Transcript Section 4.1 Systems With Two Variables

Section 4.1
Systems With Two
Variables
OBJECTIVES
Find solution of two linear
equations using:
A
The graphical method.
OBJECTIVES
Find solution of two linear
equations using:
B
The substitution
method.
OBJECTIVES
Find solution of two linear
equations using:
C
The elimination method.
OBJECTIVES
Find solution of two linear
equations using:
D
Solve applications
involving systems of
equations.
Solving Two Equations in Two
Unknowns by Elimination
1. Clear any fractions or
decimals.
Solving Two Equations in Two
Unknowns by Elimination
2. Multiply both sides of
the equations (as needed)
by numbers that make the
coefficients of one of the
variables additive inverses.
Solving Two Equations in Two
Unknowns by Elimination
3. Add the two equations.
4. Solve for the remaining
variable.
Solving Two Equations in Two
Unknowns by Elimination
5. Substitute this solution into
one of the equations and
solve for second variable.
6. Check the solution.
Chapter 4
Systems With Two Variables
Section 4.1A
Practice Test
Exercise #2
Use the graphical method to solve the system.
y – 3x = – 3
3y = 9x + 9
1. If x = 0, y – 0 = – 3,
y = –3
Plot  0, – 3 
If y = 0 – 3x = – 3,
x =1
Plot  1,0 
Use the graphical method to solve the system.
y – 3x = – 3
3y = 9x + 9
2. If x = 0, 3y = 0 + 9,
y =3
Plot  0, 3 
If y = 0 = 9x  9,
–9x = 9,
x = –1
Plot  –1,0 
Use the graphical method to solve the system.
y – 3x = – 3
3y = 9x + 9
There is no solution.
System is inconsistent.
Lines are parallel.
Use the graphical method to solve the system.
y – 3x = – 3
3y = 9x + 9
5
x
-5
5
y -5
Chapter 4
Systems With Two Variables
Section 4.1A
Practice Test
Exercise #3
Use the graphical method to solve the system.
y :  0, 3  x :  2,0 
3x + 2y = 6
x = 2 – 2 y y :  0, 3  x :  2,0 
3
y
x
5
-5
Use the graphical method to solve the system.
Infinitely many solutions
3x + 2y = 6
x = 2– 2y
3
y
x
5
-5
Chapter 4
Systems With Two Variables
Section 4.1B
Practice Test
Exercise #5
Use the substitution method to solve the system.
2x – 3y = 6
4x = 6y + 7


6y + 7
x =
4
1 6y + 7
2
– 3y = 6
4
2
6y + 7
2
– 2  3y  = 2  6 
2
6y + 7– 6y = 12


7  12
Use the substitution method to solve the system.
2x – 3y = 6
4x = 6y + 7

NO solution

1 6y + 7
2
– 3y = 6
4
2
6y + 7
2
– 2  3y  = 2  6 
2
6y + 7– 6y = 12


7  12
Chapter 4
Systems With Two Variables
Section 4.1C
Practice Test
Exercise #9
Solve the system.
x y
+ = 2
2 3
x y
+
= 1
4 8
Multiply by 6.
Multiply by 8.
3x + 2y = 12
2x + y = 8
– 4x – 2y = – 16
Multiply by –2.
Solve the system.
3x + 2y = 12
– 4x – 2y = – 16
Solve the system.
3x + 2y = 12
– 4x – 2y = – 16
– x
= –4
x = 4
Substitute x = 4 in
x y
+
= 1
4 8
4 y
+
= 1
4 8
Solve the system.
4 y
+
= 1
4 8
1+
y
= 1
8
y
= 0
8
y = 0
Solve the system.
x y
+ = 2
2 3
x y
+
= 1
4 8
x = 4, y = 0
Solution is (4, 0).
Section 4.2
Systems with Three
Variables
OBJECTIVES
A
Solve a system of three
equations and three
unknowns by the
elimination method.
OBJECTIVES
B
Determine if a system of
three equations in three
unknowns is consistent,
inconsistent, or
dependent.
