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In this 6-lesson unit, students solve linear equations using computer graphing calculator software

Subjects:

Computers & Internet, Math  

Grades:

8, 9, 10, 11, 12  



Title – Solving Equations using the Graphing Calculator
By – Sonny Mooketsi
Primary Subject – Math
Secondary Subjects – Computers/Internet
Grade Level – 8 – 12

Lesson One

Specific objective:

      Students should be able to:
  • Log-on to the computer
  • Develop the skills of using the mouse
  • Type, highlight/select simple text
  • Load and quit the graphing calculator application
  • Log-off from the computer

Switching-on the computer
& loading the graphing calculator

  1. Logging on to the computer: teacher shows students how to log-on [Window version]
  2. Once the computer is on, use the mouse to select the apple symbol on the menu bar.
  3. While the pointer is on the “apple symbol” press the mouse, hold down and drag until you get to the command ” graphing calculator”.
  4. Release the mouse button.
  5. Next use the menu bar to select “equation”.
  6. Use the procedure in steps 3 and 4 to show “full keypad”. What do you notice with the effect of moving the mouse?
  7. On the menu bar select “Demo”.
  8. Select “mouse control” using the dragging skills.
  9. Repeat the process until you get it right.
  10. Quit this application by pressing the “enter” key.

Typing simple text

    Highlight any text which appears on the screen by using the mouse. Delete any text after highlighting it by pressing the back space key. Now you are ready to type any text. Type the following; “my name is..”. The teacher will show you how to leave spaces between the words. To delete some of the words in case you have made a mistake simply use the back space key.

Quitting the graphing calculator
and switching the computer off

  1. Select the “quit” from the menu bar by: holding the mouse down, dragging until the “quit” is highlighted. What do you observe?
  2. On the menu bar, select “special”
  3. Log-off the computer by highlighting “shut down” on the menu bar.
  4. Release the mouse button. You have now switched off your computer.

Lesson Two

Specific objective

    Students should be able to solve simple linear equations in one variable.

Load graphing calculator.
Solving simple linear equations [Theory base]

    You have been taught how to solve simple equations using the method of balancing. The simple equations which we are going to consider here are those which have one variable only. To begin this exercise we shall use the variable x, though other variables are equally applicable. However in this case, the graphic calculator will accept the x-value only, so you need to change these variables to x.

Load the graphing calculator

      To solve the equation x + 2 = 7, follow the following steps:
      1. Type : x + 2 = 7.

      2. Press enter key. You will have a red line and a green line (one of these lines will be a horizontal line) normally, otherwise it will be just two lines bearing different colors. Sometimes you need to pull the screen up and down by using the mouse as soon as the hand symbol is seen in order to see the two graphs. Are the two graphs visible on screen? What do you notice when you click on any of these lines?

      3. While pressing down the mouse button go to the intersection of these lines by using the dragging skills.

      4. Release the mouse button if a popping sound is heard. What do you think this sound mean? Can you see the solutions?

To clear the screen, highlight what you have typed and clear this by pressing the backspace button.

Try these exercises:

Solve the following using the computer.

      1. x – 7 = 14

      2. 3 – x = 9

      3. 3 = 4 – 7x

      4. 2x + 5 = 7

      5. 3 = 4x

      6. 4(x – 2)/7 = 5

      7. 2a – 3 = 6

    8. -3 + 4b =9

Solutions

      1. x = 21, y = 14

      2. x = -6, y = 9

      3. x = 0.142857, y = 3

      4. x = 1, y = 7

      5. x= 0.75, y = 3

      6. x = 10.75, y = 5

      7. a = 4.5, y = 6

    8. b = 3, y = 9
Question for today! For each of the equations you solved above, you obtained two solutions: namely an x-solution and a y-solution. Which of the two solutions is correct? Explain your answer.


Lesson Three

Specific objective

    Students should be able to solve two linear simultaneous equations.

Solving two simultaneous linear equations

      You might have been introduced to various methods which are used to solve simultaneous equations in two variables. The most likely methods which you might have been exposed to are: substitution, elimination and graphically.
      When solving two simultaneous equations graphically the solutions will be given by the intersection of the graphs if the equations have a solution. This can be shown graphical as follows.

        In figure (a), there is one unique solution Explain why there is only one unique solution in figure (a).
        In figure (b) there is no solution, explain why this is the case.
        In figure (c) there are infinitely many solutions, give a reason for this.
      We shall be solving linear equations through graphical methods using the idea of substitution.

Load the graphing calculator

      We shall use the following examples for this exercise:
      Example 1
      To solve x + 3 = y, y = 3x – 2, use the following steps:
      1. Make y the subject in each equation. This is very important when using the graphing calculator. Which of the methods you have learned use this technique? Can you suggest a way of solving equations which have variables which are not in x and y, for example, 2a + 3 = b and a + 3b = -2)
      2. Equate the two equations to get: x +3= 3x + 2 since they both have y being the subject.
      3. Type : x + 3 = 3x + 2
      4. Press enter
      5. To get the solutions accurately move your arrow to one of the line and click on it to get a vertical line, do not release the mouse button yet.
      6. Drag this line towards the intersection of these two lines, and a popping indicates the intersection.
      7. Now release the mouse button. The solutions will appear on the screen as: x = 0.5 and y = 3.5. Are these the results you obtained?

