Transcript MTH55_Lec-13_sec_3-3a_3Var_Lin_Sys

Chabot Mathematics
§3.3a 3-Var
Linear Systems
Bruce Mayer, PE
Licensed Electrical & Mechanical Engineer
[email protected]
Chabot College Mathematics
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[email protected] • MTH55_Lec-13_sec_3-3a_3Var_Lin_Sys.ppt
Review § 3.2
MTH 55
 Any QUESTIONS About
• §3.2b → More System Applications
 Any QUESTIONS About HomeWork
• §3.2 → HW-09
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Systems of 3-Variables
 A linear equation in three variables is
an equation equivalent to form
Ax + By + Cz = D
• Where A, B, C, and D are real numbers
and A, B, and C are not all 0
 A solution of a system of three
equations in three variables is an
ORDERED TRIPLE (x1, y1, z1) that
makes all three equations true.
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Example  Ordered Triple Soln
 Determine whether
(2, −1, 3) is a solution
of the system
x  y  z  4,
2 x  2 y  z  3,
4 x  y  2 z  3.
 SOLUTION: In all three equations,
replace x with 2, y with −1, and z with 3.
x+y+z=4
2 + (–1) + 3 = 4
4=4

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2x – 2y – z = 3
– 4x + y + 2z = –3
2(2) – 2(–1) – 3 = 3 – 4(2) + (–1) + 2(3) = –3
3=3
–3 = –3


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Example  Ordered Triple Soln
 Determine whether
(2, −1, 3) is a solution
of the system
x  y  z  4,
2 x  2 y  z  3,
4 x  y  2 z  3.
 SOLUTION: Because (2, −1, 3) satisfies
all three equations in the system, this
triple IS a solution for the system.
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Solve 3Var Sys by Elimination
1. Write each equation in the standard
form of Ax + By+ Cz = D.
2. Eliminate one variable from one PAIR
of equations using the elimination
method.
3. If necessary, eliminate the same
variable from another PAIR of
equations to produce a system of
TWO varialbles in TWO Equations
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Solve 3Var Sys by Elimination
4. Steps 2 and 3 result in two equations
with the same two variables. Solve
these equations using the elimination
method.
5. To find the third variable, substitute
the values of the variables found in
step 4 into any of the three original
equations that contain the 3rd variable
6. Check the ordered triple in all
three of the original equations
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Example  Solve by Elimination
 Solve System
of Equations
x  y  z  6, (1)
x  2 y  z  2, (2)
x  y  3 z  8. (3)
 SOLUTION: select any two of the three
equations and work to get one equation
in two variables. Let’s add eqs (1) & (2)
x yz6
x  2y  z  2
2x + 3y
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=8
(1)
(2)
(4)
Adding to
eliminate z
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Example  Solve by Elimination
 Next, select a different pair of equations
and eliminate the same variable. Use
(2) & (3) to again eliminate z.
x  2y  z  2
x  y  3z  8
Multiplying
equation (2)
by 3
3x + 6y – 3z = 6
x – y + 3z = 8
4x + 5y
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= 14.
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(5)
Example  Solve by Elimination
 Now solve the resulting
system of eqns (4) & (5). 2x + 3y = 8
That will give us two of
4x + 5y = 14
the numbers in the
solution of the original system
 Multiply both
sides of eqn (4)
by −2 and then
add to eqn (5):
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–4x – 6y = –16,
4x + 5y = 14
–y = –2
y=2
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(4)
(5)
Example  Solve by Elimination
 Substituting into either equation (4) or
(5) we find that x = 1
 Now we have x = 1 and y = 2. To find
the value for z, we use any of the three
original equations and substitute to find
the third number z.
 Let’s use eqn (1)
x+y+z=6
and substitute our 1 + 2 + z = 6
two numbers in it:
z = 3.
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Example  Solve by Elimination
 At this point we have solved for All
Three variables in the System:
x=1
y=2
z=3
 Thus We have obtained the ordered
triple (1, 2, 3). It should be checked in
all three equations
 Finally the Solution Set: (1, 2, 3)
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Example  Solve by Elimination
 Solve System 3x  9 y  6 z  3 (1)
of Equations
2 x  y  z  2 (2)
x yz2
(3)
 SOLUTION: The equations are in
standard form and do not contain
decimals or fractions.
2 x  y  z  2 (2)
 Eliminate z from
(3)
x

