5.1
Systems of Linear Equations
in Two Variables
5.1
OBJECTIVES
1.
2.
3.
4.
Find ordered pairs associated with two equations
Solve a system by graphing
Solve a system by the addition method
Solve a system by the substitution method
Our work in this chapter focuses on systems of equations and the various solution techniques available for your work with such systems. First, let’s consider what we mean by a
system of equations.
In many applications, you will find it helpful to use two variables when labeling the
quantities involved. Often this leads to a linear equation in two variables. A typical equation might be
x 2y 6
NOTE Of course, there are an
infinite number of solutions for
an equation of this type. You
might want to verify that (2, 2)
and (6, 0) are also solutions.
A solution for such an equation is any ordered pair of real numbers (x, y) that satisfies
the equation. For example, the ordered pair (4, 1) is a solution for the equation because
substituting 4 for x and 1 for y results in a true statement.
4 2(1) 6
426
66
True
Whenever two or more equations are considered together, they form a system of
equations. If the equations of the system are linear, the system is called a linear system.
Our work here involves finding solutions for such systems. We present three methods for
solving such systems: the graphing method, the addition method, and the substitution
method.
We begin our discussion with a definition.
Definitions: Solution
A solution for a linear system of equations in two variables is an ordered pair of
real numbers (x, y) that satisfies both equations in the system.
For instance, given the linear system
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x 2y 1
2x y 8
the pair (3, 2) is a solution because after substituting 3 for x and 2 for y in the two equations
of the system, we have the two true statements
NOTE Both equations are
satisfied by (3, 2).
3 2(2) 1
and
2(3) 2 8
1 1
and
88
The solution set for a linear equation in two variables may be graphed as a line. Because
a solution to a system of equations represents a point on both lines, one approach to finding
305
306
CHAPTER 5
SYSTEMS OF LINEAR RELATIONS
NOTE It is helpful at this point
to review Section 4.1 on
graphing linear equations.
the solution for a system is to graph each equation on the same set of coordinate axes and
then identify the point of intersection. This is shown in Example 1.
Example 1
Solving a System by Graphing
Solve the system by graphing.
NOTE Solve each equation for
y and then graph.
y 2x 4
and
2x y 4
xy5
We graph the lines corresponding to the two equations of the system.
yx5
We can approximate the
solution by tracing the curves
near their intersection.
y
xy5
x
(3, 2)
2x y 4
Each equation has an infinite number of solutions (ordered pairs) corresponding to points
on a line. The point of intersection, here (3, 2), is the only point lying on both lines, and
so (3, 2) is the only ordered pair satisfying both equations. Thus, (3, 2) is the solution
for the system.
CHECK YOURSELF 1
Solve the system by graphing.
3x y 2
In Example 1, the two lines are nonparallel and intersect at only one point.
Definitions: Consistent System
A system of equations that has a unique solution is called a consistent system.
In Example 2, we examine a system representing two lines that have no point of
intersection.
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xy6
SYSTEMS OF LINEAR EQUATIONS IN TWO VARIABLES
SECTION 5.1
307
Example 2
Solving a System by Graphing
Solve the system by graphing.
2x y 4
6x 3y 18
The lines corresponding to the two equations are graphed below.
y
2x y 4
x
6x 3y 18
The lines are distinct and parallel. There is no point at which they intersect, so the system
has no solution.
Definitions: Inconsistent System
A system of equations with no solution is called an inconsistent system.
CHECK YOURSELF 2
Solve the system, if possible.
3x y 1
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6x 2y 3
Sometimes the equations in a system have the same graph.
Example 3
Solving a System by Graphing
Solve the system by graphing.
2x y 2
4x 2y 4
308
CHAPTER 5
SYSTEMS OF LINEAR RELATIONS
The equations are graphed as follows.
y
2x y 2
x
4x 2y 4
The lines have the same graph, so they have an infinite number of solutions in common.
Definitions: Dependent System
A system with an infinite number of solutions is called a dependent system.
CHECK YOURSELF 3
Solve the system by graphing.
6x 3y 12
y 2x 4
A second method for solving systems of linear equations in two variables is by the addition
method.