OBJECTIVES
C
Solve applications
involving systems of
three equations.
PROCEDURE FOR SOLVING
Three Equations in Three
Unknowns by Elimination
1. Select a pair of equations
and eliminate one variable
from this pair.
PROCEDURE FOR SOLVING
Three Equations in Three
Unknowns by Elimination
2. Select a different pair of
equations and eliminate the
same variable as in step 1.
PROCEDURE FOR SOLVING
Three Equations in Three
Unknowns by Elimination
3. Solve the pair of equations
resulting from step 1 and 2.
PROCEDURE FOR SOLVING
Three Equations in Three
Unknowns by Elimination
4. Substitute the values found
in the simplest of original
equations. Solve for third
variable.
PROCEDURE FOR SOLVING
Three Equations in Three
Unknowns by Elimination
5. Check by substituting the
values in each of the
original equations.
Solving Three Equations in
Three Unknowns by
Elimination
1. The system is consistent &
independent; it has one
solution consisting of an
ordered triple (x, y, z).
Solving Three Equations in
Three Unknowns by
Elimination
2. The system is inconsistent.
It has no solution.
Solving Three Equations in
Three Unknowns by
Elimination
3. The system is consistent
and dependent. It has
infinitely many solutions.
Chapter 4
Systems With Two Variables
Section 4.2A
Practice Test
Exercise #11
Solve the system.
x + y + z = 2 (1)
x=1
2x + y – z = 5 (2)
x + y – z = 4 (3)
(1) + (2)
3x + 2y = 7 (4)
(1) + (3)
2x + 2y = 6 (5)
(4) – (5)
x
= 1
Solve the system.
x + y + z = 2 (1)
2x + y – z = 5 (2)
x + y – z = 4 (3)
Substitute x = 1 in (4)
3(1) + 2y = 7
3 + 2y = 7
2y = 4
y = 2
x=1
y= 2
Solve the system.
x + y + z = 2 (1)
2x + y – z = 5 (2)
x + y – z = 4 (3)
x=1
y= 2
z = –1
Substitute x = 1, y = 2 in (1)
1+ 2 + z = 2
3+ z = 2
z = –1
Solution (1, 2, – 1)
Section 4.3
Coin, Distance-RateTime, Investment and
Geometry Problems
OBJECTIVES
A
Solve coin problems
with two or more
unknowns.
OBJECTIVES
B
Solve general problems
with two or more
unknowns.
OBJECTIVES
C
Solve rate, time and
distance problems with
two or more unknowns.
OBJECTIVES
D
Solve investment
problems with two or
more unknowns.
OBJECTIVES
E
Solve geometry
problems with two or
more unknowns.
Chapter 4
Systems With Two Variables
Section 4.3C
Practice Test
Exercise #16
A motorboat can go 10 mi downstream on a river in 20
min. It takes 30 min for this boat to go back upstream
the same 10 mi. Find the speed of the current.
Distance
Time =
Rate
Let r = rate of boat
c = rate of current
r + c = rate downstream
r – c = rate upstream
10 = distance each way
A motorboat can go 10 mi downstream on a river in 20
min. It takes 30 min for this boat to go back upstream
the same 10 mi. Find the speed of the current.
Distance
Time =
Rate
20 1
= hr
20 min. =
60 3
30 1
= hr
30 min. =
60 2
A motorboat can go 10 mi downstream on a river in 20
min. It takes 30 min for this boat to go back upstream
the same 10 mi. Find the speed of the current.
10
1
=
(1)
r +c 3
10
1
=
r –c 2
(2)
30 = r + c (3)
20 = r – c (4)
50 = 2r
A motorboat can go 10 mi downstream on a river in 20
min. It takes 30 min for this boat to go back upstream
the same 10 mi. Find the speed of the current.
50 = 2r
25 = r
Substitute 25 for r in (3)
30 = 25 + c
5 =c
The rate of the current is 5 mi/hr.
Section 4.4
Matrices
OBJECTIVES
A
Perform elementary
operations on systems
of equations.
