  2 ( 3)  16
 4 x  2  6   16
 4 x  8
x2
Solution: x = 2, y = -1, z = 3
Matrix Methods for Linear Equations
A matrix is defined as a rectangular array
of numbers,
 a11 a12
a
a22
21


 
a
ai 2
i1


 
a
 m1 am2
 a1 j
 a2 j

 aij

 amj
 a1n 
 a2 n 

 

 ain

 

 amn 
 x  3 y  2 z  17

  2 x  y  3z  19
3x  2 y  z  12

Augmented Matrix:
2
17 
 1 3
 2
1  3  19 


1
12 
 3  2
Row Operations on an Augmented
Matrix
1. Interchange any two rows.
2. Replace a row by a nonzero multiple
of that row.
3. Replace a row by the sum of that row
and a constant multiple of some other row.
Echelon Form of an Augmented Matrix
1 a b 
0 1 c 


1 a b d 
0 1 c e 


0 0 1 f 
2
17 
 1 3
 2
1  3  19 


1
12 
 3  2
R2  2r1  r2
R3  3r1  r3
 1  3 2 17
0  5 1 15


 3  2 1 12
2
17
1  3
0  5
1
15


7  5  39
0
2
17
1  3
0  5
1
15


7  5  39
0
1
R2   r2
5
2
17
1  3
0
1 1
 3
5


7
 5  39
0
2
17
1  3
0
1 1
 3
5


7
 5  39
0
R3  7r2  r3
1  3
2
17 


1
0
1

 3

5
0


18
0

18


5
1  3
2
17 


1
0
1

 3

5
0


18
0

18


5
5
R3   r3
18
1  3
0
1 

0
0
17 
1
 3
5

1
5
2
2 17 
1  3
0
1 1
 3
5


0
1
5
0
 x  3 y  2 z  17

1
y  z  3

5

z5

 x  3 y  2 z  17

1
y  z  3

5

z5

Let z = 5 in (2):
1
y  5  3
5
y  1  3
y  2
(1)
(2)
(3)
Let y = -2, z = 5 in (1):
x  3  2   25  17
x  6  10  17
x 1
Solution: x = 1, y = -2, z = 5
Solve
 x  3 y  2 x  17

 2 x  y  3z  19
3x  2 y  z  12

using a graphing utility.
1
1 1
0 1  2
5

0
0 0
5
7 
5
0 
Dependent system: Infinitely many solutions
1
1 1
0 1  2
5

0
0 0
5
7 
5
0 
R1   r2  r1
18 
1 0
5
5


2
7
0
1

5
5
0 0
0
0 


7
7 18
x z
5
5
2
7
y z
5
5
z is any real number
3x  4 y  z  10

Solve:  x  y  z  3
3x  11y  5z  5

1
3
1  1
0 1  4
1 
7
7

0
 6
0 0
The system is inconsistent. No solution.
Cramer’s Method for Linear Equations
If a, b, c, and d are four real numbers, the
symbol
a b
D
c d
is called a 2 by 2 determinant. Its value is
the number ad - bc; that is,
a b
D
 ad  bc
c d
Theorem: Cramer’s Rule
The solution to the system of equations
ax  by  s

 cx  dy  t
is given by
s b
t d
x
D
where D  0
a s
c t
y
D
2 4
 2( 3)  ( 1)( 4)  6  4  2
D
1
3
22  4
Dx 
6
 15
3
Dx 6
x
 3
D 2
2
22
Dy 
 8
 1  15
Dy  8
y

 4
D
2
If D = 0, then the system has
infinitely many solutions or is
inconsistent and therefore has no
solution.
A 3 by 3 determinant is symbolized by
a11 a12
a21 a22
a31 a32
a13
a23
a33
in which a11 , a12 , are real numbers.
a11 a12
a21 a22
a31 a32
a22
 a11
a32
a13
a23
a33
a23
a21 a23
a21 a22
 a12
 a13
a33
a31 a33
a31 a32
When evaluating a determinant, you can expand
across any row or down any column you choose.
a11 a12
a21 a22
a31 a32
a13
a12 a13
a23  a21
a32 a33
a33
a11 a12
a21 a22
a31 a32
a13
a11 a13
a23  a22
a31 a33
a33
a11 x  a12 y  a13z  c 1

a21 x  a22 y  a23z  c 2
a x  a y  a z  c
 31
32
33
3
a11 a12
D  a21 a22
a31 a32
a13
a23
a33
c1
Dx  c 2
c3
a12
a22
a32
a13
a23
a33
a11 c 1
Dy  a21 c 2
a31 c 3
a13
a23
a33
a11 a12
Dz  a21 a22
a31 a32
c1
c2
c3
Cramer’s Rule for Three Equations
Containing Three Variables
Dy
Dx
Dz
x
, y
, z
D
D
D
provided D  0.
2 1
1
D 2
3
1
4
2 3
3 1
2
1 2 3
 36
2
 ( 1)
1
2 3
4 3
4 2
8 1
1
Dx   4
3
1  72
3 2 3
2
8
1
Dy   2  4
1  36
4 3 3
2 1
8
Dz   2
3  4  108
4
2 3
Dx  72
x

