Download Systems of Linear Equations

10TH
EDITION
COLLEGE
ALGEBRA
LIAL
HORNSBY
SCHNEIDER
5.1 - 1
5.1
Systems of Linear Equations
Linear Systems
Substitution Method
Elimination Method
Special Systems
Applying Systems of Equations
Solving Linear Systems with Three
Unknowns (Variables)
Using Systems of Equations to Model
Data
5.1 - 2
Linear Systems
The definition of a linear equation given in
Chapter 1 can be extended to more variables;
any equation of the form
a1x1  a2 x2 
 an xn  b
for real numbers a1, a2, …, an (not all of which
are 0) and b, is a linear equation or a firstdegree equation in n unknowns.
5.1 - 3
Linear Systems
A set of equations is called a system of
equations. The solutions of a system of
equations must satisfy every equation in the
system. If all the equations in a system are
linear, the system is a system of linear
equations, or a linear system.
5.1 - 4
Linear Systems
The possible graphs of a
linear system in two
unknowns are as follows.
1. The graphs intersect at
exactly one point, which
gives the (single) ordered
pair solution of the system.
The system is consistent
and the equations are
independent.
5.1 - 5
Linear Systems
2. The graphs are
parallel lines, so
there is no solution
and the solution set is
ø. The system is
inconsistent and the
equations are
independent.
5.1 - 6
Linear Systems
3. The graphs are the
same line, and there
is an infinite number
of solutions. The
system is consistent
and the equations
are dependent.
5.1 - 7
Substitution Method
In a system of two equations with two
variables, the substitution method
involves using one equation to find an
expression for one variable in terms of
the other, and then substituting into the
other equation of the system.
5.1 - 8
SOLVING A SYSTEM BY
SUBSTITUTION
Solve the system.
Example 1
3 x  2y  11
x  y  3
(1)
(2)
Solution
Begin by solving one of the equations for one of
the variables. We solve equation (2) for y.
x  y  3
y  x 3
(2)
Add x.
5.1 - 9
SOLVING A SYSTEM BY
SUBSTITUTION
Now replace y with x + 3 in equation (1), and
solve for x.
Example 1
Note the careful
use of
parentheses.
3 x  2y  11
3 x  2( x  3)  11
3 x  2 x  6  11
5 x  6  11
5x  5
(1)
Let y = x + 3 in (1).
Distributive property
Combine terms.
Subtract.
x 1
5.1 - 10
SOLVING A SYSTEM BY
SUBSTITUTION
Replace x with 1 in equation (3) to obtain
y = 1 + 3 = 4. The solution of the system is the
ordered pair (1, 4). Check this solution in both
equations (1) and (2).
Example 1
Check:
3 x  2y  11
(1)
x  y  3
(2)
3(1)  2( 4)  11
?
1  4  3
?
11  11
True
33
True
Both check; the solution set is {(1, 4)}.
5.1 - 11
Elimination Method
Another way to solve a system of two equations,
called the elimination method, uses
multiplication and addition to eliminate a variable
from one equation. To eliminate a variable, the
coefficients of that variable in the two equations
must be additive inverses. To achieve this, we
use properties of algebra to change the system
to an equivalent system, one with the same
solution set. The three transformations that
produce an equivalent system are listed here.
5.1 - 12
TransformationS of a Linear
System
1. Interchange any two equations of the
system.
2. Multiply or divide any equation of the
system by a nonzero real number.
3. Replace any equation of the system by
the sum of that equation and a multiple of
another equation in the system.
5.1 - 13
Example 2
SOLVING A SYTEM BY
ELIMINATION
Solve the system.
Solution
3 x  4y  1
(1)
2 x  3 y  12
(2)
One way to eliminate a variable is to use the
second transformation and multiply both sides of
equation (2) by –3, giving the equivalent system
3 x  4y  1
6 x  9 y  36
(1)
Multiply (2) by –3
(3)
5.1 - 14
SOLVING A SYTEM BY
ELIMINATION
Now multiply both sides of equation (1) by 2, and
use the third transformation to add the result to
equation (3), eliminating x. Solve the result for y.