      Example 2
      To Solve y = 2x + 1,and 2x + 3y = 4 use the following steps:
      1. Make y the subject of the formula for the second equation (Explain why this is necessary).
      2. Delete the previous equations appearing on the screen. The second equation will be
        y = (4 – 2x)/3. Did you get this result?
      3. Now press the shift key and while still pressing, press 9 to get (|).
      4. Type: 4 – 2x inside these brackets.
      5. Move the mouse so as the arrow is just outside the brackets (i.e on the right hand side of brackets)
      6. Click on the mouse and release it.
      7. Type: “/ “outside brackets.
      8. Type: 3.
        Now you are ready to solve the two equations.
      9. Equate 2x + 1 to (4 – 2x)/3. 
      10. To get solutions accurately, move your arrow to one of the lines by using the mouse, to get another vertical line. Do not release mouse button yet! What happens when you release the mouse button?
      11. Drag the vertical line you obtained above towards the intersection of these two lines and release mouse button when a popping sound is heard.
        Do the two solutions which appear on the screen x = 0.125 and y = 1.25? If not what might have gone wrong?

Try these exercises

Solve the following simultaneous equations

      i)

      x + y = 2, 2y = 6x + 3
      ii)
y = x – 2, y = 2x + 1
(3x + 4)/2 = (2 – x)/5
2q – 4, 3q – p = 6

Solutions

      i)

      x = 0.125, y = 1.875
      ii)
x = -3.02344… -3.00781, y = -5.088235… -5.00781
x = -0.941176, 0.588235.
p = 0, q = 2

Lesson Four

Specific objective

    Students should be able to solve quadratic equations

Solving quadratic equations

      The solution of a quadratic equation is the two x-values where it cuts the x-axes. Otherwise if it does not cut the x-axes then it will not have any solution. There are three cases which can be shown graphically as follows:

      In figure (a) there is no solution, explain why this is the case.

      In figue (b) there are two distinct solutions, How can you tell this from this graph?

    In figure (c) there is only one repeated solution. Why is this the case?

Example 1

      To solve x

2

      - 5x – 6 = 0 use the following steps:
      1. Log on to the computer.
      2. Load graphing calculator application.
      3. Type: x 2 – 5x – 6 = 0 .
        To raise to any power on the graphing calculator, use the following steps:
        • press the command key and rest you finger on it
        • press the letter h to get a blinking bar
        • Type: 2 in this case
        • Click on the mouse.
      4. Once the graph appears on the screen, click on the left part of this graph to get a vertical line.
      5. Drag this vertical line until a popping sound is heard.
      6. Release the mouse botton.
        Write the two solutions you get:
        Are x = -1 and y = 0 the results you got?
      7. Use your mouse to drag the vertical line to the other intersection on the right of your graph. Did you get x= 6 and y= 0?

        This procedure will be the standard way of solving quadratic equations of the form ax 2 + bx + c = 0, where a is not equal to 0; a , b, c are constants variables.

      Try these exercises
      Solve the following quadratic equations

        i)    x

2

        + 3x + 4 = 0

        ii)   2x

2

        - x – 6 = 0

        iii)  3x

2

        + 5x + 1 = 0

        iv)   x

2

        + x – 4 = 0

        v)    a

2

        - 3a = 5

      Solutions

        i)     No solutions

        ii)    x = 3, y = 0 and x = -2, y= 0. Note that some graphs may not be immediately visible( e.g. ii) & iii) so to see them you have to move the screen up and down, sometimes even from left to the right.

        iii) x = -1.43426, y = 0; x=-0.232408, y=0

        iv)   x = x=-3, y = 0; x=2, y=0

        v)    a = -1.19258, y = 0, x=4.19258, y=0

Questions for the day! When you solved the equations from this previous exercise you got two solutions. One of these solutions was always y = 0. Give a reason why this has been the case.

A quadratic equation can be represented as
a x 2 + b x + c = 0, where a , b , and c are constant variables, why must a not be equal to 0 in this case?


Lesson Five

Specific objective

    Students should be able to solve two simultaneous equations, one being quadratic and the other being linear.

Solving one quadratic and one linear equations simultaneously.

      The type of simultaneous equations we shall be dealing with are of the form ax

2

      + bx + c = y and y = mx + c where a, b, c and m are constant variables and a not equal to 0 for ax

2

      + bx + c = y.

