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College Algebra

7.1 Systems of Linear Equations: Two Variables

College Algebra7.1 Systems of Linear Equations: Two Variables
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  1. Preface
  2. 1 Prerequisites
    1. Introduction to Prerequisites
    2. 1.1 Real Numbers: Algebra Essentials
    3. 1.2 Exponents and Scientific Notation
    4. 1.3 Radicals and Rational Exponents
    5. 1.4 Polynomials
    6. 1.5 Factoring Polynomials
    7. 1.6 Rational Expressions
    8. Key Terms
    9. Key Equations
    10. Key Concepts
    11. Review Exercises
    12. Practice Test
  3. 2 Equations and Inequalities
    1. Introduction to Equations and Inequalities
    2. 2.1 The Rectangular Coordinate Systems and Graphs
    3. 2.2 Linear Equations in One Variable
    4. 2.3 Models and Applications
    5. 2.4 Complex Numbers
    6. 2.5 Quadratic Equations
    7. 2.6 Other Types of Equations
    8. 2.7 Linear Inequalities and Absolute Value Inequalities
    9. Key Terms
    10. Key Equations
    11. Key Concepts
    12. Review Exercises
    13. Practice Test
  4. 3 Functions
    1. Introduction to Functions
    2. 3.1 Functions and Function Notation
    3. 3.2 Domain and Range
    4. 3.3 Rates of Change and Behavior of Graphs
    5. 3.4 Composition of Functions
    6. 3.5 Transformation of Functions
    7. 3.6 Absolute Value Functions
    8. 3.7 Inverse Functions
    9. Key Terms
    10. Key Equations
    11. Key Concepts
    12. Review Exercises
    13. Practice Test
  5. 4 Linear Functions
    1. Introduction to Linear Functions
    2. 4.1 Linear Functions
    3. 4.2 Modeling with Linear Functions
    4. 4.3 Fitting Linear Models to Data
    5. Key Terms
    6. Key Concepts
    7. Review Exercises
    8. Practice Test
  6. 5 Polynomial and Rational Functions
    1. Introduction to Polynomial and Rational Functions
    2. 5.1 Quadratic Functions
    3. 5.2 Power Functions and Polynomial Functions
    4. 5.3 Graphs of Polynomial Functions
    5. 5.4 Dividing Polynomials
    6. 5.5 Zeros of Polynomial Functions
    7. 5.6 Rational Functions
    8. 5.7 Inverses and Radical Functions
    9. 5.8 Modeling Using Variation
    10. Key Terms
    11. Key Equations
    12. Key Concepts
    13. Review Exercises
    14. Practice Test
  7. 6 Exponential and Logarithmic Functions
    1. Introduction to Exponential and Logarithmic Functions
    2. 6.1 Exponential Functions
    3. 6.2 Graphs of Exponential Functions
    4. 6.3 Logarithmic Functions
    5. 6.4 Graphs of Logarithmic Functions
    6. 6.5 Logarithmic Properties
    7. 6.6 Exponential and Logarithmic Equations
    8. 6.7 Exponential and Logarithmic Models
    9. 6.8 Fitting Exponential Models to Data
    10. Key Terms
    11. Key Equations
    12. Key Concepts
    13. Review Exercises
    14. Practice Test
  8. 7 Systems of Equations and Inequalities
    1. Introduction to Systems of Equations and Inequalities
    2. 7.1 Systems of Linear Equations: Two Variables
    3. 7.2 Systems of Linear Equations: Three Variables
    4. 7.3 Systems of Nonlinear Equations and Inequalities: Two Variables
    5. 7.4 Partial Fractions
    6. 7.5 Matrices and Matrix Operations
    7. 7.6 Solving Systems with Gaussian Elimination
    8. 7.7 Solving Systems with Inverses
    9. 7.8 Solving Systems with Cramer's Rule
    10. Key Terms
    11. Key Equations
    12. Key Concepts
    13. Review Exercises
    14. Practice Test
  9. 8 Analytic Geometry
    1. Introduction to Analytic Geometry
    2. 8.1 The Ellipse
    3. 8.2 The Hyperbola
    4. 8.3 The Parabola
    5. 8.4 Rotation of Axes
    6. 8.5 Conic Sections in Polar Coordinates
    7. Key Terms
    8. Key Equations
    9. Key Concepts
    10. Review Exercises
    11. Practice Test
  10. 9 Sequences, Probability, and Counting Theory
    1. Introduction to Sequences, Probability and Counting Theory
    2. 9.1 Sequences and Their Notations
    3. 9.2 Arithmetic Sequences
    4. 9.3 Geometric Sequences
    5. 9.4 Series and Their Notations
    6. 9.5 Counting Principles
    7. 9.6 Binomial Theorem
    8. 9.7 Probability
    9. Key Terms
    10. Key Equations
    11. Key Concepts
    12. Review Exercises
    13. Practice Test
  11. Answer Key
    1. Chapter 1
    2. Chapter 2
    3. Chapter 3
    4. Chapter 4
    5. Chapter 5
    6. Chapter 6
    7. Chapter 7
    8. Chapter 8
    9. Chapter 9
  12. Index