Example 4 illustrates the addition method of solution.
Solving a System by the Addition Method
Solve the system by the addition method.
NOTE The addition method is
sometimes called solution by
elimination for this reason.
5x 2y 12
(1)
3x 2y 12
(2)
In this case, adding the equations will eliminate variable y, and we have
8x 24
x3
(3)
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Example 4
SYSTEMS OF LINEAR EQUATIONS IN TWO VARIABLES
SECTION 5.1
309
Now equation (3) can be paired with either of the original equations. We let x 3 in equation (1):
NOTE The solution should be
checked by substituting these
values into equation (2). Here
3(3) 2
2 12
3
5(3) 2y 12
15 2y 12
2y 3
9 3 12
y
12 12
is a true statement.
3
2
2 is the solution for our system.
3
The solution set is 3, .
2
and 3,
3
CHECK YOURSELF 4
Solve the system by the addition method.
4x 3y 19
4x 5y 25
Example 4 and Check Yourself 4 were straightforward because adding the equations of
the system immediately eliminated one of the variables. Example 5 illustrates a common
situation in which we must multiply one or both of the equations by a nonzero constant before the addition method is applied. This multiplication results in a new system that is
equivalent to the original system.
Procedure:
Equivalent System
An equivalent system is formed whenever
1. One of the equations is multiplied by a nonzero number.
2. One of the equations is replaced by the sum of a constant multiple of
another equation and that equation.
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Example 5
NOTE All these solutions can
be approximated by graphing
the lines and tracing near the
intersection.This is particularly
useful when the solutions are
not integers (the technical term
for such solutions is “ugly”).
Solving a System by the Addition Method
Solve the system by the addition method.
3x 5y 19
(4)
5x 2y 11
(5)
It is clear that adding the equations of the given system will not eliminate one of the
variables. Therefore, we must use multiplication to form an equivalent system. The choice
of multipliers depends on which variable we decide to eliminate. Here we have decided to
eliminate y. We multiply equation (4) by 2 and equation (5) by 5. We then have
NOTE Note that the
coefficients of y are now
opposites of each other.
6x 10y 38
25x 10y 55
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CHAPTER 5
SYSTEMS OF LINEAR RELATIONS
Adding now eliminates y and yields
31x 93
x3
(6)
Pairing equation (6) with equation (4) gives an equivalent system, and we can substitute
3 for x in equation (4):
3 3 5y 19
9 5y 19
5y 10
y 2
The solution set for the system is (3, 2).
CHECK YOURSELF 5
Solve the system by the addition method.
2x 3y 18
6x 10y 22
The following algorithm summarizes the addition method of solving linear systems of
two equations in two variables.
Step by Step: Solving by the Addition Method
Step 1 If necessary, multiply one or both of the equations by a constant so
that one of the variables can be eliminated by addition.
Step 2 Add the equations of the equivalent system formed in step 1.
Step 3 Solve the equation found in step 2.
Step 4 Substitute the value found in step 3 into either of the equations of
the original system to find the corresponding value of the remaining
variable. The ordered pair formed is the solution to the system.
Step 5 Check the solution by substituting the pair of values found in step 4
into the other equation of the original system.
Example 6 illustrates two special situations you may encounter while applying the addition method.
Example 6
Solving a System by the Addition Method
Solve each system by the addition method.
(a) 4x 5y 20
8x 10y 19
(7)
(8)
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NOTE Again, the solution
should be checked by
substitution in equation (5).
SYSTEMS OF LINEAR EQUATIONS IN TWO VARIABLES
SECTION 5.1
311
Multiply equation (7) by 2. Then
8x 10y 40
8x 10y 19
0 21
4x 5y 20
We add the two left sides to get 0 and the two
right sides to get 21.
y
x
8x 10y 19
The result 0 21 is a false statement, which means that there is no point of intersection.
Therefore, the system is inconsistent, and there is no solution.
(b) 5x 7y 9
15x 21y 27
(9)
(10)
Multiply equation (9) by 3. We then have
NOTE The solution set could
be written (x, y)5x 7y 9.
