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Math Help - Solving systems of equations

  1. #1
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    Solving systems of equations

    Hi,

    I'm looking for help with solving these two systems of equations (any method can be used):

    1.

    -4x - 5y - 8z = 268
    -10x + 7y + 6z = -786
    -12x + 9y - 2z = -548

    2.

    -5x - 4y + 12z = -140
    9x - 8y + z = -649
    3x - 2y + 6z = -290

    Thanks for your help.
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  2. #2
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    What method have you chosen? Where are you stuck?

    Please be complete. Thank you!
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  3. #3
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    Quote Originally Posted by stapel View Post
    What method have you chosen? Where are you stuck?

    Please be complete. Thank you!
    I'm pretty sure what I used was substitution when trying to solve them. Here's my work so far for number 2:

    -5x - 4y + 12z = -140
    9x - 8y + z = -649
    3x - 2y + 6z = -290

    I multiplied row 1 by -1

    5x + 4y - 12z = 140

    Then I multiplied row 3 by 2

    6x - 4y + 12z = -580

    Added them together to get:

    11x = -440, which means x = -40

    What I did next was substitute -40 in for x in one of the equations,

    9(-40) 8y + z = -649
    -360 8y + z = -649
    -8y + z = - 289
    z = 8y 289

    is what I got. From there I'm a little unsure as to how I should continue.
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  4. #4
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    Once you get past two equations in two variables, it's often better to add equations together. For instance, you have:

    R_1:\, -5x\, -\, 4y\, +\, 12z\, =\, 140

    R_2:\, 9x\, -\, 8y\, +\, 1z\, =\, -649

    R_3:\, 3x\, -\, 2y\, +\, 6z\, =\, -290

    I see right away that I can multiply R_3 by -3 and add this to R_2 to get:

    R_1:\, -5x\, -\, 4y\, +\, 12z\, =\, 140

    R_2:\, 0x\, -\, 3y\, -\, 17z\, =\, 221

    R_3:\, 3x\, -\, 2y\, +\, 6z\, =\, -290

    Because I really would rather leave fractions until the end, and since twice three is close to five, I'll multiply R_3 by 2 and add the result to R_1. This gives me:

    R_1:\, 1x\, -\, 8y\, +\, 24z\, =\, -440

    R_2:\, 0x\, -\, 3y\, -\, 17z\, =\, 221

    R_3:\, 3x\, -\, 2y\, +\, 6z\, =\, -290

    Now I'll multiply R_1 by -3 and add the result to R_3. This gives me:

    R_1:\, 1x\, -\, 8y\, +\, 24z\, =\, -440

    R_2:\, 0x\, -\, 3y\, -\, 17z\, =\, 221

    R_3:\, 0x\, +\, 22y\, -\, 66z\, =\, 1030

    For simplicity's sake, I'll divide R_3 by 2 to get:

    R_1:\, 1x\, -\, 8y\, +\, 24z\, =\, -440

    R_2:\, 0x\, -\, 3y\, -\, 17z\, =\, 221

    R_3:\, 0x\, +\, 11y\, -\, 33z\, =\, 515


    The last two rows are now a system of two equations in two unknowns. Solve that, plug the values for y and z into the first equation, and simplify to find the value of x.
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