Lesson: Systems of Equations



Lesson: Systems of Equations

Molly Bordenkircher, Jane Buehler, Stephanie Gonzi, Lester McCurdy

Prior Knowledge: This lesson is based on the idea that students can graph linear equations with and without technology and are able to solve multi-step equations for a given variable.

Lesson Objective(s):

This lesson is designed to enable students to:

• Solve and interpret the meaning of 2 by 2 systems of linear equations graphically, by substitution and by elimination, with and without technology. (Patterns, Functions, and Algebra Standard, Grade 9, Indicator 9)

• Solve real-world problems that can be modeled, using systems of linear equations. (Patterns, Functions, and Algebra Standard, Grade 10, Indicator 11)

• Model and solve problems using matrices. (Patterns, Functions, and Algebra Standard, Grade 11, Indicator 7)

• Solve 3 by 3 systems of linear equations by elimination and using technology, and interpret graphically what the solution means (a point, line, plane, or no solution). (Patterns, Functions, and Algebra Standard, Grade 11, Indicator 9)

• Set up and solve systems of equations using matrices and graphs, with and without technology. (Patterns, Functions, and Algebra Standard, Grade 12, Indicator 5)

Definition(s):

• A system of equations is a set of equations dealt with simultaneously for which a common solution, if possible, is sought.

• A solution to system of equations is a point that lies on the graph of each equation in the system.

Part I: Identifying the solution(s) to a system of equations

1. The equations y = x + 1 and y = ½x + 2 are graphed below to the right. Do the graphs intersect? If so, name the point of intersection.

2. Verify your intersection point is correct by substituting the coordinates of the intersection point in for x and y into the original equations. Show your work and explain how your work verifies that the intersection point is correct

3. The intersection point (2, 3) is the solution to the system of equations in question 1. To solve a system of equations graphically, graph both equations and see where they intersect.  The intersection point is the solution.

4. Solve Graphically. Graph the following system of equations on your graphing calculator. Paste a screen shot of your graphs in the space below. (Hint: You will need to solve both equations in terms of y to graph the equations on your graphing calculator. Show your work for completing this.)

4x-6y=12

2x+2y=6

y1 = _______________________

y2= _______________________

5. What does the solution to the system of equations from question 4 appear to be? Verify your intersection point is correct by substituting the coordinates of the intersection point in for x and y into the original equations. Show your work and explain how your work verifies that the intersection point is correct.

6. On your graphing calculator, graph the following system of equations and paste a screen shot of your graphs in the space below.

f1(x) = 2x - 5

f2(x) = -¼x - 1

7. Is the exact solution to the system obvious? Why or why not?

8. When the solution contains non-integer values, the solution can be calculated graphically using the intersection point(s) tool. Select 6: Points and Lines from the graphing menu and arrow down to select 3: Intersection Point(s). Select the graph of f1(x) = 2x - 5 by hitting “enter” when the line is flashing. Next, select the graph of f2(x) = -¼x – 1 by hitting “enter” when the line is flashing. The solution will appear. Record the solution and paste a screen shot of graphs with the solution shown below.

9. On a new screen, graph the following system of equations and paste of screen shot of the graphs below. (Hint: you will need to solve the second equation for y to enter the equation on your graphing calculator.)

y = 2x + 4

-4x + 2y = 6

10. What relationship exists between the lines? Does the system appear to have a solution? Why or why not?

The system of equations from question 9 represents one of the three possible cases of systems of equations, called an inconsistent system.

11. Graph the following systems of equations on your graphing calculator and complete the first two columns of the chart below. If the system has a solution, give the coordinates of the solution. (Caution: Lines that appear to be parallel are not always parallel. You may need to change your viewing window to see a point of intersection.)

|System of Equations |Number of Solutions |Coordinates of solution, if one|Case |Name of System |

| | |exists | | |

|1. | | | | |

|g(x) = 2x – 0.6 | | | | |

|h(x) = -¼x | | | | |

|2. | | | | |

|i(x) = 0.5x + 9 | | | | |

|j(x) = ½ x + (18/2) | | | | |

|3. | | | | |

|y = x – 1 | | | | |

|-x + y = -1 | | | | |

|4. | | | | |

|k(x) = 1.5x + 6 | | | | |

|l(x) = 3/2x - 4 | | | | |

|5. | | | | |

|-5x – y = 4 | | | | |

|10x + 2y = 7 | | | | |

|6. | | | | |

|m(x) = 4x + 9 | | | | |

|n(x) = 3.8x - 8 | | | | |

Three Possible Cases: When we graph a system of two linear equations, one of three things may happen.