Learning Objectives

In this section, you will:
  • Solve systems of equations by graphing.
  • Solve systems of equations by substitution.
  • Solve systems of equations by addition.
  • Identify inconsistent systems of equations containing two variables.
  • Express the solution of a system of dependent equations containing two variables.
Figure 1 (credit: Thomas Sørenes)

A skateboard manufacturer introduces a new line of boards. The manufacturer tracks its costs, which is the amount it spends to produce the boards, and its revenue, which is the amount it earns through sales of its boards. How can the company determine if it is making a profit with its new line? How many skateboards must be produced and sold before a profit is possible? In this section, we will consider linear equations with two variables to answer these and similar questions.

Introduction to Systems of Equations

In order to investigate situations such as that of the skateboard manufacturer, we need to recognize that we are dealing with more than one variable and likely more than one equation. A system of linear equations consists of two or more linear equations made up of two or more variables such that all equations in the system are considered simultaneously. To find the unique solution to a system of linear equations, we must find a numerical value for each variable in the system that will satisfy all equations in the system at the same time. Some linear systems may not have a solution and others may have an infinite number of solutions. In order for a linear system to have a unique solution, there must be at least as many equations as there are variables. Even so, this does not guarantee a unique solution.

In this section, we will look at systems of linear equations in two variables, which consist of two equations that contain two different variables. For example, consider the following system of linear equations in two variables.

2x+y= 15 3xy= 5 2x+y= 15 3xy= 5

The solution to a system of linear equations in two variables is any ordered pair that satisfies each equation independently. In this example, the ordered pair (4, 7) is the solution to the system of linear equations. We can verify the solution by substituting the values into each equation to see if the ordered pair satisfies both equations. Shortly we will investigate methods of finding such a solution if it exists.

2(4)+(7)=15True 3(4)(7)=5True 2(4)+(7)=15True 3(4)(7)=5True

In addition to considering the number of equations and variables, we can categorize systems of linear equations by the number of solutions. A consistent system of equations has at least one solution. A consistent system is considered to be an independent system if it has a single solution, such as the example we just explored. The two lines have different slopes and intersect at one point in the plane. A consistent system is considered to be a dependent system if the equations have the same slope and the same y-intercepts. In other words, the lines coincide so the equations represent the same line. Every point on the line represents a coordinate pair that satisfies the system. Thus, there are an infinite number of solutions.

Another type of system of linear equations is an inconsistent system, which is one in which the equations represent two parallel lines. The lines have the same slope and different y-intercepts. There are no points common to both lines; hence, there is no solution to the system.

Types of Linear Systems

There are three types of systems of linear equations in two variables, and three types of solutions.

  • An independent system has exactly one solution pair ( x,y ). ( x,y ). The point where the two lines intersect is the only solution.
  • An inconsistent system has no solution. Notice that the two lines are parallel and will never intersect.
  • A dependent system has infinitely many solutions. The lines are coincident. They are the same line, so every coordinate pair on the line is a solution to both equations.

Figure 2 compares graphical representations of each type of system.

Figure 2

How To

Given a system of linear equations and an ordered pair, determine whether the ordered pair is a solution.

  1. Substitute the ordered pair into each equation in the system.
  2. Determine whether true statements result from the substitution in both equations; if so, the ordered pair is a solution.