y

z

2
eqns (2) & (3).
3x + 2y = 4
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(4)
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Example  Solve by Elimination
 Eliminate z from equations (1) and (2).
3x  9 y  6 z  3
2x  y  z  2
Multiplying
equation (2) by 6
3x  9 y  6 z  3
12 x  6 y  6 z  12
15x + 15y = 15
Adding
Eqns
 Eliminate x using above and eqn (4)
3x + 2y = 4
15x + 15y = 15
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Multiplying top
by 5
15x – 10y = 20
15x + 15y = 15
5y = 5
y = 1
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Adding
Eqns
Example  Solve by Elimination
 Using y = −1, find x from equation 4 by
substituting 3x + 2y = 4
3x + 2(1) = 4
 Substitute x = 2
and y = −1 to find z
 Thus The solution
is the ordered
triple (2, −1, 1)
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x=2
x+y+z=2
2–1+z=2
1+z=2
z=1
3x  9 y  6 z  3
2x  y  z  2
x yz2
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Example  Solve by Elimination
Solve the system.
2x + 3y – z = 5
(1)
–3x + 2y – 4z = 2
(2)
x – 4y + 3z = –9
7x + 5y
(3)
= 6
(4)
Step 1 Eliminate a variable from the sum of two equations. The choice of
the variable to eliminate is arbitrary. We will eliminate z.
6x + 9y – 3z = 15
x – 4y + 3z = –9
7x + 5y
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= 6
Multiply each side of (1) by 3.
(3)
Add.
(4)
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Example  Solve by Elimination
Solve the system.
2x + 3y – z = 5
(1)
–3x + 2y – 4z = 2
(2)
x – 4y + 3z = –9
7x + 5y
–5x – 10y
(3)
= 6
(4)
= –30
(5)
Step 2 Eliminate the same variable, z, from any other equations.
–9x + 6y – 12z = 6
Multiply each side of (2) by 3.
4x – 16y + 12z = –36
Multiply each side of (3) by 4.
–5x – 10y
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= –30
Add.
(5)
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Example  Solve by Elimination
Solve the system.
2x + 3y – z = 5
(1)
–3x + 2y – 4z = 2
(2)
x – 4y + 3z = –9
7x + 5y
–5x – 10y
(3)
= 6
(4)
= –30
(5)
Step 3 Eliminate a different variable and solve.
14x + 10y = 12
Multiply each side of (4) by 2.
–5x – 10y = –30
(5)
= –18
9x
Add.
(6)
x = –2
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Example  Solve by Elimination
Solve the system.
2x + 3y – z = 5
(1)
–3x + 2y – 4z = 2
(2)
x – 4y + 3z = –9
7x + 5y
–5x – 10y
(3)
= 6
(4)
= –30
(5)
Step 4 Find a second value by substituting –2 for x in (4) or (5).
7x + 5y = 6
(4)
7(–2) + 5y = 6
–14 + 5y = 6
Recall x = –2.
5y = 20
y = 4
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Example  Solve by Elimination
Solve the system.
2x + 3y – z = 5
(1)
–3x + 2y – 4z = 2
(2)
x – 4y + 3z = –9
7x + 5y
–5x – 10y
(3)
= 6
(4)
= –30
(5)
Step 5 Find a third value by substituting –2 for x and 4 for y into any of the
three original equations.
2x + 3y – z = 5
2(–2) + 3(4) – z = 5
8– z = 5
(1)
From Before x = –2 and y = 4
z = 3
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Example  Solve by Elimination
Solve the system.
2x + 3y – z = 5
(1)
–3x + 2y – 4z = 2
(2)
x – 4y + 3z = –9
(3)
Step 6 Check that the ordered triple (–2, 4, 3) is the solution of the system.
The ordered triple must satisfy all three original equations.
(1)
(2)
(3)
2x + 3y – z = 5
–3x + 2y – 4z = 2
x – 4y + 3z = –9
2(–2) + 3(4) – 3 = 5
–3(–2) + 2(4) – 4(3) = 2
(–2) – 4(4) + 3(3) = –9
–4 + 12 – 3 = 5
6 + 8 – 12 = 2
–2 – 16 + 9 = –9
5=5
2=2
–9 = –9



