Transcript Document

CHAPTER 7
Algebra: Graphs,
Functions, and Linear
Systems
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7.3
Systems of Linear Equations in Two
Variables
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Objectives
1. Decide whether an ordered pair is a solution of a
linear system.
2. Solve linear systems by graphing.
3. Solve linear systems by substitution.
4. Solve linear systems by addition.
5. Identify systems that do not have exactly one
ordered-pair solution.
6. Solve problems using systems of linear equations.
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Systems of Linear Equations & Their
Solutions
• Two linear equations are called a system of linear
equations or a linear system.
• A solution to a system of linear equations in two
variables is an ordered pair that satisfies both
equations in the system.
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Example 1: Determining Whether an
Ordered Pair is a Solution of a System
Determine whether (1,2) is a solution of the system:
2x – 3y = −4
2x + y = 4
Solution: Because 1 is the x-coordinate and 2 is the
y-coordinate of (1,2), we replace x with 1 and y with 2.
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Example 1 continued
2x – 3y = − 4
2(1) – 3(2) = −4
?
2–6=−4
?
− 4 = − 4, TRUE
2x + y = 4
2(1) + 2 = 4
?
2+2=4
?
4 = 4, TRUE
The pair (1,2) satisfies both equations; it makes each
equation true. Thus, the pair is a solution of the
system.
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Graphing 2 lines
• 2x – 3y = -4
2x + 4 = 3y
y = (2/3)x + (4/3)
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• 2x + y = 4
y = -2x + 4
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Solving Linear Systems by Graphing
• For a system with one solution, the coordinates of
the point of intersection of the lines is the system’s
solution.
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Example 2: Solving Linear Systems by
Graphing
Solve by graphing:
x + 2y = 2
x – 2y = 6.
Solution:
We find the solution by graphing both x + 2y = 2 and
x – 2y = 6 in the same rectangular coordinate system.
We will use intercepts to graph each equation.
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Example 2 Continued
x + 2y = 2:
x-intercept: Set y = 0.
x+2·0=2
x=2
The line passes
through (2,0).
y-intercept: Set x = 0.
0 + 2y = 2
2y = 2
y=1
The line passes
through (0,1).
We will graph x + 2y = 2 as a blue line.
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Example 2 Continued
x – 2y = 6:
x-intercept: Set y = 0.
x–2·0=6
x=6
The line passes
through (6,0).
y-intercept: Set x = 0.
0 – 2y = 6
−2y = 6
y = −3
The line passes
through (0,−3).
We will graph x – 2y = 6 as a red line.
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Example 2 Continued
We see the two graphs
intersect at (4,−1). Hence,
this is the solution to the
system.
We can check this
by substituting in (4,−1) into
each equation and verifying
That the solution is true for both
equations.
We leave this to the student.
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Solving Linear Systems by the Substitution
Method
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Example 3: Solving a System by
Substitution
Solve by the substitution method:
y = −x – 1
4x – 3y = 24.
Solution:
Step 1 Solve either of the equations for one variable
in terms of the other. This step has been done for us.
The first equation, y = −x – 1, is solved for y in terms
of x.
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Example 3 continued
Step 2 Substitute the expression from step 1 into the
other equation.
This gives us an equation in one variable, namely
4x – 3(−x – 1) = 24.
The variable y has been eliminated.
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Example 3 continued
Step 3 Solve the resulting equation containing one
variable.
4x – 3(-x – 1) = 24
4x + 3x + 3 = 24
7x + 3 = 24
7x = 21
x=3
Step 4 Back-substitute the obtained value into the
equation from step1. Since we found x = 3 in step 3,
then we back-substitute the x-value into the equation
from step 1 to find the y-coordinate.
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Example 3 continued
Step 4 (cont.)
With x = 3 and y = −4, the proposed solution is (3,−4).
Step 5 Check. Use this ordered pair to verify that this
solution makes each equation true. We leave this to
the student.
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Your Turn
• Solve the following system of equations by the
substitution method.
