Lesson: Systems of Equations

Project AMP Dr. Antonio R. Quesada, Director Project AMP

Lesson: Systems of Equations – Answer Document

Molly Bordenkircher, Jane Buehler, Stephanie Gonzi, Lester McCurdy

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 1: 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.

The graphs intersect at the point (2, 3)

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 = 2+ 1 3 = ½*2 + 2

3 = 3 3 = 1 + 2

3 = 3

The work above verifies that the point of intersection is (2, 3) because when a 2 is substituted in for x in both equations, the corresponding y value is 3. Therefore, the point (2, 3) lies on both graphs and is the intersection point.

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 → -6y = -4x + 12

y = 2/3x - 2

2x+2y=6 → x + y = 3

y = -x + 3

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.

The solution to the system of equations from question 4 appears to be (3, 0)

4(3) – 6(0) = 12 → 12 – 0 = 12 → 12 = 12

2(3) + 2(0) = 6 → 6 + 0 = 6 → 6 = 6

The work verifies that the intersection point is correct because when 3 is substituted in for x and 0 is substituted in for y inn each equation, a true equality exists. The point (3, 0) is a point on both graphs and is therefore the intersection point.

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?

The exact solution is not obvious because it does not appear to contain integer values.

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.

Solution: (1.77778, -1.44444)

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

-4x + 2y = 6 → 2y = 4x + 6 → y = 2x + 3

10. Does the system appear to have a solution? Why or why not?

The system does not appear to have a solution because the two lines do not appear to intersect.

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

The system of equations from question 9 does not have a solution because the two equations are parallel lines. This system is an example of an inconsistent system of equations in which the system contains zero solutions. Parallel lines represent one of the three possible cases of systems of equations.

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 exists / Case / Name of System
1.
g(x) = 2x – 0.6
h(x) = -¼x / One / (0.266667,-0.066667) / Lines Intersect / Consistent, independent
2.
i(x) = 0.5x + 9
j(x) = ½ x + (18/2) / Infinitely many / Lines Coincide / Consistent, dependent
3.
y = x – 1
-x + y = -1 / Infinitely many / Lines Coincide / Consistent, dependent
4.
k(x) = 1.5x + 6
l(x) = 3/2x - 4 / Zero / Parallel Lines / Inconsistent
5.
-5x – y = 4
10x + 2y = 7 / Zero / Parallel Lines / Inconsistent
6.
m(x) = 4x + 9
n(x) = 3.8x - 8 / One / (-85, -331) / Lines Intersect / Consistent, independent

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 http://www.jcoffman.com/Algebra2/ch3_1.htm

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.

· 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.

· To solve a system of equations graphically, graph both equations and see where they intersect. The intersection point is the solution.

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

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

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.

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. Try this for the following system of equations:

3x + 4y = -1

6x – 2y = 3

Solve the first equation for x. Show your work. 3x + 4y = -1→ 3x = -1 – 4y→x = (-1 -4y)/3

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. 6(-1 -4y)/3 – 2y = 3→ -2 – 8y – 2y = 3→

-10 y = 5

y = -1/2

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

3x + 4(-1/2) = -1

3x -2 = -1

3x = 1

x = 1/3

The solution to this system of equations is (x, y) = (1/3,-1/2)

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? yes

16.

a) Solve each of the following systems.

b) Graph each system of equations.

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

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

e) 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

-12x -18y = -18 -2x – 6y = 0 0 = 0

12x – 15y = -4 2x + 6y =7 infinitely many

-33y = -22 0 = 7, no solutions

y = 2/3 no solution

x = ½

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.

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? z

What is your result? z = 9 – 3x + 5y

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) = ______

x – 3y – 2(9 – 3x + 5y) = -8→x – 3y – 18 +6x -10y = -8 → 7x -13y = 10 →28x-52y=40

5x – 6y + 3(9 – 3x + 5y) = 15→5x – 6y+27-9x+15y = 15→-4x + 9y = -12→-28x+63y=-84

11y= -44, y=-4

x = -6, z = 7

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. By what could you multiply the second equation so that when you would add the first equation to the new second equation, the sum of the y variables would be 0? 2

2( 6x – 2y = 3) ® 12x-4y=6

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

12x – 4y = 6

15x = 5

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?

x=1/3; yes

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

Yes multiply by -2

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