Example 1

Determining Whether an Ordered Pair Is a Solution to a System of Equations

Determine whether the ordered pair ( 5,1 ) ( 5,1 ) is a solution to the given system of equations.

x+3y=8 2x9=y x+3y=8 2x9=y

Analysis

We can see the solution clearly by plotting the graph of each equation. Since the solution is an ordered pair that satisfies both equations, it is a point on both of the lines and thus the point of intersection of the two lines. See Figure 3.

Figure 3
Try It #1

Determine whether the ordered pair ( 8,5 ) ( 8,5 ) is a solution to the following system.

5x−4y=20 2x+1=3y 5x−4y=20 2x+1=3y

Solving Systems of Equations by Graphing

There are multiple methods of solving systems of linear equations. For a system of linear equations in two variables, we can determine both the type of system and the solution by graphing the system of equations on the same set of axes.

Example 2

Solving a System of Equations in Two Variables by Graphing

Solve the following system of equations by graphing. Identify the type of system.

2x+y=−8 xy=−1 2x+y=−8 xy=−1
Try It #2

Solve the following system of equations by graphing.

  2x5y=−25 4x+5y=35   2x5y=−25 4x+5y=35

Q&A

Can graphing be used if the system is inconsistent or dependent?

Yes, in both cases we can still graph the system to determine the type of system and solution. If the two lines are parallel, the system has no solution and is inconsistent. If the two lines are identical, the system has infinite solutions and is a dependent system.

Solving Systems of Equations by Substitution

Solving a linear system in two variables by graphing works well when the solution consists of integer values, but if our solution contains decimals or fractions, it is not the most precise method. We will consider two more methods of solving a system of linear equations that are more precise than graphing. One such method is solving a system of equations by the substitution method, in which we solve one of the equations for one variable and then substitute the result into the second equation to solve for the second variable. Recall that we can solve for only one variable at a time, which is the reason the substitution method is both valuable and practical.

How To

Given a system of two equations in two variables, solve using the substitution method.

  1. Solve one of the two equations for one of the variables in terms of the other.
  2. Substitute the expression for this variable into the second equation, then solve for the remaining variable.
  3. Substitute that solution into either of the original equations to find the value of the first variable. If possible, write the solution as an ordered pair.
  4. Check the solution in both equations.

Example 3

Solving a System of Equations in Two Variables by Substitution

Solve the following system of equations by substitution.

 x+y=−5  2x5y=1  x+y=−5  2x5y=1
Try It #3

Solve the following system of equations by substitution.

x=y+3 4=3x−2y x=y+3 4=3x−2y

Q&A

Can the substitution method be used to solve any linear system in two variables?

Yes, but the method works best if one of the equations contains a coefficient of 1 or –1 so that we do not have to deal with fractions.

Solving Systems of Equations in Two Variables by the Addition Method

A third method of solving systems of linear equations is the addition method. In this method, we add two terms with the same variable, but opposite coefficients, so that the sum is zero. Of course, not all systems are set up with the two terms of one variable having opposite coefficients. Often we must adjust one or both of the equations by multiplication so that one variable will be eliminated by addition.

How To

Given a system of equations, solve using the addition method.

  1. Write both equations with x- and y-variables on the left side of the equal sign and constants on the right.
  2. Write one equation above the other, lining up corresponding variables. If one of the variables in the top equation has the opposite coefficient of the same variable in the bottom equation, add the equations together, eliminating one variable. If not, use multiplication by a nonzero number so that one of the variables in the top equation has the opposite coefficient of the same variable in the bottom equation, then add the equations to eliminate the variable.
  3. Solve the resulting equation for the remaining variable.
  4. Substitute that value into one of the original equations and solve for the second variable.
  5. Check the solution by substituting the values into the other equation.

Example 4

Solving a System by the Addition Method

Solve the given system of equations by addition.

x+2y=−1 x+y=3 x+2y=−1 x+y=3

Analysis

We gain an important perspective on systems of equations by looking at the graphical representation. See Figure 5 to find that the equations intersect at the solution. We do not need to ask whether there may be a second solution because observing the graph confirms that the system has exactly one solution.

Figure 5

Example 5

Using the Addition Method When Multiplication of One Equation Is Required

Solve the given system of equations by the addition method.

