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Sec 5.2 - 1
Chapter 5
Systems of Linear Equations
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Sec 5.2 - 2
5.2
Systems of Linear Equations
in Three Variables
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Sec 5.2 - 3
5.2 Systems of Linear Equations in Three Variables
Objectives
1.
Understand the geometry of systems of three
equations in three variables.
2.
Solve linear systems (with three equations and
three variables) by elimination.
3.
Solve linear systems (with three equations and
three variables) in which some of the equations
have missing terms.
4.
Solve special systems.
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Sec 5.2 - 4
5.2 Systems of Linear Equations in Three Variables
An Equation in Three Variables
A solution of an equation in three variables, such as
4x + y – 5z = 1
is called an ordered triple and is written (x, y, z). For example, the ordered
triple (3, –6, 1) is a solution to the equation, because
4(3) + (–6) – 5(1) = 12 – 6 – 5 = 1.
Verify that another solution of this equation is (–2, 4, –1).
In the rest of this chapter, the term linear equation is extended to equations of the form
Ax + By + Cz + . . . + Dw = K,
where not all the coefficients A, B, C, . . . D equal zero.
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Sec 5.2 - 5
5.2 Systems of Linear Equations in Three Variables
Possible Solutions
1.
The three planes may meet at a single, common point that is the
solution of the system.
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Sec 5.2 - 6
5.2 Systems of Linear Equations in Three Variables
Possible Solutions
2.
The three planes may have the points of a line in common so that the
infinite set of points that satisfy the equation of the line is the solution
of the system.
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Sec 5.2 - 7
5.2 Systems of Linear Equations in Three Variables
Possible Solutions
3.
The three planes may coincide so that the solution of the system is the
set of all points on a plane.
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Sec 5.2 - 8
5.2 Systems of Linear Equations in Three Variables
Possible Solutions
4.
The planes may have no points common to all three so that there is
no solution of the system.
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Sec 5.2 - 9
5.2 Systems of Linear Equations in Three Variables
Solving a Linear System in Three Variables
Step 1 Select a variable and an equation. A good choice for the variable,
which we call the focus variable, is one that has coefficient 1 or −1.
Then select an equation, usually the one that contains the focus
variable, as the working equation.
Step 2 Eliminate the focus variable. Use the working equation and
one of the other two equations of the original system. The result
is an equation in two variables.
Step 3 Eliminate the focus variable again. Use the working equation and
the remaining equation of the original system. The result is another
equation in two variables.
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Sec 5.2 - 10
5.2 Systems of Linear Equations in Three Variables
Solving a Linear System in Three Variables
Step 4 Write the equation in two variables that result from Steps 2
and 3 as a system, and solve it. Doing this gives the values of
two of the variables.
Step 5 Find the value of the remaining variable. Substitute the values of
the two variables found in Step 4 into the working equation to obtain
the value of the focus variable.
Step 6 Check the ordered-triple solution in each of the original equations
of the system. Then write the solution set.
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Sec 5.2 - 11
5.2 Systems of Linear Equations in Three Variables
EXAMPLE 1
Solve the system.
Solving a System in Three Variables
2x + 3y – z = 5
(1)
–3x + 2y – 4z = 2
(2)
x – 4y + 3z = –9
7x + 5y
(3)
= 6
(4)
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
= 6
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Multiply each side of (1) by 3.
(3)
Add.
(4)
Sec 5.2 - 12
5.2 Systems of Linear Equations in Three Variables
EXAMPLE 1
Solve the system.
Solving a System in Three Variables
2x + 3y – z = 5
(1)
–3x + 2y – 4z = 2
(2)
x – 4y + 3z = –9
7x + 5y
–5x – 10y
(3)
= 6
(4)
= –30
(5)
Eliminate the same variable, z, from any other two 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
= –30
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Add.
(5)
Sec 5.2 - 13
5.2 Systems of Linear Equations in Three Variables
EXAMPLE 1
Solve the system.
Solving a System in Three Variables
2x + 3y – z = 5
(1)
–3x + 2y – 4z = 2
(2)
x – 4y + 3z = –9
7x + 5y
–5x – 10y
(3)
= 6
(4)
= –30
(5)
Eliminate a different variable and solve.
14x + 10y = 12
Multiply each side of (4) by 2.
–5x – 10y = –30
(5)
9x
= –18
Add.
(6)
x = –2
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Sec 5.2 - 14
5.2 Systems of Linear Equations in Three Variables
EXAMPLE 1
Solve the system.