OBJECTIVES
B
Solve systems of linear
equations using
matrices.
OBJECTIVES
C
Solve applications using
matrices.
DEFINITION
Matrix
A rectangular array of
numbers enclosed in
brackets.
PROCEDURE
Elementary Operations on
Systems of Equations
1. The order of equations may
be changed. This clearly
cannot affect the solutions.
PROCEDURE
Elementary Operations on
Systems of Equations
2. Any of the equations may
be multiplied by any
nonzero real number.
PROCEDURE
Elementary Operations on
Systems of Equations
3. Any equation of a system
may be replaced by the sum
of itself and any other
equation of the system.
PROCEDURE
Elementary Row Operations
on Matrices
1. Change the order of the
rows.
PROCEDURE
Elementary Row Operations
on Matrices
2. Multiply all elements of a
row by any nonzero number.
PROCEDURE
Elementary Row Operations
on Matrices
3. Replace any row by the
element-by-element sum of
itself and any other row.
Chapter 4
Systems With Two Variables
Section 4.4A
Practice Test
Exercise #18
Use matrices to solve the system.
x +y +z = –2
2x + y – z = 3
–x +y +z =0
R1
R2
R3
1 1 1 –2
2 1 –1 3
–1 1 1 0
Use matrices to solve the system.
x +y +z = –2
2x + y – z = 3
–x +y +z =0
R1
R2
R3
1 1 1 –2
2 1 –1 3
–1 1 1 0
Use matrices to solve the system.
x +y +z = –2
2x + y – z = 3
–x +y +z =0
R1
R2
R3
1 1 1 –2
2 1 –1 3
–1 1 1 0
Use matrices to solve the system.
x +y +z = –2
2x + y – z = 3
–x +y +z =0
R1
R2
R3
1 1 1 –2
2 1 –1 3
–1 1 1 0
Use matrices to solve the system.
x +y +z = –2
2x + y – z = 3
–x +y +z =0
1 1 1 –2
2 1 –1 3
–1 1 1 0
1 1 1 –2
= 0 –1 – 3 7
–1 1 1 0
– 2R 1 + R 2  R 2
R1 + R 3  R 3
Use matrices to solve the system.
x +y +z = –2
2x + y – z = 3
–x +y +z =0
1 1 1 –2
= 0 –1 – 3 7
0 2 2 –2
1 1 1 –2
= 0 –1 – 3 7
0 0 – 4 12
2R 2 + R 3  R 3
Use matrices to solve the system.
x + y + z = – 2 x = –1
 –1, 2, – 3 
2x + y – z = 3
y =2
z = –3
–x +y +z =0
1 1 1 –2 x +y+z= –2
–y – 3z = 7
= 0 –1 – 3 7
0 0 – 4 12
– 4z = 12
z = –3
Section 4.5
Determinants and
Cramer’s Rule
OBJECTIVES
A
Evaluate a 2 by 2
determinant.
OBJECTIVES
B
Use Cramer’s rule to
solve a system of two
equations in two
unknowns.
OBJECTIVES
C
Use minors to evaluate 3
by 3 determinants.
OBJECTIVES
D
Use Cramer’s rule to
solve a system of three
equations.
Determinant
The determinant of the matrix
a1 b1 is denoted by a1 b1 and
a2 b2
a2 b2
+
–
a
b
is definded as 1 1 = a 1b 2 -a 2b1
a1 b2
Cramer’s Rule - 2 Equations
The system a1x + b1y = d1
a2 x + b2 y = d2
1. Has the unique solution
Dy
Dx
x=
and y =
, where D  0
D
D
Cramer’s Rule - 2 Equations
The system a1x + b1y = d1
a2 x + b2 y = d2
d
b
and Dx = 1 1
d2 b2
Cramer’s Rule - 2 Equations
The system a1x + b1y = d1
a2 x + b2 y = d2
a
d
and Dy = 1 1
a2 d 2
Cramer’s Rule - 2 Equations
The system a1x + b1y = d1
a2 x + b2 y = d2
a
b
and D = 1 1
a2 b2
Cramer’s Rule - 2 Equations
The system a1x + b1y = d1
a2 x + b2 y = d2
2. Is inconsistent and has no
solution if D = 0 and any one of
Dx or Dy is different from 0.