3x+5y=−11 x2y=11 3x+5y=−11 x2y=11
Try It #4

Solve the system of equations by addition.

2x−7y=2 3x+y=−20 2x−7y=2 3x+y=−20

Example 6

Using the Addition Method When Multiplication of Both Equations Is Required

Solve the given system of equations in two variables by addition.

2x+3y=−16 5x−10y=30 2x+3y=−16 5x−10y=30

Example 7

Using the Addition Method in Systems of Equations Containing Fractions

Solve the given system of equations in two variables by addition.

x 3 + y 6 =3 x 2 y 4 =1 x 3 + y 6 =3 x 2 y 4 =1
Try It #5

Solve the system of equations by addition.

2x+3y=8 3x+5y=10 2x+3y=8 3x+5y=10

Identifying Inconsistent Systems of Equations Containing Two Variables

Now that we have several methods for solving systems of equations, we can use the methods to identify inconsistent systems. Recall that an inconsistent system consists of parallel lines that have the same slope but different y y -intercepts. They will never intersect. When searching for a solution to an inconsistent system, we will come up with a false statement, such as 12=0. 12=0.

Example 8

Solving an Inconsistent System of Equations

Solve the following system of equations.

        x=9−2y x+2y=13         x=9−2y x+2y=13

Analysis

Writing the equations in slope-intercept form confirms that the system is inconsistent because all lines will intersect eventually unless they are parallel. Parallel lines will never intersect; thus, the two lines have no points in common. The graphs of the equations in this example are shown in Figure 8.

Figure 8
Try It #6

Solve the following system of equations in two variables.

2y−2x=2 2y−2x=6 2y−2x=2 2y−2x=6

Expressing the Solution of a System of Dependent Equations Containing Two Variables

Recall that a dependent system of equations in two variables is a system in which the two equations represent the same line. Dependent systems have an infinite number of solutions because all of the points on one line are also on the other line. After using substitution or addition, the resulting equation will be an identity, such as 0=0. 0=0.

Example 9

Finding a Solution to a Dependent System of Linear Equations

Find a solution to the system of equations using the addition method.

x+3y=2 3x+9y=6 x+3y=2 3x+9y=6

Analysis

If we rewrote both equations in the slope-intercept form, we might know what the solution would look like before adding. Let’s look at what happens when we convert the system to slope-intercept form.

  x+3y=2         3y=x+2           y= 1 3 x+ 2 3 3x+9y=6         9y=−3x+6           y= 3 9 x+ 6 9           y= 1 3 x+ 2 3   x+3y=2         3y=x+2           y= 1 3 x+ 2 3 3x+9y=6         9y=−3x+6           y= 3 9 x+ 6 9           y= 1 3 x+ 2 3

See Figure 9. Notice the results are the same. The general solution to the system is ( x, − 1 3 x+ 2 3 ). ( x, − 1 3 x+ 2 3 ).

Figure 9
Try It #7

Solve the following system of equations in two variables.

    y−2x=5 −3y+6x=−15     y−2x=5 −3y+6x=−15

Using Systems of Equations to Investigate Profits

Using what we have learned about systems of equations, we can return to the skateboard manufacturing problem at the beginning of the section. The skateboard manufacturer’s revenue function is the function used to calculate the amount of money that comes into the business. It can be represented by the equation R=xp, R=xp, where x= x= quantity and p= p= price. The revenue function is shown in orange in Figure 10.

The cost function is the function used to calculate the costs of doing business. It includes fixed costs, such as rent and salaries, and variable costs, such as utilities. The cost function is shown in blue in Figure 10. The x x -axis represents quantity in hundreds of units. The y-axis represents either cost or revenue in hundreds of dollars.

Figure 10

The point at which the two lines intersect is called the break-even point. We can see from the graph that if 700 units are produced, the cost is $3,300 and the revenue is also $3,300. In other words, the company breaks even if they produce and sell 700 units. They neither make money nor lose money.

The shaded region to the right of the break-even point represents quantities for which the company makes a profit. The shaded region to the left represents quantities for which the company suffers a loss. The profit function is the revenue function minus the cost function, written as P(x)=R(x)C(x). P(x)=R(x)C(x). Clearly, knowing the quantity for which the cost equals the revenue is of great importance to businesses.

