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Math Help - Solutions for linear equations

  1. #1
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    Solutions for linear equations

    I am trying to understand how to determine the difference between no solutions for a system and infinitely many solutions.

    I have the set of equations:

    x + 2y - z = 1
    2x + y + z = 1
    x - y + 2z = 1

    To start, I make sure the determinants do not equal zero.

    They do indeed equal zero so there are no solutions.

    My next system is :

    u + 2v -w = 2
    2u + v + w = 1
    u - v + 2w = - 1

    Again, I check and the determinants equal zero. So, I assume that there are no solutions.

    However, my textbook states that for this second system there are infinitely many solutions where u = - w, v = w + 1 and w can have any value. Please explain the difference to me!!!! Thanks Frostking
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  2. #2
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    Quote Originally Posted by Frostking View Post
    My next system is :

    u + 2v -w = 2
    2u + v + w = 1
    u - v + 2w = - 1

    Again, I check and the determinants equal zero. So, I assume that there are no solutions.

    However, my textbook states that for this second system there are infinitely many solutions where u = - w, v = w + 1 and w can have any value. Please explain the difference to me!!!! Thanks Frostking
    note that the first equation = 2nd equation - 3rd equation

    so the system really consists of just two equations w/ three variables, yielding an infinite number of solutions.
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  3. #3
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    Re: linear eqns with 3 variables

    See this sample question,,,,it will really help u


    Sample: Solve for x, y and z.

    2x - y + 6z = 1 (equation 1)
    x - y + z = 2 (equation 2)
    x + y + z = 1 (equation 3)

    Steps:

    1) Solve equation 1 for y.

    After applying algebra, equation 1 becomes y = 2x + 6z - 1.

    2) Plug the value of y or (2x + 6z -1) into equations 2 and 3.

    Equation 2:

    x - y + z = 2

    x -(2x + 6z -1) + z = 2

    x -2x - 6z + 1 = 2

    -x - 6z = 2-1

    -x -6z = 1

    Equation 3:

    x + y + z = 1

    x + 2x + 6z -1 + z = 1

    3x + 7z -1 = 1

    3x + 7z = -1+1

    3x + 7z = 0

    We now have come down to a system of two equations and two variables.

    Here are the two equations and two variables:

    -x - 6z = 1 (equation A)
    3x + 7z = 0 (equation B)

    3) Solve equation A for x.

    -x - 6z = 1
    -x = 6z + 1
    x = -6z - 1

    4) Plug the value for x into equation B to solve for z.

    3x + 7z = 0

    3(-6z - 1) + 7z = 0

    -18z - 3 + 7z = 0

    -11z - 3 = 0

    -11z = 3

    z = - 11/3

    5) Now plug the value for z into equation A to find x.

    -x - 6z = 1

    -x -6(-11/3) = 1

    -x + 22 = 1

    -x = -22 + 1

    -x = -21

    x = 21

    Lastly or step 6, go back to ANY of the original 3 equations to find the value of y by substituting what you found for x and z. I will select equation 3.

    Equation 3:
    x + y + z = 1

    21 + y -11/3 = 1

    52/3 + y = 1

    y = -52/3 + 1

    y = -49/3

    Final answer: x = 21, y = -49/3 and z = -11/3
    Last edited by CaptainBlack; October 11th 2009 at 01:24 PM.
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