• y = 2x + 7
2x – y = -5
• -4x + y = -11
2x – 3y = 3
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Solving Linear Systems by the Addition
Method
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Example 5: Solving a System by the Addition
Method
Solve by the addition method:
3x + 2y = 48
9x – 8y = −24.
Solution:
Step 1 Rewrite both equations in the form Ax +
By = C. Both equations are already in this form.
Variable terms appear on the left and constants
appear on the right.
Step 2 If necessary, multiply either equation or
both equations by appropriate numbers so that
the sum of the x-coefficients or the sum of the ycoefficients is 0.
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Example 5 continued
3x + 2y = 48
9x – 8y = −24
Multiply by −3
No Change
Step 3 Add the equations.
− 9x – 6y = − 144
9x – 8y = − 24
−14y = −168
Step 4 Solve the equation in one variable. We solve
−14y = −168 by dividing both sides by −14.
14 y 168

14
14
y  12
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Example 5 continued
Step 5 Back-substitute and find the value for the
other variable.
3x + 2y = 48
3x + 2(12) = 48
3x + 24 = 48
3x = 24
x=8
Step 6 Check. The solution to the system is (8,12). We
can check this by verifying that the solution is true for
both equations. We leave this to the student.
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Linear Systems Having No Solution or
Infinitely Many Solutions
The number of solutions to a system of two linear
equations in two variables is given by one of the
following:
Number of Solutions
What This Means Graphically
Exactly one ordered-pair solution
The two lines intersect at one point.
No Solution
The two lines are parallel.
Infinitely many solutions
The two lines are identical.
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Example 7: A System with no Solution
Solve the system:
4x + 6y = 12
6x + 9y = 12.
Solution: Because no variable is isolated, we will use
the addition method.
4x + 6y = 12 Multiply by 3
6x + 9y = 12 Multiply by -2
Add:
The false statement 0 = 12 indicates that the system has
no solution. The solution is the empty set, Ø.
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Example 8: A System with Infinitely Many
Solutions
Solve the system:
y = 3x – 2
15x – 5y = 10.
Solution: Because the variable y is isolated in y = 3x – 2,
the first equation, we will use the substitution method.
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Example 8 continued
The statement 10 = 10 is true. Hence, this indicates that
the system has infinitely many solutions.
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Modeling with Systems of Equations:
Making Money (and Losing It)
Revenue and Cost Functions
A company produces and sells x units of a product.
• Revenue Function R(x) = (price per unit sold)x
• Cost Function C(x) = fixed cost + (cost per unit
produced)x
The point of intersection of the graphs of the revenue and
cost functions is called the break-even point.
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Example 9: Finding a Break-Even Point
A company is planning to manufacture radically
different wheelchairs. Fixed cost will be $500,000
and it will cost $400 to produce each wheelchair.
Each wheelchair will be sold for $600.
a. Write the cost function, C, of producing x
wheelchairs.
b. Write the revenue function, R, from the sale of x
wheelchairs.
c. Determine the break-even point. Describe what this
means.
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Example 9 Continued
Solution:
a. The cost function is the sum of the fixed cost and the
variable cost.
b. The revenue function is the money generated from
the sale of x wheelchairs.
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Example 9 continued
c. The break even point occurs where the graphs of C
and R intersect. Thus, we find this point by solving
the system
C(x) = 500,000 + 400x
R(x) = 600x,
Or
y = 500,000 + 400x
y = 600x.
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Example 9 continued
Using substitution, we substitute 600x in for y in the
first equation:
600x = 500,000 + 400x
200x = 500,000
x = 2500
Back-substituting 2500 for x in either of the system’s
equations (or functions), we obtain
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Example 9 continued
The break-even point is (2500, 1,500,000). This
means that the company will break even if it produces
and sells 2500 wheelchairs for $1,500,000.
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The Profit Function
• The profit, P(x), generated after producing and selling
x units of a product is given by the profit function
P(x) = R(x) – C(x),
where R and C are the revenue and cost, respectively.
The profit function, P(x), for the
previous example is
P(x) = R(x) – C(x)
= 600x – (500,000 + 400x)
= 200x – 500,000.
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