Example 10

Finding the Break-Even Point and the Profit Function Using Substitution

Given the cost function C(x)=0.85x+35,000 C(x)=0.85x+35,000 and the revenue function R(x)=1.55x, R(x)=1.55x, find the break-even point and the profit function.

Analysis

The cost to produce 50,000 units is $77,500, and the revenue from the sales of 50,000 units is also $77,500. To make a profit, the business must produce and sell more than 50,000 units. See Figure 11.

Figure 11

We see from the graph in Figure 12 that the profit function has a negative value until x=50,000, x=50,000, when the graph crosses the x-axis. Then, the graph emerges into positive y-values and continues on this path as the profit function is a straight line. This illustrates that the break-even point for businesses occurs when the profit function is 0. The area to the left of the break-even point represents operating at a loss.

Figure 12

Example 11

Writing and Solving a System of Equations in Two Variables

The cost of a ticket to the circus is $25.00 $25.00 for children and $50.00 $50.00 for adults. On a certain day, attendance at the circus is 2,000 2,000 and the total gate revenue is $70,000. $70,000. How many children and how many adults bought tickets?

Try It #8

Meal tickets at the circus cost $4.00 $4.00 for children and $12.00 $12.00 for adults. If 1,650 1,650 meal tickets were bought for a total of $14,200, $14,200, how many children and how many adults bought meal tickets?

Media

Access these online resources for additional instruction and practice with systems of linear equations.

7.1 Section Exercises

Verbal

1.

Can a system of linear equations have exactly two solutions? Explain why or why not.

2.

If you are performing a break-even analysis for a business and their cost and revenue equations are dependent, explain what this means for the company’s profit margins.

3.

If you are solving a break-even analysis and get a negative break-even point, explain what this signifies for the company?

4.

If you are solving a break-even analysis and there is no break-even point, explain what this means for the company. How should they ensure there is a break-even point?

5.

Given a system of equations, explain at least two different methods of solving that system.

Algebraic

For the following exercises, determine whether the given ordered pair is a solution to the system of equations.

6.

5xy=4 x+6y=2 5xy=4 x+6y=2 and (4,0) (4,0)

7.

−3x5y=13  x+4y=10 −3x5y=13  x+4y=10 and (−6,1) (−6,1)

8.

3x+7y=1 2x+4y=0 3x+7y=1 2x+4y=0 and (2,3) (2,3)

9.

−2x+5y=7   2x+9y=7 −2x+5y=7   2x+9y=7 and (−1,1) (−1,1)

10.

x+8y=43 3x−2y=−1 x+8y=43 3x−2y=−1 and (3,5) (3,5)

For the following exercises, solve each system by substitution.

11.

  x+3y=5 2x+3y=4   x+3y=5 2x+3y=4

12.

 3x−2y=18 5x+10y=−10  3x−2y=18 5x+10y=−10

13.

4x+2y=−10 3x+9y=0 4x+2y=−10 3x+9y=0

14.

2x+4y=−3.8 9x−5y=1.3 2x+4y=−3.8 9x−5y=1.3

15.

2x+3y=1.2 3x6y=1.8 2x+3y=1.2 3x6y=1.8

16.

   x−0.2y=1 −10x+2y=5    x−0.2y=1 −10x+2y=5

17.

    3x+5y=9 30x+50y=−90     3x+5y=9 30x+50y=−90

18.

 −3x+y=2 12x−4y=−8  −3x+y=2 12x−4y=−8

19.

1 2 x+ 1 3 y=16 1 6 x+ 1 4 y=9 1 2 x+ 1 3 y=16 1 6 x+ 1 4 y=9

20.

1 4 x+ 3 2 y=11 1 8 x+ 1 3 y=3 1 4 x+ 3 2 y=11 1 8 x+ 1 3 y=3

For the following exercises, solve each system by addition.

21.

−2x+5y=−42   7x+2y=30 −2x+5y=−42   7x+2y=30

22.

6x−5y=−34 2x+6y=4 6x−5y=−34 2x+6y=4

23.

    5xy=−2.6 −4x−6y=1.4     5xy=−2.6 −4x−6y=1.4

24.

7x−2y=3 4x+5y=3.25 7x−2y=3 4x+5y=3.25

25.

