3

							3.6 Solving Systems of Linear
Equations in Three Variables
Warm-Up


   No Solution



   Infinitely many
   solutions
 Here is a system of three linear
  equations in three variables:
 x  2 y  3 z  3
                         The ordered triple (2,-1,1)
                         is a solution to this

 2 x  5 y  4 z  13
                         system since it is a
                         solution to all three
5 x  4 y  z  5
                        equations.


2  2(1)  3(1)  2  2  3  3

2(2)  5(1)  4(1)  4  5  4  13
5(2)  4(1)  1  10  4  1  5

    The graph of a linear equation in
    three variables is a plane. Three
     planes in space can intersect in
         different ways (pg 152).
                The planes could               The planes could
                intersect in a                 intersect in a line.
                single point.                  The system has
                The system has                 infinitely many
                exactly one                    solutions
                solution
The planes could have NO point of intersection. The left figure shows planes that
intersect pairwise, but all 3 do not have a common point of intersection. The right
figure shows parallel planes. Each system has NO solution.
The linear combination method
in lesson 3.2 can be extended
  to solve a system of linear
 equations in three variables.
                 Solve this system
3 x  2 y  4 z  11   Our strategy will be to use two
                       of the equations to eliminate
2 x  y  3z  4       one of the variables.
5 x  3 y  5 z  1

                        We will then use two other
                        equations to eliminate the
                        same variable.




                        Once we have two equations
                        with two variables, we can
                        use the technique we learned
                        in lesson 3.2
                 Solve this system
3 x  2 y  4 z  11   Our strategy will be to use two
                       of the equations to eliminate
2 x  y  3z  4       one of the variables.
5 x  3 y  5 z  1

                        We will then use two other
                        equations to eliminate the
                        same variable.




                        Once we have two equations
                        with two variables, we can
                        use the technique we learned
                        in lesson 3.2
                           Solve this system
      3 x  2 y  4 z  11     Equation 1
      
-3
2     2 x  y  3z  4         Equation 2

      5 x  3 y  5 z  1
                               Equation 3

     Multiply Eq. 2 by 2 and add it to Eq. 1.
                                                      Solve this new system of linear
     Save this result
                                                      equation in two variables.
     3x  2 y  4 z  11
                                                     Multiply the bottom eq. by 7 and
     4 x  2 y  6 z  8                             add it to the top eq.

      7x         +10z = 19                            7 x  10 z  19
                                                      
                                                      7 x  28z  91
 Now multiply Eq. 2 by -3 and
 add it to Eq. 3. Save this
 result.
 6 x  3 y  9 z  12                                       -18z=-72 or z = 4
                                  Substituting z=4 into either of the new equations will
 ...5 x  3 y  5 z  1          give x = -3……finally substituting these values into any
                                   of the original equations give y = 2.
        -x       -4z = -13                      Our final solution is (-3,2,4)
   Here is a system with No
Solution. Watch what happens
    when we try to solve it.
x  y  z  2          Equation 1

3x  3 y  3z  14     Equation 2
x  2 y  z  4
                       Equation 3




3x  3 y  3z  6       Add -3 times Eq 1 to Eq 2.

..3x  3 y  3z  14          Since this is a false equation, you can
                               conclude the original system of equations
               0=8             has no solution.
        Here is a system with MANY
       solutions. Watch what happens
            when we try to solve it.
  x  y  z  2     Equation 1
  
  x  y  z  2     Equation 2
  2 x  2 y  z  4 Equation 3
                                       Solving this new system of two equation
                                        by adding -3 times the first eq. to 2 times
 x  y  z  2    Add Eq. 1 to Eq. 2   the second eq. produces the identity 0 =
                                       0. So, the system has infinitely many
 x  y  z  2                         solution.
   2x + 2y    =4    New EQ. 1            You could describe the solution this way:
                                         divide New Eq 1 by 2 to get x+y=2, or
 x  y  z  2 Add Eq 2 to Eq 3         y=-x+2. Substituting this into the
                                        orignial Equation 1 produces z = 0. So
2 x  2 y  z  4                       any ordered triple of the form (x, -x+2,0) is
                                         a solution. For example (0,2,0) and
 3x +3y      =6                          (2,0,0) are solutions.
                      New EQ 2
                  Substitution Method
 x  y  z  24

5 x  3 y  z  56
x  y  z

Since x+y=z, substitute this for z in the
first two equations

 x  y  ( x  y)  24

5 x  3 y  ( x  y)  56
 Simplify

                                Finally, solve this linear system of two equations
2 x  2 y  24                 and two variables to get x = 4 and y =8

6 x  4 y  56                Since z=x+y, z = 12. Our final solution is (4,8,12)
    Assignment 3.6

Pg 157, 1-16 (1st column)
        19 & 20
         22-24