Example 2
6 x  8y  2
6 x  9 y  36
17 y  34
y 2
Multiply (1) by 2
(3)
Add.
Solve for y.
5.1 - 15
Example 2
SOLVING A SYTEM BY
ELIMINATION
Substitute 2 for y in either of the original equations
and solve for x.
3 x  4y  1
(1)
3 x  4(2)  1
3x  8  1
3x  9
Let y = 2 in (1).
Multiply.
Add 8.
x 3
5.1 - 16
Example 2
SOLVING A SYTEM BY
ELIMINATION
A check shows that
(3, 2) satisfies both
equations (1) and (2);
the solution set is
{(3, 2)}.
5.1 - 17
SOLVING AN INCONSISTENT
SYSTEM
Solve the system.
Example 3
3 x  2y  4
(1)
6 x  4 y  7
(2)
Solution
To eliminate the variable x, multiply both sides of
equation (1) by 2.
6 x  4y  8
6 x  4 y  7
0  15
Multiply (1) by 2
(2)
False.
5.1 - 18
Example 3
SOLVING AN INCONSISTENT
SYSTEM
Since 0 = 15 is false, the
system is inconsistent
and has no solution. As
suggested here by the
graph, this means that
the graphs of the
equations of the system
never intersect. (The
lines are parallel.) The
solution set is the empty
set.
5.1 - 19
SOLVING A SYSTEM WITH
Example 4
INFINITELY MANY SOLUTIONS
Solve the system.
8 x  2y  4
(1)
 4x  y  2
(2)
Solution
Divide both sides of equation (1) by 2, and add
the result to equation (2).
4 x  y  2
4 x  y  2
00
Divide (1) by 2.
(2)
True.
5.1 - 20
Example 4
SOLVING A SYSTEM WITH
INFINITELY MANY SOLUTIONS
The result, is a true statement, which indicates that
the equations of the original system are equivalent.
Any ordered pair that satisfies either equation will
satisfy the system. From equation (2),
4 x  y  2
(2)
y  2  4 x.
5.1 - 21
Example 4
SOLVING A SYSTEM WITH
INFINITELY MANY SOLUTIONS
The solutions of the system can be
written in the form of a set of
ordered pairs (x, 2 + 4x), for any
real number x. Some ordered pairs
in the solution set are
(0, 2 + 4 • 0) = (0, 2), (1, 2 + 4 • 1) =
(1, 6), (3, 14), and (–2, –6). As
shown here, the equations of the
original system are dependent and
lead to the same straight line graph.
The solution set is written
{(x, 2 + 4x)}.
5.1 - 22
Note In the algebraic solution for Example
4, we wrote the solution set with the variable x
arbitrary. We could write the solution set with y
arbitrary:
 y  2  

 , y .
 4  
By selecting values for y and solving for x in this
ordered pair, we can find individual solutions.
Verify again that (0, 2) is a solution by letting
y = 2 and solving for x to obtain 2  2  0.
4
5.1 - 23
Applying Systems of Equations
Many applied problems involve more than one
unknown quantity. Although some problems with
two unknowns can be solved using just one
variable, it is often easier to use two variables.
To solve a problem with two unknowns, we must
write two equations that relate the unknown
quantities. The system formed by the pair of
equations can then be solved using the methods
of this chapter.
5.1 - 24
Solving An Applied Problem By
Writing A System of Equations
Step 1 Read the problem carefully until you understand
what is given and what is to be found.
Step 2 Assign variables to represent the unknown values,
using diagrams or tables as needed. Write down
what each variable represents.
Step 3 Write a system of equations that relates the
unknowns.
Step 4 Solve the system of equations.
Step 5 State the answer to the problem. Does it seem
reasonable?
Step 6 Check the answer in the words of the original
problem.
5.1 - 25
Example 5
USING A LINEAR SYSTEM TO
SOLVE AN APPLICATION
Salaries for the same position can vary
depending on the location. In 2006, the
average of the salaries for the position of
Accountant I in San Diego, California, and
Salt Lake City, Utah, was $41,175.50. The
salary in San Diego, however, exceeded
the salary in Salt Lake City by $3697.