Categorizing Systems by Names – Consistent, Inconsistent, Dependent, and Independent

1. The lines have one point of intersection.  The point of intersection is the only solution of the system. The system is consistent and independent.

2. The lines are parallel.  If this is the case, there is no point that satisfies both equations.  The system has no solution. The system is inconsistent.

3. The lines coincide. Therefore, the equations have the same graph and every solution of one equation is a solution of the other.  There is an infinite number of solutions. The system is consistent and dependent.

This information is summarized in the chart below.

|Case |Number of Solutions |Name of System |

|1. lines intersect |One |Consistent, independent |

|2. parallel lines |Zero |Inconsistent |

|3. lines coincide |Infinitely many |Consistent, dependent |

Information in question 11 is taken from

12. Using the information from above, complete the last two columns of the chart in question 12.

13. Summarize all definitions and concepts concerning systems of linear equations you have learned thus far.

Part II: Solving Systems of Equations by the Substitution Method

You can solve a system of equations by solving one equation for one of the variables; then substitute this result into the second equation.

Here is an example of solving a system of equations by substitution.

Find the solution to the following system of equations. x = y + 3

y + 3x = 1

Step 1: The first equation is solved for x. Substitute (y + 3) in for x in the second equation.

y + 3x = 1 becomes y + 3(y + 3) = 1

Step 2: Solve for y.

y + 3y + 9 = 1

4y = -8

y = -2

Step 3: Substitute -2 in for y in the first equation to solve for x.

x = y + 3 becomes

x = -2 + 3

x = 1

Step 4: Write the solution as an ordered pair.

The solution to the system of equations is (1, -2).

15. Use substitution to solve the following system of equations:

3x + 4y = -1

6x – 2y = 3

Solve for x in the first equation. Show your work.

Substitute the value you got for x into the second equation, rewrite the equation, and solve for y. (Hint: Remember to use parentheses.) Show your work.

Now, substitute the value you got for y into either equation and solve for x.

The solution to this system of equations is (x, y) = __________

Check your work by graphing the system of equations on your graphing calculator. Find the intersection of the two lines. Does it agree with your answer above?

16. Solve each of the following systems.

a. Graph each system of equations.

b. Classify each system as consistent independent, inconsistent, or consistent dependent. (State the solution if one exists.)

c. Solve the systems by substitution. Be sure to show your work.

d. Notice when solving by substitution in (b) and (c), either an untrue statement results or an identity results. Compare these answers with the type of system you named in bullet 2.

a. 2x + 3y = 3 b. x + 3y = 0 c. 1.4x - .3y = 20

12x – 15y = -4 2x + 6y = 7 2.8x –. 6y = 40

You can classify a system of linear equations by the number of solutions.  Systems that have a unique solution are independent systems.  However, not every system has a unique solution.  Systems with an untrue statement are inconsistent.  Systems with an identity are consistent and dependent.  Summarize what types of solution you will get when solving a system by substitution by providing examples of each type. 

17. What will happen if you have a system of 3 equations with 3 variables?

3x – 5y + z = 9

x – 3y – 2z = -8

5x – 6y + 3z = 15

What variable would you solve for in the first equation? _____ What is your result?_____________________

Now, use this result and substitute it into the second and the third equations. Do you notice that you now have two equations with two variables? Solve this new system. Substitute the results into an equation to get the value of the variable that you chose first. The solution to this system of equations is (x, y, z) = _______________

Verify by substituting the solution into each of the equations.

Part III: Solving Systems of Equations by the Elimination Method

18. Let’s go back to the same system of equations that we used in #15 and find another method to solve them. 3x + 4y = -1

6x – 2y = 3

Look at the second equation. What value could you multiply the second equation by so that when you add the first equation to the new second equation, the sum of the y variables would be 0? ___

2( 6x – 2y = 3)(______________________________________

Now add the two equations: 3x + 4y = -1

12x – 4y = 6

Solve for x. _____ Substitute this value into either equation and solve for y. Does your answer correspond to the answer you got by the substitution method, and by graphing?

