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Math Help - extra credit problem i got :(

  1. #1
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    im stuck - its for some extra credits

    x+y+z+a = 16480

    x=y+ 1\2z
    y=a-z
    z=3x
    a=y+z
    find a,x,y,z

    See i try subbing into the top but i always have too many variable and anything else at the point undos something. I think i am missing a place where i can simplify. Please help. Any tips or anything will be appreciated.
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  2. #2
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    extra credit problem i got :(

    x+y+z+a = 16480

    x=y+ 1\2z
    y=a-z
    z=3x
    a=y+z
    find a,x,y,z

    See i try subbing into the top but i always have too many variable and anything else at the point undos something. I think i am missing a place where i can simplify. Please help. Any tips or anything will be appreciated.
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  3. #3
    Senior Member OReilly's Avatar
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    Quote Originally Posted by spiderz
    x+y+z+a = 16480

    x=y+ 1\2z
    y=a-z
    z=3x
    a=y+z
    find a,x,y,z

    See i try subbing into the top but i always have too many variable and anything else at the point undos something. I think i am missing a place where i can simplify. Please help. Any tips or anything will be appreciated.

    Can you please verify is x:
     <br />
 x = y + \frac{1}{{2z}} <br />
    or
     <br />
 x = \frac{{y + 1}}{{2z}}  <br />
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  4. #4
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    Quote Originally Posted by OReilly
    Can you please verify is x:
     <br />
 x = y + \frac{1}{{2z}} <br />
    or
     <br />
 x = \frac{{y + 1}}{{2z}}  <br />
    i meant one half times Z or .5z

    so it should say Y plus half of Z
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  5. #5
    Senior Member OReilly's Avatar
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    Quote Originally Posted by spiderz
    i meant one half times Z or .5z

    so it should say Y plus half of Z
    You mean
    <br />
x = y + \frac{z}{2}<br />
    ?
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  6. #6
    Senior Member OReilly's Avatar
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    If
    <br />
x = y + \frac{z}{2}<br />
    is correct than this is solution:

    \begin{array}{l}<br />
 x = y + \frac{z}{2} \\ <br />
 x = y + \frac{{3x}}{2} \\ <br />
 x = y + \frac{{3x}}{2} \\ <br />
 x = \frac{{2y + 3x}}{2} \\ <br />
 2x = 2y + 3x \\ <br />
 2x - 3x = 2y \\ <br />
  - x = 2y \\ <br />
 x =  - 2y \\ <br />
 y =  - \frac{x}{2} \\ <br />
 \end{array}<br />

    \begin{array}{l}<br />
 a =  - \frac{x}{2} + z \\ <br />
 a =  - \frac{x}{2} + 3x \\ <br />
 a = \frac{{5x}}{2} \\ <br />
 \end{array}<br />

    \begin{array}{l}<br />
 x + \frac{{5x}}{2} + \frac{{5x}}{2} = 16480 \\ <br />
 6x = 16480 \\ <br />
 x = \frac{{16480}}{6} = \frac{{8240}}{3} \\ <br />
 \end{array}<br />

    \begin{array}{l}<br />
 z = 3\frac{{8240}}{3} \\ <br />
 z = 8240 \\ <br />
 \end{array}<br />

    \begin{array}{l}<br />
 y =  - \frac{{8240}}{6} \\ <br />
  \\ <br />
 a =  - \frac{{8240}}{6} + 8240 = \frac{{41200}}{6} = \frac{{20600}}{3} \\ <br />
 \end{array}<br />

    So finally we can fill all variables in equation x+y+z+a=16480:
    \frac{{8240}}{3} - \frac{{8240}}{6} + 8240 + \frac{{20600}}{3} = 16480

    \frac{{16480 - 8240 + 49440 + 41200}}{6} = 16480
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  7. #7
    Senior Member OReilly's Avatar
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    Please, don't post twice same message.
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  8. #8
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    Quote Originally Posted by OReilly
    If
    <br />
x = y + \frac{z}{2}<br />
    is correct than this is solution:

    \begin{array}{l}<br />
 x = y + \frac{z}{2} \\ <br />
 x = y + \frac{{3x}}{2} \\ <br />
 x = y + \frac{{3x}}{2} \\ <br />
 x = \frac{{2y + 3x}}{2} \\ <br />
 2x = 2y + 3x \\ <br />
 2x - 3x = 2y \\ <br />
  - x = 2y \\ <br />
 x =  - 2y \\ <br />
 y =  - \frac{x}{2} \\ <br />
 \end{array}<br />

    \begin{array}{l}<br />
 a =  - \frac{x}{2} + z \\ <br />
 a =  - \frac{x}{2} + 3x \\ <br />
 a = \frac{{5x}}{2} \\ <br />
 \end{array}<br />

    \begin{array}{l}<br />
 x + \frac{{5x}}{2} + \frac{{5x}}{2} = 16480 \\ <br />
 6x = 16480 \\ <br />
 x = \frac{{16480}}{6} = \frac{{8240}}{3} \\ <br />
 \end{array}<br />

    \begin{array}{l}<br />
 z = 3\frac{{8240}}{3} \\ <br />
 z = 8240 \\ <br />
 \end{array}<br />

    \begin{array}{l}<br />
 y =  - \frac{{8240}}{6} \\ <br />
  \\ <br />
 a =  - \frac{{8240}}{6} + 8240 = \frac{{41200}}{6} = \frac{{20600}}{3} \\ <br />
 \end{array}<br />

    So finally we can fill all variables in equation x+y+z+a=16480:
    \frac{{8240}}{3} - \frac{{8240}}{6} + 8240 + \frac{{20600}}{3} = 16480

    \frac{{16480 - 8240 + 49440 + 41200}}{6} = 16480
    thank you *bows*
    does this show that y = z?
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  9. #9
    Senior Member OReilly's Avatar
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    Quote Originally Posted by spiderz
    thank you *bows*
    does this show that y = z?
    No,
    <br />
y = - \frac{{8240}}{6}<br />
    and
    <br />
z = 8240<br />
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  10. #10
    Super Member malaygoel's Avatar
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    India
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    Quote Originally Posted by spiderz
    x+y+z+a = 16480

    x=y+ 1\2z
    y=a-z
    z=3x
    a=y+z
    find a,x,y,z

    See i try subbing into the top but i always have too many variable and anything else at the point undos something. I think i am missing a place where i can simplify. Please help. Any tips or anything will be appreciated.
    Equations y=a-z and a=y+z are just the csame equations.
    The trick is to express everything in terms of a variable you want. I choose z.
    z=3x hence x=z/3
    find the value of using the equation x= y + 1/2z
    find a =y+z
    then substitute value of x,y and a (in terms of z) in the first equation and the value of z.
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  11. #11
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    (Moderator Note: I merged the threads, I was going to delete it but malaygoel made a post and I considered it to mean to delete it and have his post gone forever).
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