This means the set of all
ordered pairs (x, y) that make
5x 7y 9 a true statement.
15x 21y 27
15x 21y 27
0 0
We add the two equations.
y
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x
Both variables have been eliminated, and the result is a true statement. The two lines
coincide, and there are an infinite number of solutions, one for each point on that line. We
have a dependent system.
CHECK YOURSELF 6
Solve each system by the addition method.
(a) 3x 2y 8
9x 6y 11
(b) x 2y 8
3x 6y 24
312
CHAPTER 5
SYSTEMS OF LINEAR RELATIONS
The results of Example 6 can be summarized as follows.
Rules and Properties: Solving a System of Two Linear
Equations
When a system of two linear equations is solved:
1. If a false statement such as 3 4 is obtained, then the system is inconsistent
and has no solution.
2. If a true statement such as 8 8 is obtained, then the system is dependent
and has an infinite number of solutions.
A third method for finding the solutions of linear systems in two variables is called the substitution method. You may very well find the substitution method more difficult to apply
in solving certain systems than the addition method, particularly when the equations involved in the substitution lead to fractions. However, the substitution method does have important extensions to systems involving higher degree equations, as you will see in later
mathematics classes.
To outline the technique, we solve one of the equations from the original system for one
of the variables. That expression is then substituted into the other equation of the system to
provide an equation in a single variable. That equation is solved, and the corresponding
value for the other variable is found as before, as Example 7 illustrates.
Example 7
Solving a System by the Substitution Method
(a) Solve the system by the substitution method.
2x 3y 3
(11)
y 2x 1
(12)
Because equation (12) is already solved for y, we substitute the expression 2x 1 for y
in equation (11).
NOTE We now have an
equation in the single
variable x.
2x 3(2x 1) 3
Solving for x gives
2x 6x 3 3
x
NOTE To check this result, we
We now substitute
substitute these values in
equation (11) and have
2
3
3 2 3
2
3 6 3
3
2
y2
3
for x in equation (12).
2
2 1
3
312
3 3
A true statement!
The solution set for our system is
3
,2 .
2
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4x 6
SYSTEMS OF LINEAR EQUATIONS IN TWO VARIABLES
SECTION 5.1
313
(b) Solve the system by the substitution method.
NOTE Why did we choose to
solve for y in equation (14)? We
could have solved for x, so that
x
y2
3
We simply chose the easier case
to avoid fractions.
2x 3y 16
(13)
3x y 2
(14)
We start by solving equation (14) for y.
3x y 2
y 3x 2
(15)
y 3x 2
Substituting in equation (13) yields
2x 3(3x 2) 16
2x 9x 6 16
11x 22
x2
We now substitute 2 for x in equation (15).
y322
624
NOTE The solution should be
checked in both equations of
the original system.
The solution set for the system is (2, 4). We leave the check of this result to you.
CHECK YOURSELF 7
Solve each system by the substitution method.
(a) 2x 3y 6
x 3y 6
(b) 3x 4y 3
x 4y 1
The following algorithm summarizes the substitution method for solving linear systems
of two equations in two variables.
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Step by Step: Solving by the Substitution Method
Step 1 If necessary, solve one of the equations of the original system for one
of the variables.
Step 2 Substitute the expression obtained in step 1 into the other equation of
the system to write an equation in a single variable.
Step 3 Solve the equation found in step 2.
Step 4 Substitute the value found in step 3 into the equation derived in step 1
to find the corresponding value of the remaining variable. The ordered
pair formed is the solution for the system.
Step 5 Check the solution by substituting the pair of values found in step 4
into both equations of the original system.
A natural question at this point is, How do you decide which solution method to use?
First, the graphical method can generally provide only approximate solutions. When exact
solutions are necessary, one of the algebraic methods must be applied. Which method to
use depends totally on the given system.
CHAPTER 5
SYSTEMS OF LINEAR RELATIONS
If you can easily solve for a variable in one of the equations, the substitution method
should work well. However, if solving for a variable in either equation of the system leads
to fractions, you may find the addition approach more efficient.
CHECK YOURSELF ANSWERS
1.
y
2.