Cramer’s Rule - 2 Equations
The system a1x + b1y = d1
a2 x + b2 y = d2
3. Is inconsistent and has no
solution if D = 0 and any one of
Dx or Dy is different from 0.
DEFINITION
Minor
The determinant that remains
after deleting the row and
column in which the element
appears.
Minor
In the determinant
a1 b1 c1
a2 b2 c2
a3 b3 c3
The minor of a1 is
b2 c2
b3 c3
DEFINITION
Sign Array
For a 3  3 determinant :
+–+
–+–
+–+
Cramer’s Rule - 3 Equations
The System
a1 x + b1y + c1 z = d1
a2 x + b2 y + c2 z = d2
a3 x + b3 y + c3 z = d3
Cramer’s Rule - 3 Equations
1. Has the unique solution
Dy
Dx
Dz
x= ,y= ,z=
D
D
D
where D  0
Cramer’s Rule - 3 Equations
and
d1 b1 c1
a1 b1 c1
D = a2 b2 c2 , Dx = d2 b2 c2
a3 b3 c3
d3 b3 c3
Cramer’s Rule - 3 Equations
a1 d1 c1
a1 b1 d1
Dy = a2 d2 c2 , Dz = a2 b2 d2
a3 d3 c3
a3 b3 d3
D 0
Cramer’s Rule - 3 Equations
2. Is inconsistent and has no
solution if:
D = 0 and any one of Dx , Dy
or Dz is different from 0.
Cramer’s Rule - 3 Equations
3. Has no unique solution if:
D = 0 and Dx = Dy = Dz = 0.
Chapter 4
Systems With Two Variables
Section 4.5A
Practice Test
Exercise #19a
Evaluate.
4 –2
=  4  – 6  –  5  – 2 
5 –6
= – 24 –  – 10 
= – 24 + 10
= – 14
Chapter 4
Systems With Two Variables
Section 4.5A
Practice Test
Exercise #19b
Evaluate.
b. 4 – 2 =  4  – 3  –  6  – 2 
6 –3
= – 12 –  – 12 
= – 12 + 12
=0
Chapter 4
Systems With Two Variables
Section 4.5B
Practice Test
Exercise #20
Solve by Cramer’s rule.
2x – 3y = 5 D = 10
6x – 4y = 7
D = 2 – 3 =  2  – 4  –  6  – 3 
6 –4
= – 8 –  – 18 
= – 8 + 18
D = 10
Solve by Cramer’s rule.
2x – 3y = 5 D = 10
6x – 4y = 7 D x = 1
D x = 5 – 3 =  5  – 4  –  7  – 3 
7 –4
= – 20 –  – 21
= – 20 + 21
Dx = 1
Solve by Cramer’s rule.
2x – 3y = 5 D = 10 Dy = – 16
6x – 4y = 7 D x = 1
Dy = 2 5
6 7
=  2  7  –  6  5 
= 14 – 30
Dy = – 16
Solve by Cramer’s rule.
2x – 3y = 5 D = 10 Dy = – 16
6x – 4y = 7 D x = 1
x=
y=
Dx
D
Dy
D
1
=
10
– 16
–8
=
=
10
5
1
8

Solution:  , – 
5
 10
Chapter 4
Systems With Two Variables
Section 4.5C
Practice Test
Exercise #21
Evaluate.
–2 4 –2
1 0 3
4 5 2
=  – 2
0 3
1 3
1 0
–4
+  – 2
5 2
4 2
4 5
Evaluate.