Determine the salary for the Accountant I
position in San Diego and in Salt Lake City.
5.1 - 26
Example 5
USING A LINEAR SYSTEM TO
SOLVE AN APPLICATION
Solution
Step 1 Read the problem. We must find the
salary of the Accountant I position in
San Diego and in Salt Lake City.
Step 2 Assign variables. Let x represent the
salary of the Accountant I position in
San Diego and y represent the salary
for the same position in Salt Lake City.
5.1 - 27
Example 5
USING A LINEAR SYSTEM TO
SOLVE AN APPLICATION
Solution
Step 3 Write a system of equations. Since
the average salary for the Accountant I
position in San Diego and Salt Lake
City was $41,175.50, one equation is
xy
 41,175.50
2
Multiply both sides of the equation by 2 to get
the equation
(1)
x  y  82,351
5.1 - 28
Example 5
USING A LINEAR SYSTEM TO
SOLVE AN APPLICATION
The salary in San Diego exceeded the salary in
Salt Lake City by $3697. Thus, which gives the
system of equations x – y = 3697, which gives
the system of equations
x  y  82,351
(1)
x  y  3697
(2)
5.1 - 29
USING A LINEAR SYSTEM TO
SOLVE AN APPLICATION
Step 4 Solve the system. To eliminate y, add
the two equations.
(1)
x  y  82,351
Example 5
x  y  3697
2x
 86,048
x  43,024
(2)
Add.
Solve for x.
To find y, substitute 43,024 for x in equation (2).
43,024  y  3697
 y  39,327
y  39,327
Let x = 43,024 in (2).
Subtract 43,024.
Multiply by –1.
5.1 - 30
Example 5
USING A LINEAR SYSTEM TO
SOLVE AN APPLICATION
Step 5 State the answer. The salary for the
position of Accountant I was $43,024 in
San Diego and $39,327 in Salt Lake City.
Step 6 Check. The average of $43,024 and
$39,327 is
$43,024  $39,327
 $41,175.50
2
Also, $43,024 is $3697 more than $39,327, as
required. The answer checks.
5.1 - 31
Solving Linear Equations with three
Unknowns (Variables)
Earlier, we saw that the graph of a linear
equation in two unknowns is a straight
line. The graph of a linear equation in
three unknowns requires a threedimensional coordinate system. The
three number lines are placed at right
angles. The graph of a linear equation in
three unknowns is a plane.
5.1 - 32
Solving Linear Equations with three
Unknowns (Variables)
Some possible intersections of planes representing
three equations in three variables are shown here.
5.1 - 33
Solving Linear Equations with three
Unknowns (Variables)
Some possible intersections of planes representing
three equations in three variables are shown here.
5.1 - 34
Solving Linear Equations with three
Unknowns (Variables)
Some possible intersections of planes representing
three equations in three variables are shown here.
5.1 - 35
Solving a System
To solve a linear system with three unknowns,
first eliminate a variable from any two of the
equations. Then eliminate the same variable
from a different pair of equations. Eliminate a
second variable using the resulting two
equations in two variables to get an equation
with just one variable whose value you can now
determine. Find the values of the remaining
variables by substitution. The solution of the
system is written as an ordered triple.
5.1 - 36
Example 6
SOLVING A SYSTEM OF EQUATIONS
WITH THREE VARIABLES
Solve the system.
3 x  9 y  6z  3
(1)
2x  y  z  2
xy z2
(2)
(3)
Solution
Eliminate z by adding equations (2) and (3) to get
3 x  2y  4
(4)
5.1 - 37
Example 6
SOLVING A SYSTEM OF EQUATIONS
WITH THREE VARIABLES
To eliminate z from another pair of equations,
multiply both sides of equation (2) by 6 and
add the result to equation (1).
12 x  6 y  6z  12
3 x  9 y  6z  3
15 x  15 y
 15
Multiply (2) by 6.
(1)
(5)
Make sure
equation (5) has
the same two
variables as
equation (4).