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Example  Missing Term(s)
Solve the system.
5x + 4y = 23
(1)
–5y – 2z = 3
(2)
–8x + 9z = –2
(3)
25x – 8z = 127
(4)
Since equation (3) is missing the variable y, a good way to begin the
solution is to eliminate y again by using equations (1) and (2).
25x + 20y
= 115
–20y – 8z = 12
25x
– 8z = 127
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Multiply each side of (1) by 5.
Multiply each side of (2) by 4.
Add.
(4)
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Example  Missing Term(s)
5x + 4y = 23
Solve the system.
(1)
–5y – 2z = 3
(2)
–8x + 9z = –2
(3)
25x – 8z = 127
(4)
Now use equations (3) and (4) to eliminate z and solve.
–64x + 72z = –16
Multiply each side of (3) by 8.
225x – 72z = 1143
Multiply each side of (4) by 9.
161x
Add.
= 1127
(5)
x = 7
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Bruce Mayer, PE
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Example  Missing Term(s)
Solve the system.
5x + 4y = 23
(1)
–5y – 2z = 3
(2)
–8x + 9z = –2
(3)
25x – 8z = 127
(4)
Substituting into equation (1) gives
Substituting into equation (2) gives
5x + 4y = 23
–5y – 2z = 3
5(7) + 4y = 23
–5(–3) – 2z = 3
35 + 4y = 23
15 – 2z = 3
4y = –12
–2z = –12
y = –3.
z = 6.
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Bruce Mayer, PE
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Example  Missing Term(s)
Solve the system.
5x + 4y = 23
(1)
–5y – 2z = 3
(2)
–8x + 9z = –2
(3)
 Thus, x = 7, y = −3, and z = 6. Check these
values in each of the original equations of the
system to verify that the solution set of the
system is the triple (7, −3, 6)
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Inconsistent System
 If, in the process of performing
elimination on a linear system, an
equation of the form 0 = a occurs,
where a ≠ 0, then the system has
no solution and is inconsistent.
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Bruce Mayer, PE
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Example  Inconsistent Sys
 Solve the System  x  y  2z  5 (1)

of Equations
 2x  y  z  7 (2)
 3x  2y  5z  20 (3)

 SOLUTION: To eliminate x from Eqn
(2), add −2 times Eqn (1) to Eqn (2)
2x  2y  4z  10
2x  y  z  7 (3)
3y  3z  3
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(4)
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Example  Inconsistent Sys
 Next add −3 times 3x  3y  6z  15
Eqn (1) to Eqn (3)
3x  2y  5z  20 (3)
to eliminate x
y  z  5 (5)
from Eqn (3)
 We now have the following system
(1)
 x  y  2z  5

(4)
 3y  3z  3

y z  5
(5)

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Example  Inconsistent Sys
 Now Multiply Eqn (4)
by 1/3 to obtain
y  z  1
 To eliminate y from Eqn (5), add −1
times Eqn (6) to Eqn (5)
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y  z  1
yz  5
(5)
06
(7)
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(6)
Example  Inconsistent Sys
 We now have the
system in simplified
elimination form:
 x  y  2z  5 (1)

 y  z  1 (5)

0  6 (7)

 This system is equivalent to the original
system. Since equation (7) is false
(a contradiction), we conclude that the
solution set of the system is Ø (the NULL
Set), and the system is INconsistent.
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Dependent Systems

If, in the process of performing an
elimination solution for a linear system
i. an equation of the form 0 = a (a ≠ 0)
does not occur, but
ii. an equation of the form 0 = 0 does
occur, then the system of equations
has infinitely many solutions and the
equations are dependent.
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Example  Dependent System
 Solve the System x  y  z  7

of Equations
 3x  2y  12z  11
 4 x  y  11z  18

(1)
(2)
(3)
 SOLUTION: Eliminate x from Eqn (2) by
adding −3 times Eqn (1) to Eqn (2
3x  3y  3z  21
3x  2y  12z  11
(2)
5y  15z  10
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Bruce Mayer, PE
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Example  Dependent System
 Eliminate x from 4x  4y  4z  28
Eqn (3) by adding 4x  y  11z  18
−4 times Eqn (1)
5y  15z  10
to Eqn (3)
 We now
7
x  y  z 
have the