–2 4 –2
1 0 3
4 5 2
=  – 2
0 3
1 3
1 0
–4
+  – 2
5 2
4 2
4 5
=  – 2  0 – 15  – 4  2 – 12  – 2  5 – 0 
=  – 2  – 15  – 4  – 10  – 2  5 
= 30 + 40 – 10 = 60
Chapter 4
Systems With Two Variables
Section 4.5D
Practice Test
Exercise #23
Solve by Cramer’s rule.
x + 2y + z = 6
x +y –z =7
2x – y + 2z = – 3
D=
1 2 1
1 1 –1
2 –1 2
= 1 1 –1 –2 1 –1 +1 1 1
–1 2
2 2
2 –1
= 1 2 – 1 – 2  2 + 2  + 1 – 1 – 2 
Solve by Cramer’s rule.
x + 2y + z = 6
x +y –z =7
2x – y + 2z = – 3
D=
1 2 1
1 1 –1
2 –1 2
= 1 1 – 2  4  + 1 – 3 
=1–8–3
= – 10
Solve by Cramer’s rule.
x + 2y + z = 6
x +y –z =7
2x – y + 2z = – 3
6 2 1
Dx = 7 1 – 1
–3 –1 2
= 6 1 –1 –2 7 –1 +1 7 1
–1 2
–3 2
–3 –1
= 6  2 – 1 – 2  14 – 3  + 1 – 7 – 3 
Solve by Cramer’s rule.
x + 2y + z = 6
x +y –z =7
2x – y + 2z = – 3
6 2 1
Dx = 7 1 – 1
–3 –1 2
= 6  2 – 1 – 2  14 – 3  + 1 – 7 – 3 
= 6  1 – 2  11 + 1 – 4 
= 6 – 22 – 4 = – 20
Solve by Cramer’s rule.
x + 2y + z = 6
x +y –z =7
2x – y + 2z = – 3
1 6 1
Dy = 1 7 – 1
2 –3 2
= 1 7 –1 –6 1 –1 +1 1 7
–3 2
2 2
2 –3
= 1 14 – 3  – 6  2 + 2  – 1 – 3 – 14 
Solve by Cramer’s rule.
x + 2y + z = 6
x +y –z =7
2x – y + 2z = – 3
1 6 1
Dy = 1 7 – 1
2 –3 2
= 1 14 – 3  – 6  2 + 2  – 1 – 3 – 14 
= 1 11 – 6  4  + 1 – 17 
= 11 – 24 – 17 = – 30
Solve by Cramer’s rule.
x + 2y + z = 6
x +y –z =7
2x – y + 2z = – 3
1 2 6
Dz = 1 1 7
2 –1 –3
= 1 1 7 –2 1 7 +6 1 1
–1 –3
2 –3
2 –1
= 1 – 3 + 7  – 2  – 3 – 14  + 6  – 1– 2 
Solve by Cramer’s rule.
x + 2y + z = 6
x +y –z =7
2x – y + 2z = – 3
1 2 6
Dz = 1 1 7
2 –1 –3
= 1 – 3 + 7  – 2  – 3 – 14  + 6  – 1– 2 
= 1 4  – 2  – 17  + 6  – 3 
= 4 + 34 – 18 = 20
Solve by Cramer’s rule.
x + 2y + z = 6
x +y –z =7
2x – y + 2z = – 3
Dx
– 20
=
= 2
x=
– 10
D
Dy
– 30
=
= 3
y=
– 10
D
Dz
20
=
z=
= –2
– 10
D
Solve by Cramer’s rule.
x + 2y + z = 6
x +y –z =7
2x – y + 2z = – 3
Solution:  2, 3, – 2 
System is Consistent
Section 4.6
Systems of Linear
Inequalities
OBJECTIVES
A
Graphing systems of
two linear inequalities.
OBJECTIVES
B
Graphing systems of
inequalities.
PROCEDURE
Graphing Inequalities
1. Sketch the line corresponding
to each inequality using
dashed lines for < or > and
solid lines for  or  .
PROCEDURE
Graphing Inequalities
2. Use a test point to shade
the half-plane that is the
graph of each linear
inequality.
PROCEDURE
Graphing Inequalities
3. Graph is the intersection of
the half-planes, that is, the
region consisting of the
points satisfying all
inequalities.
Chapter 4
Systems With Two Variables
Section 4.6B
Practice Test
Exercise #25