Solving a System in Three Variables
2x + 3y – z = 5
(1)
–3x + 2y – 4z = 2
(2)
x – 4y + 3z = –9
7x + 5y
–5x – 10y
(3)
= 6
(4)
= –30
(5)
Find a second value by substituting –2 for x in (4) or (5).
7x + 5y = 6
7(–2) + 5y = 6
(4)
Let x = –2.
–14 + 5y = 6
5y = 20
y = 4
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Sec 5.2 - 15
5.2 Systems of Linear Equations in Three Variables
EXAMPLE 1
Solve the system.
Solving a System in Three Variables
2x + 3y – z = 5
(1)
–3x + 2y – 4z = 2
(2)
x – 4y + 3z = –9
7x + 5y
–5x – 10y
(3)
= 6
(4)
= –30
(5)
Find the value of the remaining variable 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
(1)
Let x = –2 and y = 4.
8– z = 5
z = 3
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Sec 5.2 - 16
5.2 Systems of Linear Equations in Three Variables
EXAMPLE 1
Solve the system.
Solving a System in Three Variables
2x + 3y – z = 5
(1)
–3x + 2y – 4z = 2
(2)
x – 4y + 3z = –9
(3)
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
True
True
True
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Sec 5.2 - 17
5.2 Systems of Linear Equations in Three Variables
EXAMPLE 1
Solve the system.
Solving a System in Three Variables
2x + 3y – z = 5
(1)
–3x + 2y – 4z = 2
(2)
x – 4y + 3z = –9
(3)
Because (–2, 4, 3) satisfies all three equations of the system, the solution set
is { (–2, 4, 3) }.
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Sec 5.2 - 18
5.2 Systems of Linear Equations in Three Variables
EXAMPLE 2
Solving a System of Equations
with Missing Terms
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)
Sec 5.2 - 19
5.2 Systems of Linear Equations in Three Variables
EXAMPLE 2
Solving a System of Equations
with Missing Terms
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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Sec 5.2 - 20
5.2 Systems of Linear Equations in Three Variables
EXAMPLE 2
Solving a System of Equations
with Missing Terms
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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Sec 5.2 - 21
5.2 Systems of Linear Equations in Three Variables
EXAMPLE 2
Solve the system.
Solving a System of Equations
with Missing Terms
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
{ (7, –3, 6) }.
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Sec 5.2 - 22
5.2 Systems of Linear Equations in Three Variables
EXAMPLE 3
Solve the system.
Solving an Inconsistent System
with Three Variables
3x – 2y +
z = 8
(1)
4x –
z = –2
(2)
y –
–6x + 4y – 2z = 3
7x – 3y
= 6
(3)
(4)
Eliminate z by adding equations (1) and (2).
3x – 2y +
z = 8
(1)
4x –
z = –2
(2)
y –
7x – 3y
= 6
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Add.
(4)
Sec 5.2 - 23
5.2 Systems of Linear Equations in Three Variables
EXAMPLE 3
Solve the system.
Solving an Inconsistent System
with Three Variables
3x – 2y +
z = 8
(1)
4x –
z = –2
(2)
y –
–6x + 4y – 2z = 3
7x – 3y
= 6
(3)
(4)
Now,
eliminate
z again
using indicates
equationsthat
(1) equations
and (3). (1) and (3) have no
The resulting
false
statement
common solution. Thus, the system is inconsistent and the solution set is Ø.
The graph of this system would show these two planes parallel to one
another.
6x – 4y + 2z = 16
–6x + 4y – 2z = 3
0 = 19
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Multiply each side of (1) by 2.
(3)
False
Sec 5.2 - 24
5.2 Systems of Linear Equations in Three Variables
Parallel Planes
NOTE
If a false statement results when adding as in Example 3, it is not
necessary to go any further with the solution. Since two of the three
planes are parallel, it is not possible for the three planes to have any
common points.
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Sec 5.2 - 25
5.2 Systems of Linear Equations in Three Variables
EXAMPLE 4
Solve the system.
Solving a System of Dependent
Equations with Three Variables
3x – 4y +
3x –
4
z = 12
y + 1z = 3
6x – 8y +
4
2z = 24
(1)
(2)
(3)
Multiplying each side of equation (1) by 2 gives equation (3). Multiplying
each side of equation (2) by 8 also gives equation (3). Because of this, the
equations are dependent. All three equations have the same graph. The
solution is written
{ (x, y, z) | 3x – 4y + z = 12 }.
Although any one of the three equations could be used to write the solution
set, we use the equation with coefficients that are integers with no common
factor (except 1), as we did in Section 5.1.
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Sec 5.2 - 26