 −x+2y=−1 5x−10y=6  −x+2y=−1 5x−10y=6

26.

      7x+6y=2 −28x−24y=−8       7x+6y=2 −28x−24y=−8

27.

5 6 x+ 1 4 y=0 1 8 x 1 2 y= 43 120 5 6 x+ 1 4 y=0 1 8 x 1 2 y= 43 120

28.

   1 3 x+ 1 9 y= 2 9 1 2 x+ 4 5 y= 1 3    1 3 x+ 1 9 y= 2 9 1 2 x+ 4 5 y= 1 3

29.

−0.2x+0.4y=0.6            x−2y=−3 −0.2x+0.4y=0.6            x−2y=−3

30.

−0.1x+0.2y=0.6      5x−10y=1 −0.1x+0.2y=0.6      5x−10y=1

For the following exercises, solve each system by any method.

31.

5x+9y=16   x+2y=4 5x+9y=16   x+2y=4

32.

6x−8y=−0.6 3x+2y=0.9 6x−8y=−0.6 3x+2y=0.9

33.

5x−2y=2.25 7x−4y=3 5x−2y=2.25 7x−4y=3

34.

  x 5 12 y= 55 12 −6x+ 5 2 y= 55 2   x 5 12 y= 55 12 −6x+ 5 2 y= 55 2

35.

7x−4y= 7 6 2x+4y= 1 3 7x−4y= 7 6 2x+4y= 1 3

36.

3x+6y=11 2x+4y=9 3x+6y=11 2x+4y=9

37.

       7 3 x 1 6 y=2 21 6 x+ 3 12 y=−3        7 3 x 1 6 y=2 21 6 x+ 3 12 y=−3

38.

1 2 x+ 1 3 y= 1 3 3 2 x+ 1 4 y= 1 8 1 2 x+ 1 3 y= 1 3 3 2 x+ 1 4 y= 1 8

39.

2.2x+1.3y=−0.1 4.2x+4.2y=2.1 2.2x+1.3y=−0.1 4.2x+4.2y=2.1

40.

 0.1x+0.2y=2 0.35x−0.3y=0  0.1x+0.2y=2 0.35x−0.3y=0

Graphical

For the following exercises, graph the system of equations and state whether the system is consistent, inconsistent, or dependent and whether the system has one solution, no solution, or infinite solutions.

41.

3xy=0.6 x−2y=1.3 3xy=0.6 x−2y=1.3

42.

x+2y=4  2x−4y=1 x+2y=4  2x−4y=1

43.

  x+2y=7 2x+6y=12   x+2y=7 2x+6y=12

44.

3x−5y=7   x−2y=3 3x−5y=7   x−2y=3

45.

  3x−2y=5 −9x+6y=−15   3x−2y=5 −9x+6y=−15

Technology

For the following exercises, use the intersect function on a graphing device to solve each system. Round all answers to the nearest hundredth.

46.

   0.1x+0.2y=0.3 −0.3x+0.5y=1    0.1x+0.2y=0.3 −0.3x+0.5y=1

47.

−0.01x+0.12y=0.62 0.15x+0.20y=0.52 −0.01x+0.12y=0.62 0.15x+0.20y=0.52

48.

0.5x+0.3y=4 0.25x−0.9y=0.46 0.5x+0.3y=4 0.25x−0.9y=0.46

49.

0.15x+0.27y=0.39 −0.34x+0.56y=1.8 0.15x+0.27y=0.39 −0.34x+0.56y=1.8

50.

−0.71x+0.92y=0.13 0.83x+0.05y=2.1 −0.71x+0.92y=0.13 0.83x+0.05y=2.1

Extensions

For the following exercises, solve each system in terms of A,B,C,D,E, A,B,C,D,E, and F F where AF AF are nonzero numbers. Note that AB AB and AEBD. AEBD.

51.

x+y=A xy=B x+y=A xy=B

52.

x+Ay=1 x+By=1 x+Ay=1 x+By=1

53.

Ax+y=0 Bx+y=1 Ax+y=0 Bx+y=1

54.

Ax+By=C x+y=1 Ax+By=C x+y=1

55.