5y

15z

10

equivalent system
 5y  15z  10

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(2)
(1)
(4)
(5)
Example  Dependent System
 To eliminate y
from Eq (5), add
−1 times Eqn (4)
to Eqn (5)
5y  15z  10
5y  15z  10
00
(5)
(6)
 We now
 x  y  z  7 (1)

have the
5y  15z  10 (4)

equivalent system

0  0 (6)

in Final Form
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Bruce Mayer, PE
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Example  Dependent System
 The equation 0 = 0 may be interpreted
as 0·z = 0, which is true for every value
of z. Solving eqn (4)
x  yz7
for y, we have y = 3z – 2.
x  3z  2   z  7
Substituting into eqn (1)
x  2z  5
and solving for x.
 Thus every triple (x, y, z) =
(2z + 5, 3z – 2, z) is a soln of the
system for each value of z.
E.g, for z = 1, the triple is (7, 1, 1).
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Geometry of 3Var Systems
 The graph of a linear equation in
three variables, such as
Ax + By + Cz = C (where A, B, and
C are not all zero), is a plane in
three-Dimensional (3D) space.
 Following are the possible situations
for a system of three linear
equations in three variables.
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Geometry of 3Var Systems
a) Three planes
intersect in a single
point. The system
has only one
solution
b) Three planes
intersect in one
line. The system
has infinitely many
solutions
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Geometry of 3Var Systems
c) Three planes
coincide with each
other. The system
has only one
solution.
d) There are three
parallel planes.
The system has
no solution.
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Geometry of 3Var Systems
e) Two parallel planes
are intersected by
a third plane. The
system has no
solution.
f) Three planes have
no point in
common. The
system has no
solution.
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Example  Quadratic Modeling
 The following table shows the higher-order
multiple birth rates in the USA since 1971.
At the right is a scatter plot of this data.
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Example  Quadratic Modeling
 We will SELECT three
arbitrary ordered pairs to
construct the model
 We Pick the Points
(1,29), (11,40) and (21,100)
and substitute the x & y
values from these ordered
pairs into the Quadratic
Function
2
y  ax  bx  c
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a0
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Example  Quadratic Modeling
 Selecting three representative ordered pairs,
we can write a system of three equations.
 Solving this 3-Variable System by Elimination:
 This produces our Model
y  0.2455x  1.856 x  30.7105
2
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Example  Quadratic Modeling
 Use to the Model to
Estimate the HiOrder Births in 1993
y  f x  
0.2455 x 2  1.856 x  30.71
 In 1993 x = 23
 Sub 23 into Model
y  0.245523  1.85623  30.71
y  0.2455529  1.85623  30.71
y  129.86  42.69  30.71
y  117.9
2
 Thus we estimate
that in 1993 the
Hi-Order BirthRate
was about 118
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WhiteBoard Work
 Problems From §3.3 Exercise Set
• 28

Quadratic
Function
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All Done for Today
A
CONSISTENT
System
z = -1.553x - 2.642y - 10.272 (darker green)
z = 1.416x - 1.92y - 10.979 (medium green)
z = -.761x - .236y - 7.184 (lighter green)
THE THREE PLANES SHARE ONE POINTBruce Mayer, PE
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Chabot Mathematics
Appendix
r  s  r  s r  s 
2
2
Bruce Mayer, PE
Licensed Electrical & Mechanical Engineer
[email protected]
–
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Bruce Mayer, PE
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Equivalent Systems

Operations That Produce Equivalent
Systems
1. Interchange the position of any two eqns
2. Multiply any eqn by a nonzero constant
3. Add a nonzero multiple of one eqn to
another
 A special type of Elimination called
Gaussian Elimination uses these steps to
solve multivariable systems
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Bruce Mayer, PE
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Solve 3Var Sys by Elimination
1. Rearrange the equations, if necessary,
to obtain an x-term with a nonzero
coefficient in the first equation. Then
multiply the first equation by the
reciprocal of the coefficient of the
x-term to get 1 as a leading coefficient.
2. By adding appropriate multiples of the
first equation, eliminate any x-terms
from the second and third equations
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Solve 3Var Sys by Elimination
2. (cont.) Multiply the resulting second
equation by the reciprocal of the
coefficient of the y-term to get 1 as the
leading coefficient.
3. If necessary by adding appropriate
multiple of the second equation from
Step 2, eliminate any y-term from the
third equation. Solve the resulting
equation for z.
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Solve 3Var Sys by Elimination
4. Back-substitute the values of z from
Steps 3 into one of the equations in
Step 3 that contain only y and z, and
solve for y.
5. Back-substitute the values of y and z
from Steps 3 and 4 in any equation
containing x, y, and z, and solve for x
6. Write the solution set (Soln Triple)
7. Check soln in the original equations
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Bruce Mayer, PE
[email protected] • MTH55_Lec-13_sec_3-3a_3Var_Lin_Sys.ppt