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Math Help - Intermediate algebra - "Solve each systems"

  1. #1
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    Unhappy Intermediate algebra - "Solve each systems"

    Hi, I've been trying to solve this for over an hour now by extending the method of elimination like the textbook says. Very frustrated.
    Please help.

    Solve each systems. Express the solution in the form (x,y,z,w).

    x+y+z-w=5
    2x+y-z+w=3
    x-2y+3z+w=18
    -x-y+z+2w=8

    The answer is (2,1,5,3)
    I don't know how to get there.
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  2. #2
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    Quote Originally Posted by Justiceapple View Post
    Hi, I've been trying to solve this for over an hour now by extending the method of elimination like the textbook says. Very frustrated.
    Please help.

    Solve each systems. Express the solution in the form (x,y,z,w).

    x+y+z-w=5
    2x+y-z+w=3
    x-2y+3z+w=18
    -x-y+z+2w=8

    The answer is (2,1,5,3)
    I don't know how to get there.
    x + y + z - w = 5
    2x + y - z + w = 3
    x - 2y + 3z + w = 18
    -x - y + z - 2w = 8.

    Apply R_2 - 2R_1 \to R_2, R_3 - R_1 \to R_3, R_4 + R_1 \to R_1.


    x + y + z - w = 5
    -y - 3z + 3w = -7
    -3y + 2z + 2w = 13
    2z - 3w = 13

    Apply -R_2 \to R_2


    x + y + z - w = 5
    y + 3z - 3w = 7
    -3y + 2z + 2w = 13
    2z - 3w = 13

    Apply R_3 + 3R_2 \to R_3


    x + y + z - w = 5
    y + 3z - 3w = 7
    11z - 7w = 34
    2z - 3w = 13

    Interchange R_3 and R_4.


    x + y + z - w = 5
    y + 3z - 3w = 7
    2z - 3w = 13
    11z - 3w = 13

    Apply \frac{1}{2}R_3


    x + y + z - w = 5
    y + 3z - 3w = 7
    z - \frac{3}{2}w = \frac{13}{2}
    11z - 3w = 13

    Apply R_4 - 11R_3


    x + y + z - w = 5
    y + 3z - 3w = 7
    z - \frac{3}{2}w = \frac{13}{2}
    \frac{27}{2}w = -\frac{117}{2}



    So w = \frac{2}{27}\left(-\frac{117}{2}\right)

    w = -\frac{13}{3}.


    z - \frac{3}{2}w = \frac{13}{2}

    z - \frac{3}{2}\left(-\frac{13}{3}\right) = \frac{13}{2}

    z + \frac{13}{2} = \frac{13}{2}

    z = 0.


    y + 3z - 3w = 7

    y + 3(0) - 3\left(-\frac{13}{3}\right) = 7

    y + 13 = 7

    y = -6.


    x + y + z - w = 5

    x - 6 + 0 + \frac{13}{3} = 5

    x - \frac{5}{3} = 5

    x = \frac{20}{3}.



    To check, sub into original Eq 2:

    2x+y-z+w = 2\left(\frac{20}{3}\right) - 6 - 0 - \frac{13}{3}

     = \frac{40}{3} - \frac{18}{3} - \frac{13}{3}

     = \frac{9}{3}

     = 3 as required.


    So (x, y, z, w) = \left(\frac{20}{3}, 6, 0, -\frac{13}{3}\right).

    The solution your text book gives is incorrect.
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  3. #3
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    Oh no, you wrote down the problem wrong

    The R4 is -x-y+z+2w=8 . It's +2w.

    But I think I kind of get what to do now. Let me work on this problem first thing in the morning tomorrow. My head hurts from studying Logic for another class right now....

    Thank you so much. When I saw that you replied, I screamed like I won a lottery or something. I appreciate your kindness very much.

    P.S. I studied Math for 6 hours today ...
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  4. #4
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    Oh hahahaha, yes I copied it down wrong.

    But you see the basic structure now.

    Try to solve yours the same way.
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  5. #5
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    Quote Originally Posted by Justiceapple View Post
    Oh no, you wrote down the problem wrong
    That's pretty funny.
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  6. #6
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    Quote Originally Posted by VonNemo19 View Post
    That's pretty funny.
    So I think the score is Prove It -1, Von Nemo -2
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