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Section 7.3, Slide *

CHAPTER 7

Algebra: Graphs, Functions, and Linear Systems

Section 7.3, Slide *

7.3

Systems of Linear Equations in Two Variables

Section 7.3, Slide *

Objectives

Decide whether an ordered pair is a solution of a linear system.
Solve linear systems by graphing.
Solve linear systems by substitution.
Solve linear systems by addition.
Identify systems that do not have exactly one ordered-pair solution.
Solve problems using systems of linear equations.

Section 7.3, Slide *

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.

Section 7.3, Slide *

Determine whether (1, 2) is a solution of the system:

2x – 3y = −4

2x + y = 4

Example: Determining Whether an Ordered Pair is a Solution of a System

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.

Section 7.3, Slide *

Example continued

2x – 3y = −4 2x + y = 4

2(1) – 3(2) = −4 2(1) + 2 = 4

2 – 6 = − 4 2 + 2 = 4

− 4 = − 4, TRUE 4 = 4, TRUE

The pair (1,2) satisfies both equations; it makes each equation true. Thus, the pair is a solution of the system.

Section 7.3, Slide *

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.

Section 7.3, Slide *

Example: 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.

Section 7.3, Slide *

Example Continued

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.

x + 2y = 2:

Section 7.3, Slide *

Example Continued

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.

x – 2y = 6:

Section 7.3, Slide *

Example 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.

Section 7.3, Slide *

Solving Linear Systems by the Substitution Method

Section 7.3, Slide *

Example: 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.

Section 7.3, Slide *

Example 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.

Section 7.3, Slide *

Example 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.

Section 7.3, Slide *

Example 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.

Section 7.3, Slide *

Solving Linear Systems by the Addition Method

Section 7.3, Slide *

Example: 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 y-coefficients is 0.

Section 7.3, Slide *

Example continued

3x + 2y = 48 − 9x – 6y = −144

9x – 8y = −24 9x – 8y = −24

Step 3 Add the equations. −14y = −168

Multiply by −3

No Change

Step 4 Solve the equation in one variable. We solve −14y = −168 by dividing both sides by −14.

Section 7.3, Slide *

Example 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.

Section 7.3, Slide *

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.

Section 7.3, Slide *

Example: 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

6x + 9y = 12

The false statement 0 = 12 indicates that the system has no solution. The solution is the empty set, Ø.

Multiply by 3

Multiply by -2

Section 7.3, Slide *

Example: 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.

Section 7.3, Slide *

Example continued

The statement 10 = 10 is true. Hence, this indicates that the system has infinitely many solutions.

Section 7.3, Slide *

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.

Section 7.3, Slide *

Example: 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.

Write the cost function, C, of producing x wheelchairs.
Write the revenue function, R, from the sale of x wheelchairs.
Determine the break-even point. Describe what this means.

Section 7.3, Slide *

Example 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.

Section 7.3, Slide *

Example continued

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,

y = 500,000 + 400x

y = 600x.

Section 7.3, Slide *

Example 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

Section 7.3, Slide *

Example 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.

Section 7.3, Slide *

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.

14168

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