5.1 - 38
Example 6
SOLVING A SYSTEM OF EQUATIONS
WITH THREE VARIABLES
To eliminate x from equations (4) and (5), multiply
both sides of equation (4) by 5 and add the result to
equation (5). Solve the resulting equation for y.
15 x  10 y  20
15 x  15 y  15
5y   5
y  1
Multiply (4) by –5.
(5)
Add.
Divide by 5.
5.1 - 39
Example 6
SOLVING A SYSTEM OF EQUATIONS
WITH THREE VARIABLES
Using y = – 1 , find x from equation (4) by
substitution.
3 x  2( 1)  4
(4) with y = –1.
x2
5.1 - 40
Example 6
SOLVING A SYSTEM OF EQUATIONS
WITH THREE VARIABLES
Substitute 2 for x and –1 for y in equation (3)
to find z.
2  ( 1)  z  2
(3) with x = 2, y = –1.
z 1
Verify that the ordered triple (2, –1, 1) satisfies
all three equations in the original system. The
solution set is {(2, –1,1)}.
5.1 - 41
Caution
Be careful not to end up
with two equations that still have three
variables. Eliminate the same variable
from each pair of equations.
5.1 - 42
Example 7
SOLVING A SYSTEM OF TWO
EQUATIONS WITH THREE VARIABLES
Solve the system.
Solution
x  2y  z  4
3 x  y  4z  9
(1)
(2)
Geometrically, the solution is the intersection of the
two planes given by equations (1) and (2). The
intersection of two different nonparallel planes is a
line. Thus there will be an infinite number of ordered
triples in the solution set, representing the points on
the line of intersection.
To eliminate x, multiply both sides of equation (1) by
–3 and add the result to equation (2). (Either y or z
could have been eliminated instead.)
5.1 - 43
Example 7
SOLVING A SYSTEM OF TWO
EQUATIONS WITH THREE VARIABLES
3 x  6 y  3z  12
3 x  y  4z  9
7 y  7z  21
7z  7 y  21
Solve this
equation for z.
z  y  3
Multiply (1) by –3.
(2)
(3)
Add 7y.
Divide each term
by –7.
5.1 - 44
Example 7
SOLVING A SYSTEM OF TWO
EQUATIONS WITH THREE VARIABLES
This gives z in terms of y. Express x also in terms of
y by solving equation (1) for x and substituting for z
in the result.
x  2y  z  4
(1)
x  2y  z  4
Solve for x.
x  2y  (  y  3)  4
Substitute (– y + 3)
for z.
x  y  1
With y arbitrary, the solution set is
of the form {(–y +1, y, –y+3)}.
Simplify.
Use parentheses
around
– y + 3.
5.1 - 45
Note
Had we solved equation (3) in
Example 7 for y instead of z, the solution would
have had a different form but would have led to
the same set of solutions. In that case we would
have z arbitrary, and the solution set would be of
the form {(–2 + z, 3 – z, z)}.
By choosing z = 2, one solution would be
(0, 1, 2), which was found above.
5.1 - 46
Modeling Data
Applications with three unknowns
usually require solving a system of
three equations. We can find the
equation of a parabola in the form
y  ax  bx  c
2
by solving a system of three
equations with three variables.
5.1 - 47
Example 8
USING CURVE FITTING TO FIND AN
EQUATION THROUGH THREE POINTS
Find the equation of the parabola y = ax2 + bx + c
that passes through the points (2,4), (–1, 1), and
(–2, 5).
Solution
Since the three points lie on the graph of the
equation y = ax2 + bx + c, they must satisfy the
equation. Substituting each ordered pair into the
equation gives three equations with three
variables.
5.1 - 48
Example 8
USING CURVE FITTING TO FIND AN
EQUATION THROUGH THREE POINTS
4  a(2)  b(2)  c, or
2
1  a( 1)2  b( 1)  c, or
5  a( 2)2  b( 2)  c, or
4  4a  2b  c
(1)
1 a  b  c
(2)
5  4a  2b  c
(3)
5.1 - 49
Example 8
USING CURVE FITTING TO FIND AN
EQUATION THROUGH THREE POINTS
To solve this system, first eliminate c using
equations (1) and (2).