Could you have first found a value in the first equation to eliminate x? Look again at the first equation. What value could it be multiplied by in order to eliminate the x variables when the two equations would be added together?

Here is the work that another student has done to solve a system of equations. Explain what the student did in each step.

5x + 4 y = 24

3x = 2 + 2y

Step 1. 5x + 4y = 24

3x – 2y = 2

Step 2. 15x + 12y = 72

-15x + 10y = -10

Step 3. 22 y = 62

Step 4. y = 31/11

Step 5. 3x = 2 + 2(31/11)

Step 6. x = 84/33

19. Solve each of the following systems by elimination. Check your answers by graphing, and paste a screen shot of each graph with the solution shown.

a. 3x – 12y = 25 b. .5x = .25y

2x – 4y = 7 x + 2y = 30

20. Systems in three variables can also be solved by using the elimination method. Work with any two equations at a time. Look at the following system.

x + 2y + 3z = 5

3x + 2y – 2z = -13

5x + 3y –z = -11

Look at the first two equations. You would probably want to eliminate y first. What value would you multiply either the first or the second equation by in order to do this?__________

Show your work, and add the two equations.

What variables remain in the sum of the two equations?

Now look at the second and third equations. By what would you multiply the second equation, and by what would you multiply the third equation, so that you could also eliminate y?

Show your work, and add these two equations.

Now you have two equations with two variables. Continue to use the elimination method and solve for x and z.

Substitute these two values into an equation and solve for x.

Check your answers by substituting the values for x, y, and z into each equation? What do you notice?

Can you graph these equations and check your results? Why, or why not? (Hint: Remember from geometry what three non-collinear points determine.) Explain what the intersection of the three equations might be.

21. Write a system of two equations with two variables in which it would be easier to solve by substitution. Explain your reasoning. Then solve by both substitution and by graphing. Show your work and your graph.

22. Write a system of three equations with three variables in which there are infinitely many solutions. Explain how you know that this will be true in the system you wrote.

23. Explain the significance of solving a system of equations by graphing

Solve using any methods for systems of equations

24. 5x-2y+4z=2

-4x+3y+8z=60

3x+2y-3z=-13

x=_______ y=________ z=________

25. 4x+z=24

-7x-6y+25z=0

12x+4y+8z=44

x=_______ y=________ z=________

26. 6x-4y+16z=9

24z-12x+2y=-1

-8y-4z-6x=0

x=_______ y=________ z=________

Part IV: Solving Systems of Linear Equations with

Row Echelon Form on Matrices

Definition(s):

• A matrix is a rectangular arrangement of numbers

• An augmented matrix is a matrix which contains the coefficients of the variables, the last column contains the numbers on the right-hand side of the equations.

• Reduced row echelon form is a matrix in row echelon form with every column that has a leading 1 having 0's in all other positions.

There is no more efficient way to solve a system of linear equations than with Row

Reduction to Echelon Form (rref) on an augmented matrix. There are a variety of

algebraic, geometric, and numerical methods available, but using the rref command

applied to an input augmented matrix handles all possible cases of solving systems of

linear equations. The parts of the following activity will show how to use the TI-Nspire

to deal with a variety of systems, some of which have unique solutions, infinitely many

solutions, or no solutions at all. (NOTE: The TI-83+/84 also have the RREF command

at MATRIX > CALC > RREF to use on an input augmented matrix.)

To get started, look at the example of a system of 3 linear equations containing 3

variables/unknowns.

GIVEN the system of 3 linear equations in standard form: [pic]

The coefficients and constants in the system are going to be entered as a 3 by 4 matrix,

one row for each equation.

On the TI-Nspire CAS, open a document

• 1: Add Calculator page

• Choose menu

• 8: Matrix & Vector

(NOTE: This is option 7 on the TI-Nspire.)

• 4: Reduced Row-Echelon Form to create an rref() command.

• With the cursor inside the parenthesis, choose control, multiplication key, Matrix

• Select 3 rows and tab to enter 4 columns, then tab to OK to create the template to receive the values for this system

The field of rows and columns are set to receive the coefficients and constants from the system in the example.

NOTE: It is best to use the tab key to move from one entry field to the next.

The tab key will move through the fields column by column and row by row.