Ax+By=C Dx+Ey=F Ax+By=C Dx+Ey=F

Real-World Applications

For the following exercises, solve for the desired quantity.

56.

A stuffed animal business has a total cost of production C=12x+30 C=12x+30 and a revenue function R=20x. R=20x. Find the break-even point.

57.

A fast-food restaurant has a cost of production C(x)=11x+120 C(x)=11x+120 and a revenue function R(x)=5x. R(x)=5x. When does the company start to turn a profit?

58.

A cell phone factory has a cost of production C(x)=150x+10,000 C(x)=150x+10,000 and a revenue function R(x)=200x. R(x)=200x. What is the break-even point?

59.

A musician charges C(x)=64x+20,000, C(x)=64x+20,000, where x x is the total number of attendees at the concert. The venue charges $80 per ticket. After how many people buy tickets does the venue break even, and what is the value of the total tickets sold at that point?

60.

A guitar factory has a cost of production C(x)=75x+50,000. C(x)=75x+50,000. If the company needs to break even after 150 units sold, at what price should they sell each guitar? Round up to the nearest dollar, and write the revenue function.

For the following exercises, use a system of linear equations with two variables and two equations to solve.

61.

Find two numbers whose sum is 28 and difference is 13.

62.

A number is 9 more than another number. Twice the sum of the two numbers is 10. Find the two numbers.

63.

The startup cost for a restaurant is $120,000, and each meal costs $10 for the restaurant to make. If each meal is then sold for $15, after how many meals does the restaurant break even?

64.

A moving company charges a flat rate of $150, and an additional $5 for each box. If a taxi service would charge $20 for each box, how many boxes would you need for it to be cheaper to use the moving company, and what would be the total cost?

65.

A total of 1,595 first- and second-year college students gathered at a pep rally. The number of freshmen exceeded the number of sophomores by 15. How many freshmen and sophomores were in attendance?

66.

276 students enrolled in a freshman-level chemistry class. By the end of the semester, 5 times the number of students passed as failed. Find the number of students who passed, and the number of students who failed.

67.

There were 130 faculty at a conference. If there were 18 more women than men attending, how many of each gender attended the conference?

68.

A jeep and BMW enter a highway running east-west at the same exit heading in opposite directions. The jeep entered the highway 30 minutes before the BMW did, and traveled 7 mph slower than the BMW. After 2 hours from the time the BMW entered the highway, the cars were 306.5 miles apart. Find the speed of each car, assuming they were driven on cruise control.

69.

If a scientist mixed 10% saline solution with 60% saline solution to get 25 gallons of 40% saline solution, how many gallons of 10% and 60% solutions were mixed?

70.

An investor earned triple the profits of what she earned last year. If she made $500,000.48 total for both years, how much did she earn in profits each year?

71.

An investor who dabbles in real estate invested 1.1 million dollars into two land investments. On the first investment, Swan Peak, her return was a 110% increase on the money she invested. On the second investment, Riverside Community, she earned 50% over what she invested. If she earned $1 million in profits, how much did she invest in each of the land deals?

72.

If an investor invests a total of $25,000 into two bonds, one that pays 3% simple interest, and the other that pays 2 7 8 % 2 7 8 % interest, and the investor earns $737.50 annual interest, how much was invested in each account?

73.

If an investor invests $23,000 into two bonds, one that pays 4% in simple interest, and the other paying 2% simple interest, and the investor earns $710.00 annual interest, how much was invested in each account?

74.

CDs cost $5.96 more than DVDs at All Bets Are Off Electronics. How much would 6 CDs and 2 DVDs cost if 5 CDs and 2 DVDs cost $127.73?

75.

A store clerk sold 60 pairs of sneakers. The high-tops sold for $98.99 and the low-tops sold for $129.99. If the receipts for the two types of sales totaled $6,404.40, how many of each type of sneaker were sold?

76.

A concert manager counted 350 ticket receipts the day after a concert. The price for a student ticket was $12.50, and the price for an adult ticket was $16.00. The register confirms that $5,075 was taken in. How many student tickets and adult tickets were sold?

77.

Admission into an amusement park for 4 children and 2 adults is $116.90. For 6 children and 3 adults, the admission is $175.35. Assuming a different price for children and adults, what is the price of the child’s ticket and the price of the adult ticket?

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