2
D  36
Dy
36
y

 1
D  36
Dz  108
z

3
D
 36
Solution: x = 2, y = -1, z = 3
Solving Non-linear Systems of Equations
Two algebraic methods:
1. Substitution
2. Elimination
The first example here is one quadratic equation with one
linear equation. The second example here is two quadratic
equations.
2 y 2  3xy  6 y  2 x  4  0 (1)
Solve: 
x  y  7  0 (2)

The hard way: Use quadratic formula on (1) to get,
2 y  ( 6  3x ) y  2 x  4  0
2
a = 2, b = 6 - 3x, c = 2x + 4
 (6  3x)  (6  3x)  4(2)(2 x  4)
y
2( 2)
2
3x  6  ( 6  3x )  8( 2 x  4)
y
4
2
The easy way: Solve linear equation (2) to get y(x).
x y70
y  x7
Compare y(x) formulas from hard and easy
ways on next slide.
3x  6  (6  3x)  8(2 x  4)
Y1 
4
2
3x  6  (6  3x)  8(2 x  4)
Y2 
4
2
Y3  x  7
 30  x  30;20  y  20
Continuing easy way solution to end:
2 y  3xy  6 y  2 x  4  0 (1)
Solve: 
x  y  7  0 (2)

2
x y70
y  x7
2( x  7)  3x ( x  7)  6( x  7)  2 x  4  0








 


y  x7
y  x7 y  x7
2


2 x 14x  49  3x  21x  6x  42  2x  4  0
2
2


2 x  14x  49  3x  21x  6x  42  2x  4  0
2
2
2 x  28x  98  3x  21x  6x  42  2 x  4  0
2
2
 x  x  60  0
2
 1  1  4( 1)(60)  1  241
x

2( 1)
2
2
 7.26 or 8.26
y  x7
 1  241
If x 
, then
2
 1  241
y
 7  14.26
2
 1  241
If x 
, then
2
 1  241
y
 7  1.26
2
Examples of two quadratic equations:
Multiply (1) by -1
  2 x 2  6 y 2  12 y  0
 2
2
2 x  3 y  4 y  4  0
Add (1) and (2)
 9 y  16 y  4  0
2
(1)
(2)
 9 y  16 y  4    9 y  2 y  2  0
2
y   or y  2 Try each y possibility in (1).
9
2
2
2 x  6 y  12 y  0 1
2
2
2
2


2 x  6    12    0
 9
 9
2 80
2x 
0
27
No Solution with y=-2/9.
2
2 x  6 y  12 y  0
2
2
1
2 x  6( 2)  12( 2)  0
2
2 x  24  24  0
2
2
2x  0
x=0
2
Solution: (0, 2)
Systems of Linear Inequalities & Linear Programming
Graph the linear inequality 2 x  y  8  0
2x  y  8  0
y  2x  8
0
10
5
2x - y -8 > 0
Test point: (1, 3)
2(1) - 3 - 8 = -9 < 0 Does not belong
to graph
Test point: (5, 1)
2(5) - 1 - 8 = 1 > 0 Belongs to graph
Shaded area but not line is graph.
y = 2x - 8
(1,3) not on graph
5
(5,1) is on graph
-10
 x y6
Graph the system: 
4 x  2 y  10
Graph x - y = 6
y=x-6
Graph 4x + 2y = 10
2y = -4x + 10
y = -2x + 5
(1)
(2)
10
y=x-6
5
0
5
10
10
y = -2x + 5
Test Point: (0, 0)
x-y>6
0-0>6
NO, shade side
opposite (0, 0).
4x + 2y < 10
4(0) + 2(0) < 10 Yes, shade side
containing (0, 0).
Shaded area including its linear boundaries is graph.
y = -2x + 5
10
y=x-6
5
0
10
5
10
A linear programming problem in two
variables x and y consists of maximizing
(or minimizing) a linear objective function
z = Ax + By, A and B are real numbers,
not both 0
subject to certain constraints expressible as
linear inequalities in x and y.
Maximize z  4 x  2 y
subject to 3x  y  16
2 x  3 y  20
x  0; y  0
 2 x  20
y
3
20
y = -3x + 16
10
(0, 20/3)
(4, 4)
(0,0)
0
5 (16/3, 0)
10
Corner Point
(0, 0)
z = 4x + 2y
4(0) + 2(0) = 0
(0, 20/3)
40/3
(16/3, 0)
21
(4, 4)
24
The maximum value of z is 24 and
occurs at the point (4, 4).
Theorem: Location of the Solution of a Linear
Programming Problem
If a linear programming problem has a solution, it
is located at a corner point of the graph of the
feasible points.
If a linear programming problem has multiple
solutions, at least one of them is located at a
corner point of the graph of the feasible points.
In either case, the corresponding value of the
objective function is unique.