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Anonymous asked in Education & ReferenceHomework Help · 7 years ago

Find the values of k?

Find the values of k such that the following system has a) no solution, or b) a unique one.

x+2y+2z=1

y+kz=1

-x+y+kz=k

2 Answers

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  • 7 years ago
    Favorite Answer

       x  +  2y  +  2z  =  1 ... equation 1 (given)

        - x  +         y  +  kz  =  k ... equation 3 (given)

       ——————————<ADD>

         3y  +  (2 + k)z  =  1 + k ... equation A

        y + kz = 1 ... equation 2 (given)

      3y + 3kz = 3 ... equation 2 (multiplied by 3)

                    3y  +  (2 + k)z  =  1 + k ... equation A

                    3y  +     3kz  =  3 ... equation 2 (×3)

                  ——————————<subtract>

                     (2 + k – 3k)z  =  k – 2

                        z  =  - (k – 2)  ⁄  [2(k – 1) ]

                          ... so: k ≠ 1

    From this "z" solution and the given equation 2, you can solve for "y" in terms of "k"

    Then "x" in terms of "k" can be found using the given equation 1 or 3

    The solutions for y, and x also have this: k ≠ 1: restriction

     Answer:

         There are NO solutions when: k  = 1

         There is a unique solution for all OTHER real "k" values

         Example: for k  =  2 ,  x = -1 , y = 1 , z = 0

         so there are an infinite number of unique solutions.

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  • 7 years ago

    y = 1 - kz

    Substitute into the other two eq'n.

    x + 2(1 - kz) + 2z = 1

    -x + 1 - kz + kz = k

    x + 2 - 2kz + 2z = 1

    -x + 1 = k

    Add

    2 + 2z (1- k) = 1

    1 = k

    Substitue

    2 + 2z ( k - k ) = 1

    2 + 2z(0) = 1

    2 + 0 = 1

    2 = 1

    Answer NO sol'n !!!!

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