Solving Systems of Linear Equations using the Personal Algebra Tutor ®

Solving Systems of Linear Equations using the Personal Algebra Tutor illustrates the versatility of the program. The following example will be solved using five different methods.

The Personal Algebra Tutor:
  • Solves Systems of 2 or 3 Linear Equations Step-by-step
  • Covers methods used from beginning algebra thru preCalculus
  • Solutions of the following System of Linear Equations produced by the Personal Algebra Tutor will be shown.
        Y + X =  5                     Equation # 1 
        Y -  2X =  -  1                Equation # 2 
    	
    • Elementary Algebraic Operations
    • Graphing Solution
    • Determinants
    • Matrices Using Gauss Elimination and Back Substitution
    • Matrices Using Gauss-Jordan Elimination
  • All methods except the Graphing method can be used for systems of three Linear Equations. An output of the Personal Algebra Tutor is shown below for a System of three Linear Equations using Gauss-Jordan Elimination

Solution Using Elementary Operations

    Y + X =  5                     Equation # 1 
    Y -  2X =  -  1                Equation # 2 
 
          |  Next ...
          |    Subtract: Equation #2  - Equation #1  to eliminate Y
 
     3X =  6                       Eq. #1  and #2  Combined
 
          |  Next ...
          |    Divide by coefficient of variable to solve for X.
 
    X =  2                         Solution for X
 
          |    Next choose one of the original equations and
          |    substitute the value of X into that equation.
          |    ... Choosing Equation # 2 
 
   Y -  4 =  -  1
 
    Y =  3                         Solution for Y
 
          |    **** Solution Set:  X = 2,   Y = 3,  

Graphical Solution

    Y + X =  5                     Equation # 1 
    Y -  2X =  -  1                Equation # 2 

Plotting Y+X =5
 
  Plot the X-Intercept(Red):  Set y=0; Solve for x: (5,0)
  Plot the Y-Intercept(Blue): Set x=0; Solve for y: (0,5)
 
 
Draw the line between the points.
 
Finding Slope-Intercept Equation of Line.
  Solve for 'Y'.
 
  Y= -1 X +5  Slope_Intercept Equation
 
  Slope = -1  The coefficient of the X term.
  Y-Intercept = 5  The constant term.
 
Done.
Plotting     2X - Y =  1
 
  Plot the X-Intercept(Red):  Set y=0; Solve for x: (0.5,0)
  Plot the Y-Intercept(Blue): Set x=0; Solve for y: (0,-1)
 
 
Draw the line between the points.
 
Finding Slope-Intercept Equation of Line.
  Solve for 'Y'.
 
  Y= 2 X -1  Slope_Intercept Equation
 
  Slope = 2  The coefficient of the X term.
  Y-Intercept = -1  The constant term.
 
Done.
          |  The lines intersect at x=2, y=3
          |    The solution is x=2, y=3
 
*****  Problem Summary  *****
  Problem
 
                        Y+X =5    Equation #1
                   2X - Y =  1    Equation #2
  Solution
 
                 Solution: x=2, y=3

Solution Using Determinants

Y+X =5                          Eq 1
           
2X-Y=1                          Eq 2
         
  Evaluate the Coefficient Determinant, D
 
  
     |  1    1   |
 D = |           |
     |  2    -1  |
 
 
 D = 1*(-1) -(1)*(2) = -3
 
  
     |  5    1   |
 Dx= |           |
     |  1    -1  |
 
 
 Dx= 5*(-1) -(1)*(1) = -6
 
 Matrix Solution

      Dx     -6
 X = ---- = ----  = 2
      D      -3
 
 
 
     |  1    5   |
 Dy= |           |
     |  2    1   |
 
 
 Dy= 1*(1) -(5)*(2) = -9
 
 
      Dy     -9
 Y = ---- = ----  = 3
      D      -3

 ****  X = 2, Y = 3  Solution is Consistent.

Matrix Solution Using Gauss Elimination and Back Substitution

Y+X =5                          Eq 1
            
2X-Y=1                          Eq 2
         
 
     |  1    1    |
 
 Mc= |  2    -1   |  Coefficient Matrix
 
 
     |  1    1     |   5    |
 
 Ma= |  2    -1    |   1    |  Augmented Matrix
 
 
 Solve using the Augmented Matrix
 
  
     |  1   1    |   5   |
 
 M = |  0   3    |   9   |    2 R1 -  R2 -> R2
 
 
Force all Row leading non-zero coefficients to 1.
 