      Where will the solutions of such equations be located on a graph?
      Now I want you to use your previous knowledge to solve the following equations:

        i)    x

2

        - 5x + 6 = y, y = x – 1

        ii)   2x

2

        - 3x – 5 = y, (2x + 2)/4 = y

        iii)  2x + y = 4, 2x

2

        + 5x – 3 = y

        iv)   a

2

        - 3a + 5 = b, a + b = 5

      Hint! y = x

2

      - 5x + 6 and y = x – 1 can be written as x

2

      - 5x + 6 = x – 1 since they are both equal to y and as such they can be equated together (think of substitution method). This idea is very useful when solving such equations, since graphing calculator can not allow you to type two equations in the form of that in (i) to (iii) above.
      Solutions

        i)    x = 4.41421 and y = 3.541421

        or x = 1.58579 and y = 0. 585786

        ii)   x = 4.63314 and y = 2.56657

        or x= -1.13314 and -0.31657

        iii) x = 0.811738 and y = 2.37652

        x = -4.31174 and 12.6235

        iv) a = 2 and b = 3

      a = 0 and b = 5

Lesson Six

Specific objective

    Students should be able to solve quadratic equations simultaneous.

Solving two quadratic equations

      The solution of these two will also be given by the intersection of the two curves.
      Try and solve the following simultaneous equations.

        i)     x

2

        - 5x + 6 = y , 2x

2

        + 3x – 4 = y

        ii)   4x 2 – x + 3 = y , y = x

2

        - 5x + 10

        iii)  a

2

        + 5b + 6 = b, b = 2a

2

        - 3a – 7

        iv)   a

2

        -3b + 5 = 3a, 5b – 3a

2

        + 8 = 4

      Solutions

        i)     x= -9.09902 and y = 134.287

        or x= 1.09902 and y = 1.71275

        ii)   x = 1 and y = 6

        or x = -2.33333 and y = 27.1111

        iii)  a = -1.38516 and b = 0.992858

        a = 9.38516 and b = 141.007

        iv)   a = 1.6979 and b = 0.929722

      a = -5.4479 and b = 17.0078
Question of the day! In some cases you might have heard two popping sounds why is this the case?


Lesson Seven

Specific objective

    Students should be able to solve two simultaneous equations one being linear and the other being non-linear.

Solving two simultaneous equations one linear and the other non-linear.

    Like in the previous cases above, the solution of these equations will be given by the intersection of the two graphs if they intersect. What do you think will happen when the two have no solutions? Care must be taken here as some of the graphs may have two branches, to see the rest of the graphs you need to use the zoom box.

Using your previous knowledge try to solve the following simultaneous equations.

      i)     x + y = 9 and xy = 8

      [make y the subject of each equation].

      ii)   xy = 2 and y = 3x

      iii)  x(y – 1) = 4 and 6 – y = 2x

      iv)   2ab + 4 = 0 and a + b = 7

    v)   k; + 2/h = 4 and 2k + h = -4

Solutions

      i)     x = 8 and y = 1 or x = 1 and y = 8

      ii)   x = 0.816497 and y = -2.44949 or x = -0.816497 and y = -2.44949

      iii)  x = -3.13746 and y = -0.274917 or x = 0.637459 and y = 7.27492

      iv)   a = -0.274917 and b= 7.27492 or x = 7.27492 and b = -0.274917

    v)    k = -12.3246 and h = 4.16228 or k = 0.324555 and h = -2.16228

Unit test

Attempt all questions through the aid of the computer.
All answers must be written on the answer sheet provided
The starred questions require thorough thought before attempting them.

Solve the following equations

1. a) x + 5 = 2 [1]
  b) 3 – x = -5 [1]
  c) 2x – 3 = 9 [1]
* d) -3x + 1/2 = 0.5 [2]
     
2. a) x 2 + 5x – 24 =0 [2]
  b) x 2 – 8x – 65 = 0 [2]
  c) x 2 – 15x + 36 = 0 [2]
  d) 2x 2 – 2x – 12 = 0 [2]
* e) n 2 – 5x + 2 = 0 [3]
     
3. a) x 2 – 3x + 6 = 2x [2]
  b) 3 – x = 3x 2 + x – 5 [2]
  c) 2x 2 – 4x + 7 = (x – 6)/3 [2]
  d) 4x 2 + 3x – 2 = x – 4 [2]
* e) p 2 – 3p = 32p + 5 [3]
     
4. a) x 2 – 3x + 6 = 2x 2 – 4x + 1 [2]
  b) 5x 2 -x + 5 = 2x 2 + 7x = 4 [2]
  c) x 2 – 5x + 7 = x 2 [2]
  d) 2x 2 + 8x = -15x 2 [2]
* e) 3a 2 – 4x = 12x 2 + 1 [3]
     
5. a) 12x 2 – 5x – 2 = xy – 5 [2]
  b) (2yx – 3)/5 = -x 2 – 3x [2]
  c) xy – 2x 2 + 5 = 2z + 7 [2]
  d) -xy = x 2 [2]
* e) ab = 2a 2 – 3a + 6 [3]
     
6. a) 5x = 16 + 2y , 2x = 3y + 2 [2]
  b) 4y = x + 17 , 3y – 4x = 3 [2]
  c) xy – 2x 2 + 5 = 2z + 7 [2]
* d) 3v + 3w = 40 , 5v + 2w = 34 [3]
* e) 3a + 2b = 3, 6a + 10b = 24 [3]

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