Download Section 4.2

Chapter 4
Systems of
Linear
Equations and
Inequalities
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CHAPTER
4
Systems of Equations and
Inequalities
4.1
4.2
4.3
Solving Systems of Linear Equations in
Two Variables
Solving Systems of Linear Equations in
Three Variables
Solving Applications Using Systems of
Equations
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4.2
Solving Systems of Linear
Equations in Three Variables
1. Determine if an ordered triple is a solution for
a system of equations.
2. Understand the types of solution sets for
systems of three equations.
3. Solve a system of three linear equations using
the elimination method.
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Example
Determine whether (2, –1, 3) is a solution of the
x  y  z  4,
system
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
2x – 2y – z = 3
– 4x + y + 2z = –3
2(2) – 2(–1) – 3 = 3 – 4(2) + (–1) + 2(3) = –3
3=3
–3 = –3
Because (2, 1, 3) satisfies all three equations in
the system,
for the system. TRUE
TRUE it is a solution
TRUE
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Types of Solution Sets
A Single Solution:
If the planes intersect at a
single point, that ordered
triple is the solution to the
system.
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Types of Solution Sets
Infinite Number of Solutions:
If the three planes intersect
along a line, the system has an
infinite number of solutions,
which are the coordinates of
any point along that line.
Infinite Number of Solutions:
If all three graphs are the same
plane, the system has an
infinite number of solutions,
which are the coordinates of
any point in the plane.
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Types of Solution Sets
No Solution:
If all of the planes are
parallel, the system has
no solution.
No Solution:
Pairs of planes also can
intersect, as shown.
However, because all three
planes do not have a
common intersection, the
system has no solution.
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Example
Solve the system using elimination.
x  y  z  6,
x  2 y  z  2,
x  y  3z  8.
(1)
(2)
(3)
Solution
We select any two of the three equations and
work to get one equation in two variables. Let’s
add equations (1) and (2):
x yz6
(1)
(2)
x  2y  z  2
Adding to
(4)
2x + 3y
=8
eliminate z
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Next, we select a different pair of equations and
eliminate the same variable. Let’s use (2) and (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
= 14.
(5)
Now we solve the resulting system of equations
(4) and (5). That will give us two of the numbers
in the solution of the original system,
(4)
2x + 3y = 8
(5)
4x + 5y = 14
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We multiply both sides of equation (4) by –2 and
then add to equation (5):
–4x – 6y = –16,
4x + 5y = 14
–y = –2
y=2
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.
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Let’s use equation (1) and substitute our two
numbers in it:
x+y+z=6
1+2+z=6
z = 3.
We have obtained the ordered triple (1, 2, 3). It
should check in all three equations.
The solution is (1, 2, 3).
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Solving Systems of Three Linear Equations Using Elimination
1. Write each equation in the form 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.
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 third variable.
6. Check the ordered triple in all three of the original equations.
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Example
Solve the system using elimination.
3x  9 y  6 z  3 (1)
2 x  y  z  2 (2)
x  y  z  2 (3)
Solution
The equations are in standard form.
Eliminate z from equations (2) and (3).
2 x  y  z  2 (2)
x  y  z  2 (3)
(4)
Adding
3x + 2y = 4
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3x  9 y  6 z  3 (1)
2 x  y  z  2 (2)
Eliminate z from equations (1) and (2). x  y  z  2 (3)
continued
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
Eliminate x from equations (4) and (5).
3x + 2y = 4
15x + 15y = 15
Multiplying top
by 5
15x – 10y = 20
15x + 15y = 15
5y = 5
y = 1
Adding
Using y = 1, find x from equation 4 by substituting.
3x + 2y = 4
3x + 2(1) = 4
x=2
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continued
3x  9 y  6 z  3 (1)
2 x  y  z  2 (2)
x  y  z  2 (3)
Substitute x = 2 and y = 1 to find z.
x+y+z=2
2–1+z=2
1+z=2
z=1
The solution is the ordered triple (2, 1, 1).
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Example
Solve the system using elimination.
x  3y  z  1
(1)
(2)
2x  y  2z  2
x  2 y  3z  1 (3)
Solution
The equations are in standard form.
Eliminate x from equations (1) and (2).
2 x  6 y  2 z  2 (1)
2 x  y  2 z  2 (2)
(4)
Adding
5y  4z = 0
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x  3 y  z  1 (1)
2 x  y  2 z  2 (2)
Eliminate x from equations (1) and (3). x  2 y  3z  1(3)
continued
x  3y  z  1
 x  2 y  3z  1
5y + 4z = 2
(5)
Eliminate y from equations (4) and (5).
5y  4z = 0
5y + 4z = 2
0=2
All variables are eliminated and the resulting equation
is false, which means that this system has no solution;
it is inconsistent.
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