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Math Help - One more I need help linear systems

  1. #1
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    One more I need help linear systems

    On this problem, I think I'm more confused about the possible answers than the problem itself. How would you suggest I go about it?

    Solve the system
    {x+2y-2z=-1
    {x+3y+z=10
    {2x+6y+2z=20

    solutions:
    a. (8c-23, -3c+10, c) for any real number c.
    b. (8c-22, -3c+10, c) for any real number c.
    c. (8c-22, -3c+11, c) for any real number c.
    d. (8c-23, -3c+11, c) for any real number c.
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  2. #2
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    Quote Originally Posted by mike1 View Post
    On this problem, I think I'm more confused about the possible answers than the problem itself. How would you suggest I go about it?

    Solve the system
    {x+2y-2z=-1
    {x+3y+z=10
    {2x+6y+2z=20
    .
    [ 1 2 -2 -1]
    [ 1 3 1 10]
    [ 2 6 2 20]
    Divide the third row by 2,
    [ 1 2 -2 -1]
    [ 1 3 1 10]
    [ 1 3 1 10]
    Subtract 2nd from 3rd and subtract 1st from 2nd
    [ 1 2 -2 -1]
    [ 0 1 3 11]
    [ 0 0 0 0]
    Multiply 2nd by (-2) add to 1st,
    [1 0 -8 -23]
    [0 1 3 11]
    Thus,
    x-8z=-23
    y+3z=11
    Solve,
    x=8z-23
    y=-3z+11
    And "z" can be anything thus z=c
    Thus,
    (x,y,z)=(8c-23,-3c+11,c)
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  3. #3
    MHF Contributor
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    Quote Originally Posted by mike1 View Post
    On this problem, I think I'm more confused about the possible answers than the problem itself. How would you suggest I go about it?

    Solve the system
    {x+2y-2z=-1
    {x+3y+z=10
    {2x+6y+2z=20

    solutions:
    a. (8c-23, -3c+10, c) for any real number c.
    b. (8c-22, -3c+10, c) for any real number c.
    c. (8c-22, -3c+11, c) for any real number c.
    d. (8c-23, -3c+11, c) for any real number c.
    I am trained to look first for what is being asked for in a problem before I start thinking for the solution.
    So I applied that in your question/problem above.
    Being asked for is Solve the system, and possible answers are in terms of "c".

    Where is "c" in the system?

    ------------------
    Forget about "c".
    The system has 3 unknowns, x,y,z, and 3 equations.
    Good.

    But the 2nd and 3rd equations are one and the same--only wearing different clothes--so in reality there are only 2 independent equations with 3 unknowns.
    No good.

    For a system to be "solvable", or with unique answers/solution, the number of independent equations must equal the number of unknowns.
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  4. #4
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    Hello, Mike!

    Solve the system : . \begin{array}{ccc}x+2y-2z &= \;-1 \\x+3y+z & =\;10 \\2x+6y+2z& = \:20\end{array}

    \begin{array}{cc}a.\;(8c-23, -3c+10, c) & b.\;(8c-22, -3c+10, c) \\  c.\;(8c-22, -3c+11, c) & d.\;(8c-23, -3c+11, c)\end{array} . for any real number c.

    Since the second and third equations are equivalent,
    . . we have the system: . \begin{array}{cc}x + 2y - 2x & =\:-1 \\ x + 3y + z & =\:10\end{array}\begin{array}{cc}(1)\\(2)\end{arra  y}

    Subtract (1) from (2): . y + 3z\:=\:11\quad\Rightarrow\quad y \:=\:-3z + 11

    Substitute into (1): . x + 2(-3z + 11) - 2x \:=\:-1\quad\Rightarrow\quad x \:=\:8x - 23

    We have x,\,y,\,z expressed as functions of z: . \begin{array}{ccc}x \:= & 8z - 23 \\ y \:= & -3z + 11 \\ z \:= & z\end{array}

    On the right, replace z with a parameter c.

    And we have: . \begin{Bmatrix}x \:= & 8c - 23 \\ y \:= & -3z + 11 \\ z \:= & c\end{Bmatrix} . . . answer d

    This is a parametric representation of all the solutions to the system.

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