4  4a  2b  c
1  a  b  c
3  3a  3b
(1)
Multiply (2) by –1.
(4)
5.1 - 50
Example 8
USING CURVE FITTING TO FIND AN
EQUATION THROUGH THREE POINTS
Now, use equations (2) and (3) to eliminate the
same variable, c.
1
a bc
5   4a  2b  c
(2)
 4  3a  b
(5)
Multiply (3) by –1.
Equation (5) must have
the same two variables as
equation (4).
5.1 - 51
Example 8
USING CURVE FITTING TO FIND AN
EQUATION THROUGH THREE POINTS
Solve the system of equations (4) and (5) in two
variables by eliminating a.
3  3a  3b
 4   3a  b
1 
1
 b
4
4b
(4)
(5)
Add.
Divide by 4.
5.1 - 52
Example 8
USING CURVE FITTING TO FIND AN
EQUATION THROUGH THREE POINTS
Find a by substituting –¼ for b in equation (4).
1 a  b
1
1 a 
4
5
a
4
(4)
Let b = –¼.
Add ¼ .
5.1 - 53
Example 8
USING CURVE FITTING TO FIND AN
EQUATION THROUGH THREE POINTS
5
1
Finally, find c by substituting a  and b   in
4
4
equation (2).
(2)
1 a  b  c
5  1
1      c
4  4
6
1  c
4
1
 c
2
1
5
Let a  , b   .
4
4
Add.
Subtract
6
.
4
5 2 1
1
The required equation is y  x  x  , or
4
4
2
2
y  1.25 x  .25 x  .5.
5.1 - 54
Example 9
SOLVING AN APPLICATION USING A
SYSTEM OF THREE EQUATIONS
An animal feed is made from three ingredients: corn,
soybeans, and cottonseed. One unit of each
ingredient provides units of protein, fat, and fiber as
shown in the table. How many units of each
ingredient should be used to make a feed that
contains 22 units of protein, 28 units of fat, and 18
units of fiber?
Corn
Soybeans
Cottonseed
Total
.25
.4
.2
22
Fat
.4
.2
.3
28
Fiber
.3
.2
.1
18
Protein
5.1 - 55
Example 9
SOLVING AN APPLICATION USING A
SYSTEM OF THREE EQUATIONS
Solution
Step 1 Read the problem. We must determine
the number of units of corn, soybeans,
and cottonseed.
Step 2 Assign variables. Let x represent the
number of units of corn, y the number of units
of soybeans, and z the number of units of
cottonseed.
5.1 - 56
Example 9
SOLVING AN APPLICATION USING A
SYSTEM OF THREE EQUATIONS
Step 3 Write a system of equations. Since the
total amount of protein is to be 22 units,
the first row of the table yields
.25 x  .4 y  .2z  22.
(1)
Also, for the 28 units of fat,
.4 x  .2y  .3z  28,
(2)
and, for the 18 units of fiber,
.3 x  .2y  .1z  18.
(3)
5.1 - 57
Example 9
SOLVING AN APPLICATION USING A
SYSTEM OF THREE EQUATIONS
Multiply equation (1) on both sides by 100, and
equations (2) and (3) by 10 to get the equivalent
system
25 x  40 y  20z  2200 (4)
4 x  2y  3z  280
(5)
3 x  2y  z  180
(6)
5.1 - 58
Example 9
SOLVING AN APPLICATION USING A
SYSTEM OF THREE EQUATIONS
Step 4 Solve the system. Using the methods
described earlier in this section, we find
that x = 40, y = 15, and z = 30.
Step 5 State the answer. The feed should
contain 40 units of corn, 15 units of
soybeans, and 30 units of cottonseed.
Step 6 Check. Show that the ordered triple
(40, 15, 30) satisfies the system formed
by equations (1), (2), and (3).
5.1 - 59
Note Notice how the table in
Example 9 is used to set up the equations
of the system. The coefficients in each
equation are read from left to right. This
idea is extended in the next section,
where we introduce solution of systems
by matrices.
5.1 - 60