The arrow keys on the NAVPAD are not as automatic as the tab key for navigating the field.

Use the enter key after value entry is complete. The matrix will automatically be row reduced to echelon form.

The output translates into the following set of equations revealing the solutions to the system of linear equations for this example:

[pic]

It should be easy to see that the row reduced form gives the solution for the values of x,

y, and z in the last column. Reduction to echelon form indicates that the main

diagonal is populated with 1’s and elements off the diagonal are 0’s when the system

has a unique solution. However, two other possibilities can occur when RREF is

applied to an input augmented matrix representing some system of equations.

The first alternative is a system that has no solutions. A reduced-echelon form that

has a row of zeros in the coefficient section and a non-zero number in the augmentation

column occurs with a system that has no solution. The following is an example of a

system translated into an augmented matrix and the outcome that occurs when RREF is

applied to the system. Follow the steps above for entering this example. Remember to

use the tab key for moving through the entry fields of the matrix you created.

[pic] should be entered as rref [pic] then press enter

The output should be [pic]. The last row translates into the false statement 0x+0y+0z=1, indicating no solution.

The second alternative is a system with infinitely many solutions. The following is

an example of a system translated into an augmented matrix and the outcome that

occurs when RREF is applied to the system:

[pic] should be entered as rref [pic], then press enter

The output should be [pic]. The last row translates to the true statement 0x+0y+0z=0 indicating infinitely many solutions.

In this second case, the variable z is considered a value open to selection or an arbitrary variable. However, the values of x and y vary along with the choice of value for variable z. Consider the analogy to the social process of one person becoming engaged to marry another person. With the choice of future spouse also comes the selection of future mother- and father in-law and a whole bunch of other in-laws, but the initial selection of a significant other is a more-or-less arbitrary choice. The same is true with choice of z and the subsequent determination of the values of x and y. To generate members of the solution set, consider the row-reduced form of the previous example:

[pic]translates back into the system [pic] or [pic]

Here are some of the infinitely many solutions that exist for this system based on

random choices of z. All of the solutions are generated by the equations for x and y:

|Z |X |Y |(x,y,z) |

|0 |[pic] |[pic] |[pic] |

|1 |3 |-10 |(3,-10,1) |

|9 |2 |10 |(2,10,9) |

|-10 |[pic] |[pic] |[pic] |

EXAMPLES OF OTHER SYSTEMS OF EQUATIONS:

(NOTE: Refer back to the calculator entry instructions on the first page of this activity.)

27. Solve this system of two equations with two unknowns using the method above.

[pic]

28. Arrange the system of two equations with two unknowns into standard form. Find the solution.

[pic]

29. An extra equation is included in the system with two unknowns to be identified. Find the solution.

[pic]

30. Here is a system of 4 equations with 4 unknowns, so your augmented matrix will have 4 rows and 5 columns. Find the solution.

[pic]

Use the TI-Nspire to check the solution to this system numerically. Provide a screen capture of your work below.

Courtesy of Paul A. Trogdon, Cary High School, Cary, North Carolina

Word problems involving systems of equations.

Solve each of the following questions using any of the methods presented in this lesson. Show all work and state your reasoning for using the method for obtaining the solution.

31. A mechanic earns $15.00 profit per oil change C, and $40.00 profit per brake job j. If the mechanic wants to earn $220.00 a day and will only do as many oil changes as brake jobs, then what is the number of oil changes and brake jobs the mechanic has to do to earn $220.00?

32. The Millers just opened a carpet store. Their startup costs were $5000, and their cost of carpet is $8 per square yard. How many square yards of carpet do they need to sell in order to break even if they sell the carpet for $18 per square yard?

33. You are a human resource director. It is your job to schedule workers for your plant. There are three workers that are supposed to work a combined 110 hours per week. Worker 1 is supposed to work 5 hours more than worker 2, and worker 2 is supposed to work 15 hours more than worker 3. How long should each one work?

34. Beverly has 31 days to complete her quilt for the county fair. The green squares in the quilt can be sewn at a rate of 5 squares per day, and the red squares at a rate of 8 squares per day. The quilt can have up to 96 squares total. The green fabric g costs about $1.60 per square and the red fabric r costs about $2.40 per square. Write the function that describes the cost of the quilt, give the number of days sewing green squares and red squares, and what the final cost of the quilt?

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