6.2. TWO-VARIABLE LINEAR SYSTEMS
What You Should Learn
• Use the method of elimination to solve systems of linear equations
in two variables.
• Interpret graphically the numbers of solutions of systems of linear
equations in two variables.
• Use systems of linear equations in two variables to model and solve
real-life problems.
The Method of Elimination
We have studied two methods for solving a system of
equations: substitution and graphing. Now we will study the
method of elimination.
The key step in this method is to obtain, for one of the
variables, coefficients that differ only in sign so that adding
the equations eliminates the variable.
Equation 1
Equation 2
Add equations.
The Method of Elimination
Note that by adding the two equations, you eliminate the
and obtain a single equation in y.
x -terms
Solving this equation for y produces y = 2, which you can then back
substitute into one of the original equations to solve for x.
The Method of Elimination
Example 1 – Solving a System of Equations by Elimination
Solve the system of linear equations.
3𝑥 + 2𝑦 = 4 Equation 1
5𝑥 − 2𝑦 = 12 Equation 2
Example 1 – Solution
Add two equations vertically
3𝑥 + 2𝑦 = 4
5𝑥 − 2𝑦 = 12
8𝑥 = 16
𝑥=2
Put this in the equation 1. Then,
3 ∙ 2 = 2𝑦 = 4
6 + 2𝑦 = 4
2𝑦 = −2
𝑦 = −1
Example 1 – Solution
So, we get 2, −1 .
Check with the equation 2:
5 ∙ 2 − 2 −1 = 10 + 2 = 12 True
Example 2 – Solving a System of Equations by Elimination
Solve the system of linear equations.
Equation 1
Equation 2
Example 2 – Solution
For this system, you can obtain coefficients that differ only in sign by
multiplying Equation 2 by 4.
Write Equation 1.
Multiply Equation 2 by 4.
Add equations.
Solve for x.
Example 2 – Solution
By back-substituting
for y.
2x – 4y = –7
cont’d
into Equation 1, you can solve
Write Equation 1.
Substitute
–4y = –6
Move -1 to the right.
Solve for y.
The solution is
for x.
Example 2 – Solution
cont’d
Check this in the original system, as follows.
Check:
Write original Equation 1.
2x – 4y = –7
Substitute into Equation 1.
–1 – 6 = –7
Equation 1 checks.
5x + y = –1
Write original Equation 2.
Substitute into Equation 2.
Equation 2 checks.
Example 3
Solve the system of linear equations.
5𝑥 + 3𝑦 = 9
2𝑥 − 4𝑦 = 14
Equation 1
Equation 2
Example 3-Solution
4×Equation 1:
3×Equation 2:
4(5𝑥 + 3𝑦) = 4 ∙ 9
20𝑥 + 12𝑦 = 36
3 2𝑥 − 4𝑦 = 3 ∙ 14
6𝑥 − 12𝑦 = 42
(4×Equation 1) + (3×Equation 2): Add vertically
20𝑥 + 12𝑦 = 36
6𝑥 − 12𝑦 = 42
26𝑥 =
78
𝑥=3
Example 3-Solution
Put this in the equation 1:
5 ∙ 3 + 3𝑦 = 9
15 + 3𝑦 = 9
3𝑦 = −6
𝑦 = −2
Therefore, the solution is 3, −2 .
The Method of Elimination
In Example 2, the two systems of linear equations (the
original system and the system obtained by multiplying by
constants)
are called equivalent systems because they have
precisely the same solution set.
The Method of Elimination
The operations that can be performed on a system of linear
equations to produce an equivalent system are
(1) interchanging any two equations,
(2) multiplying an equation by a nonzero constant, and
(3) adding a multiple of one equation to any other equation
in the system.
Graphical Interpretation of Solutions
It is possible for a general system of equations to have exactly one
solution, two or more solutions, or no solution. If a system of linear
equations has two different solutions,
it must have an infinite number of solutions.
Graphical Interpretation of Solutions
A system of linear equations is consistent if it has at least
one solution.
A consistent system with exactly one solution is
independent, whereas a consistent system with infinitely
many solutions is dependent.
A system is inconsistent if it has no solution.
Example 4 – Recognizing Graphs of Linear Systems
Match each system of linear equations with its graph in
Figure 6.7. Describe the number of solutions and state
whether the system is consistent or inconsistent.
Figure 6.7
Example 4 – Solution
a. The graph of system (a) is a pair of parallel lines (ii). The
lines have no point of intersection, so the system has no
solution. The system is inconsistent.
b. The graph of system (b) is a pair of intersecting
lines (iii). The lines have one point of intersection, so
the system has exactly one solution. The system is
consistent.
c. The graph of system (c) is a pair of lines that
coincide (i). The lines have infinitely many points of
intersection, so the system has infinitely many solutions.
The system is consistent.
Applications
At this point, you may be asking the question “How can I
tell which application problems can be solved using a
system of linear equations?” The answer comes from
the following considerations.
1. Does the problem involve more than one unknown
quantity?
2. Are there two (or more) equations or conditions to be
satisfied?
If one or both of these situations occur, the appropriate
mathematical model for the problem may be a system of
linear equations.
Example 8 – An Application of a Linear System
An airplane flying into a headwind travels the 2000-mile
flying distance between Chicopee, Massachusetts and
Salt Lake City, Utah in 4 hours and 24 minutes. On the
return flight, the same distance is traveled in 4 hours.
Find the airspeed of the plane and the speed of the wind,
assuming that both remain constant.
Solution:
The two unknown quantities are the speeds of the wind and
the plane.
Example 8 – Solution
If r1 is the speed of the plane and r2 is the speed of the
wind, then
r1 – r2 = speed of the plane against the wind
r1 + r2 = speed of the plane with the wind
as shown in Figure 6.10.
Figure 6.10
cont’d
Example 8 – Solution
cont’d
Using the formula distance = (rate)(time) for these two
speeds, you obtain the following equations.
2000 = (r1 – r2)
2000 = (r1 + r2)(4)
These two equations simplify as follows.
Equation 1
Equation 2
Example 8 – Solution
cont’d
To solve this system by elimination, multiply Equation 2 by
11.
Write Equation 1.
Multiply Equation 2 by 11.
Add equations.
Example 8 – Solution
cont’d
So,
miles per hour
Speed of plane
and
miles per hour.
Speed of wind
Example 9
The demand and supply equations for a video game console are
𝑝 = 180 − 0.00001𝑥 Equation 1
𝑝 = 90 + 0.00002𝑥
Equation 2
where 𝑝 is the price per unit in dollars and 𝑥 is the number of units.
Find the equilibrium point for this market. The equilibrium point is the
price 𝑝 and number of units 𝑥 that satisfy both the demand and supply
equations.
Example 9 - Solution
Equation 2+(− Equation 1): Add vertically
𝑝 = 90 + 0.00002𝑥
−𝑝 = −180 + 0.00001𝑥
0 = −90 + 0.00003𝑥
0.00003𝑥 = 90
𝑥=
90
0.00003
= 3,000,000 : Number of units
To find the price, put the number of units to the equation 2:
𝑝 = 90 + 0.00002 ∙ 3000000 = 90 + 60 = 150 dollars per unit.
The equilibrium point is 150 dollars/unit when there are 3,000,000
video game consoles.