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Math Help - System of Equations

  1. #1
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    System of Equations

    I cant finish this problem but have to solve it today!!!
    Solve for all real and b equations


    My progress:
    to make this simple the top is (1) the midle is (2) and the bottom is (3)
    i add (2) with (1) to get 3x+2az=2a
    then i add (2) and (3) and get
    6x + az + 3z = a + b
    now im lost and cant solve it plz i need fast help :/
    Last edited by skeeter; November 4th 2012 at 10:39 AM. Reason: removed begging in the title
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  2. #2
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    Re: Need fast help!

    Quote Originally Posted by Petrus View Post
    I cant finish this problem but have to solve it today!!!
    Solve for all real and b equations


    My progress:
    to make this simple the top is (1) the midle is (2) and the bottom is (3)
    i add (2) with (1) to get 3x+2az=2a
    then i add (2) and (3) and get
    6x + az + 3z = a + b
    now im lost and cant solve it plz i need fast help :/
    Calculate 2(1) (1) is the equation number.
    Then do 2(1) - (2) = 0x + 3y + a = 0

    Notice that the result does not have an x in it. Now do the same thing for the 3rd equation:
    4(1) - (3)
    Again the x term is missing. Can you finish this process?

    And please don't beg. It's rude.

    -Dan
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  3. #3
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    Re: Need fast help!

    Quote Originally Posted by topsquark View Post
    Calculate 2(1) (1) is the equation number.
    Then do 2(1) - (2) = 0x + 3y + a = 0

    Notice that the result does not have an x in it. Now do the same thing for the 3rd equation:
    4(1) - (3)
    Again the x term is missing. Can you finish this process?

    And please don't beg. It's rude.

    -Dan
    i got confused on this Calculate 2(1) (1) is the equation number. do u mean multiplicate 3x+2az=2a with x+y+az?
    sorry but got confused on ur post, can u write the equation insted of 2(1)im so panic right now have to solve this fast :S dead line is close :/ insted of
    Last edited by Petrus; November 4th 2012 at 10:22 AM.
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  4. #4
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    Re: System of Equations

    Quote Originally Posted by Petrus View Post
    I cant finish this problem but have to solve it today!!!
    Solve for all real and b equations
    I would use a matrix inverse.
    \left[ {\begin{array}{rrr}   1 & 1 & a  \\   2 & { - 1} & a  \\   4 & 1 & 3  \\ \end{array} } \right]^{ - 1}  = \frac{1}{{9\left( {a - 1} \right)}}\left[ {\begin{array}{rrr}   { - (a + 3)} & {(a - 3)} & {2a}  \\   {2(2a - 3)} & { - (4a - 3)} & a  \\   6 & 3 & { - 3}  \\ \end{array} } \right]
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  5. #5
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    Re: System of Equations

    Quote Originally Posted by Plato View Post
    I would use a matrix inverse.
    \left[ {\begin{array}{rrr}   1 & 1 & a  \\   2 & { - 1} & a  \\   4 & 1 & 3  \\ \end{array} } \right]^{ - 1}  = \frac{1}{{9\left( {a - 1} \right)}}\left[ {\begin{array}{rrr}   { - (a + 3)} & {(a - 3)} & {2a}  \\   {2(2a - 3)} & { - (4a - 3)} & a  \\   6 & 3 & { - 3}  \\ \end{array} } \right]
    i wanna solve this without matrix
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  6. #6
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    Re: System of Equations

    Quote Originally Posted by Petrus View Post
    i wanna solve this without matrix
    That is fine. It harder without it.
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    Re: System of Equations

    Quote Originally Posted by Plato View Post
    That is fine. It harder without it.
    yeah but im not "good" in matrix i need to sit and watch and read alot about matrix but i wanna solve this so i understand can u help me :/?
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  8. #8
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    Re: System of Equations

    The first equation is x+ y+ az= a and the second equation is 2x- y+ az= a. Subtracting the second equation from the first, -x+ 2y= 0 or x= 2y. Replacing x with 2y in the equations gives 2y+ y+ az= 3y+ az= a, 2x- y+ az= 4y- y+ az or 3y+ az=a again, and 4x+ 3y+ 3z= 8y+ 3y+ 3z= 11y+ 3z= b. If you multiply 3y+ az= a by 11, you get 33y+ 11az= 11a and if you multiply 11y+ 3z= b by 3, you get 33y+ 9z= 3b.

    So now you have 33y+ 11az= 11a and 33y+ 9z= 3b. If you subtract the second of those from the first, what do you get?
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    Re: System of Equations

    Quote Originally Posted by HallsofIvy View Post
    The first equation is x+ y+ az= a and the second equation is 2x- y+ az= a. Subtracting the second equation from the first, -x+ 2y= 0 or x= 2y. Replacing x with 2y in the equations gives 2y+ y+ az= 3y+ az= a, 2x- y+ az= 4y- y+ az or 3y+ az=a again, and 4x+ 3y+ 3z= 8y+ 3y+ 3z= 11y+ 3z= b. If you multiply 3y+ az= a by 11, you get 33y+ 11az= 11a and if you multiply 11y+ 3z= b by 3, you get 33y+ 9z= 3b.

    So now you have 33y+ 11az= 11a and 33y+ 9z= 3b. If you subtract the second of those from the first, what do you get?
    9z-11az=3b-11a
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  10. #10
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    Re: System of Equations

    Quote Originally Posted by Petrus View Post
    9z-11az=3b-11a
    z(9-11a) = 3b-11a

    continue ...
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    Re: System of Equations

    z = (3b-11a)/(9-11a) what shall i do next =S ?
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    Re: System of Equations

    Quote Originally Posted by Petrus View Post
    z = (3b-11a)/(9-11a) what shall i do next =S ?
    solve for x and y ...
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  13. #13
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    Re: System of Equations

    y = (a*b-3a)/(11a-9)
    x= (2a(b-3))/(11a-9)

    is this correct?
    Last edited by Petrus; November 4th 2012 at 02:41 PM.
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  14. #14
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    Re: System of Equations

    Quote Originally Posted by Petrus View Post
    y = (a*b-3a)/(11a-9)
    x= (2a(b-3))/(11a-9)

    is this correct?
    it's your problem, you check it ...
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