 
     |  1    1     |   5    |         X +1 Y  = 5
 
 M = |  0    1     |   3    |              Y  = 3
 
 
  Back Substitution
 
  Y = 3
 
  X = 5  -(1)Y = 5 -(1)(3)
 
  X = 2
 
 ****  X = 2, Y = 3  Solution is Consistent.
===========================================

Matrices Using Gauss-Jordan Elimination

     
Y+X =5                          Eq 1
             
2X-Y=1                          Eq 2
         
 
     |  1    1    |
 
 Mc= |  2    -1   |  Coefficient Matrix
 
 
     |  1    1     |   5    |
 
 Ma= |  2    -1    |   1    |  Augmented Matrix
 
 
 Solve using the Augmented Matrix
 
 
     |  1   1    |   5   |
 
 M = |  0   3    |   9   |    2 R1 -  R2 -> R2
 
 
 
     |  -3   0     |   -6   |     R2 - 3 R1 -> R1
 
 M = |  0    3     |   9    |
 
 
Force all Row leading non-zero coefficients to 1.
 
 
     |  1   0    |   2   |  X = 2
 
 Mj= |  0   1    |   3   |  Y = 3
 
 
 ****  X = 2, Y = 3  Solution is Consistent.

System of 3 Linear Equations Example Matrices Using Gauss-Jordan Elimination

         
2X-Y+Z=44                       Eq 1 
            
-X+3Y-2Z=-53                    Eq 2
        
          
5X-6Y-Z=19                      Eq 3
         
 
     |  2     -1    1     |
 
 Mc= |  -1    3     -2    |  Coefficient Matrix
 
     |  5     -6    -1    |
 
 
     |  2     -1    1      |   44    |
 
 Ma= |  -1    3     -2     |   -53   |  Augmented Matrix
 
     |  5     -6    -1     |   19    |
 
 
 Solve using the Augmented Matrix
 
 
 
     |  2     -1    1      |   44    |
 
 M = |  -1    3     -2     |   -53   |
 
     |  0     7     7      |   182   |    5 R1 - 2 R3 -> R3
 
 
 
     |  2     -1    1      |   44    |
 
 M = |  0     5     -3     |   -62   |     R1 + 2 R2 -> R2
 
     |  0     7     7      |   182   |
 
 
 
     |  2       -1      1        |   44      |
 
 M = |  0       5       -3       |   -62     |
 
     |  0       0       -56      |   -1344   |    7 R2 - 5 R3 -> R3
 
 
 
     |  -112    56      0        |   -1120   |    - R3 + 56 R1 -> R1
 
 M = |  0       5       -3       |   -62     |
 
     |  0       0       -56      |   -1344   |
 
 
 
     |  -112    56      0        |   -1120   |
 
 M = |  0       280     0        |   560     |    -3 R3 + 56 R2 -> R2
 
     |  0       0       -56      |   -1344   |
 
 
     |  -112    56      0        |   -1120   |    R1/280 -> R1
 
 M = |  0       1       0        |   2       |
 
     |  0       0       -56      |   -1344   |
 
 
 
     |  112     0       0        |   1232    |    56 R2 -  R1 -> R1
 
 M = |  0       1       0        |   2       |
 
     |  0       0       -56      |   -1344   |
 
 
Force all Row leading non-zero coefficients to 1.
 
 
     |  1    0    0     |   11   |  X = 11
 
 Mj= |  0    1    0     |   2    |  Y = 2
 
     |  0    0    1     |   24   |  Z = 24
 
 
 ****  X = 11, Y = 2, Z = 24  Solution is Consistent.

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