3x y 2
y
6x 2y 3
(2, 4)
x
x
xy6
no solution
3x y 1
3.
4.
y
5
, 3
2
5. (3, 4)
6x 3y 12
y 2x 4
x
An infinite number of solutions.
6. (a) Inconsistent system: no solution; (b) dependent system: an infinite number of
solutions.
2
3
7. (a) 4, ; (b) 2,
3
4
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314
Name
5.1
Exercises
Section
Date
In exercises 1 to 8, solve each system by graphing. If a unique solution does not exist,
state whether the system is dependent or inconsistent.
ANSWERS
1. x y 6
2. x y 8
xy4
xy2
1.
y
y
2.
3.
x
x
4.
5.
6.
3. x 2y 4
4. x 2y 2
x y1
x 2y 6
x
x
5. 3x y 3
6. 3x 2y 12
3x y 6
y 3
y
y
x
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8.
y
y
7.
7.
x 3y 12
2x 3y 6
x
8. 3x 6y 9
x 2y 3
y
y
x
x
315
ANSWERS
9.
In exercises 9 to 22, solve each system by the addition method. If a unique solution does
not exist, state whether the system is inconsistent or dependent.
10.
9.
2x y 1
2x 3y 5
10.
11.
x 2y 2
3x 2y 12
12. 2x 3y 1
13.
x y3
3x 2y 4
14.
15.
2x y 8
4x 2y 16
16. 3x 4y 2
11.
x 3y 12
2x 3y 6
12.
13.
14.
5x 3y 16
15.
16.
x y 2
2x 3y 21
17.
18.
19.
4x y 20
20.
21.
17. 5x 2y 31
18. 2x y 4
4x 3y 11
6x 3y 10
22.
23.
19.
24.
3x 2y 7
6x 4y 15
20. 3x 4y 0
5x 3y 29
25.
21. 2x 7y 2
3x 5y 14
26.
22.
5x 2y 3
10x 4y 6
27.
28.
In exercises 23 to 34, solve each system by the substitution method. If a unique solution
does not exist, state whether the system is inconsistent or dependent.
y 2x 12
25. 3x 2y 18
x 3y 5
27. 10x 2y 4
y 5x 2
316
24. x y 4
x 2y 2
26. 3x 18y 4
x 6y 2
28. 4x 5y 6
y 2x 10
© 2001 McGraw-Hill Companies
23. x y 7
ANSWERS
29. 3x 4y 9
30. 6x 5y 27
y 3x 1
29.
x 5y 2
30.
31.
31.
x 7y 3
2x 5y 15
32. 4x 3y 11
5x y 11
32.
33.
34.
33. 4x 12y 5
x 3y 1
34. 5x 6y 21
35.
x 2y 5
36.
37.
In exercises 35 to 40, solve each system by any method discussed in this section.
35. 2x 3y 4
38.
36. 5x y 2
x 3y 6
5x 3y 6
39.
40.
41.
37. 4x 3y 0
38. 7x 2y 17
5x 2y 23
x 4y 4
42.
43.
44.
39. 3x y 17
40. 7x 3y 51
5x 3y 5
y 2x 9
In exercises 41 to 44, solve each system by any method discussed in this section.
Hint: You should multiply to clear fractions as your first step.
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41.
43.
1
x
2
1
x
3
1
y
3
2
x
3
1
x
3
3
y 3
5
2
y 3
5
8
42.
y 2
44.
1
1
x y0
5
2
3
x y4
2
3
1
x y 5
8
2
1
3
x y 4
4
2
317
Answers
1. Solution: (5, 1)
3. Solution: (2, 1)
y
y
x
x
5. Inconsistent system
7. Solution: (6, 2)
y
y
x
25.
33.
11.
5,
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41.
3
13. (2, 1)
15. Dependent system
2
(5, 3)
19. Inconsistent system
21. (8, 2)
23. (5, 2)
1
(4, 3)
27. Dependent system
29.
31. (10, 1)
,2
3
8
Inconsistent system
35.
37. (3, 4)
39. (4, 5)
2, 3
(12, 6)
43. (9, 15)
9. (2, 3)
17.
x
318