7.2 two-variable linear systems - academicsacademics.utep.edu/portals/1788/calculus material/7_2...

7.2 TWO-VARIABLE LINEAR SYSTEMS

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• 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.

What You Should Learn

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The Method of Elimination

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The Method of Elimination

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Example 2 – Solving a System of Equations by Elimination

Solve the system of linear equations.

Solution:

Multiplying Equation 2 by 4.

Equation 1

Equation 2

Write Equation 1.

Multiply Equation 2 by 4.

Add equations.

Solve for x.

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By back-substituting into Equation 1, you can solve

for y.

2x – 4y = –7

–4y = –7

The solution is

cont’d Example 2 – Solution

Write Equation 1.

Substitute for x.

Combine like terms.

Solve for y.

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Check this in the original system, as follows.

Check

2x – 4y = –7

–1 – 6 = –7

5x + y = –1

cont’d Example 2 – Solution

Write original Equation 1.

Substitute into Equation 1.

Equation 1 checks.

Write original Equation 2.

Substitute into Equation 2.

Equation 2 checks.

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The Method of Elimination

Equivalent systems: have precisely the same solution set.

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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.

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Graphical Interpretation of Solutions

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Graphical Interpretation of Solutions

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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.

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Example

5𝑥 + 3𝑦 = 92𝑥 − 4𝑦 = 14

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Example

𝑥 − 2𝑦 = 3

−2𝑥 + 4𝑦 = 1

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Example

2𝑥 − 𝑦 = 14𝑥 − 2𝑦 = 3

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In Class Assignment

0.2𝑥 − 0.5𝑦 = −27.80.3𝑥 + 0.4𝑦 = 68.7

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Example 4 – Recognizing Graphs of Linear Systems

Match each system of linear equations with its graph

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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.

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Applications

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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.

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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.

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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 7.10.

cont’d Example 8 – Solution

Figure 7.10

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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.

cont’d Example 8 – Solution

Equation 1

Equation 2

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To solve this system by elimination, multiply Equation 2 by

11.

cont’d Example 8 – Solution

Write Equation 1.

Multiply Equation 2 by 11.

Add equations.

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So,

miles per hour

and

miles per hour.

cont’d Example 8 – Solution

Speed of plane

Speed of wind