Results 1 to 11 of 11

Thread: extra credit problem i got :(

  1. #1
    Newbie
    Joined
    Jun 2006
    Posts
    4

    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.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Newbie
    Joined
    Jun 2006
    Posts
    4

    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.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Senior Member OReilly's Avatar
    Joined
    Mar 2006
    Posts
    340
    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:
    $\displaystyle
    x = y + \frac{1}{{2z}}
    $
    or
    $\displaystyle
    x = \frac{{y + 1}}{{2z}}
    $
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    Jun 2006
    Posts
    4
    Quote Originally Posted by OReilly
    Can you please verify is x:
    $\displaystyle
    x = y + \frac{1}{{2z}}
    $
    or
    $\displaystyle
    x = \frac{{y + 1}}{{2z}}
    $
    i meant one half times Z or .5z

    so it should say Y plus half of Z
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Senior Member OReilly's Avatar
    Joined
    Mar 2006
    Posts
    340
    Quote Originally Posted by spiderz
    i meant one half times Z or .5z

    so it should say Y plus half of Z
    You mean
    $\displaystyle
    x = y + \frac{z}{2}
    $
    ?
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Senior Member OReilly's Avatar
    Joined
    Mar 2006
    Posts
    340
    If
    $\displaystyle
    x = y + \frac{z}{2}
    $
    is correct than this is solution:

    $\displaystyle \begin{array}{l}
    x = y + \frac{z}{2} \\
    x = y + \frac{{3x}}{2} \\
    x = y + \frac{{3x}}{2} \\
    x = \frac{{2y + 3x}}{2} \\
    2x = 2y + 3x \\
    2x - 3x = 2y \\
    - x = 2y \\
    x = - 2y \\
    y = - \frac{x}{2} \\
    \end{array}
    $

    $\displaystyle \begin{array}{l}
    a = - \frac{x}{2} + z \\
    a = - \frac{x}{2} + 3x \\
    a = \frac{{5x}}{2} \\
    \end{array}
    $

    $\displaystyle \begin{array}{l}
    x + \frac{{5x}}{2} + \frac{{5x}}{2} = 16480 \\
    6x = 16480 \\
    x = \frac{{16480}}{6} = \frac{{8240}}{3} \\
    \end{array}
    $

    $\displaystyle \begin{array}{l}
    z = 3\frac{{8240}}{3} \\
    z = 8240 \\
    \end{array}
    $

    $\displaystyle \begin{array}{l}
    y = - \frac{{8240}}{6} \\
    \\
    a = - \frac{{8240}}{6} + 8240 = \frac{{41200}}{6} = \frac{{20600}}{3} \\
    \end{array}
    $

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

    $\displaystyle \frac{{16480 - 8240 + 49440 + 41200}}{6} = 16480$
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Senior Member OReilly's Avatar
    Joined
    Mar 2006
    Posts
    340
    Please, don't post twice same message.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Newbie
    Joined
    Jun 2006
    Posts
    4
    Quote Originally Posted by OReilly
    If
    $\displaystyle
    x = y + \frac{z}{2}
    $
    is correct than this is solution:

    $\displaystyle \begin{array}{l}
    x = y + \frac{z}{2} \\
    x = y + \frac{{3x}}{2} \\
    x = y + \frac{{3x}}{2} \\
    x = \frac{{2y + 3x}}{2} \\
    2x = 2y + 3x \\
    2x - 3x = 2y \\
    - x = 2y \\
    x = - 2y \\
    y = - \frac{x}{2} \\
    \end{array}
    $

    $\displaystyle \begin{array}{l}
    a = - \frac{x}{2} + z \\
    a = - \frac{x}{2} + 3x \\
    a = \frac{{5x}}{2} \\
    \end{array}
    $

    $\displaystyle \begin{array}{l}
    x + \frac{{5x}}{2} + \frac{{5x}}{2} = 16480 \\
    6x = 16480 \\
    x = \frac{{16480}}{6} = \frac{{8240}}{3} \\
    \end{array}
    $

    $\displaystyle \begin{array}{l}
    z = 3\frac{{8240}}{3} \\
    z = 8240 \\
    \end{array}
    $

    $\displaystyle \begin{array}{l}
    y = - \frac{{8240}}{6} \\
    \\
    a = - \frac{{8240}}{6} + 8240 = \frac{{41200}}{6} = \frac{{20600}}{3} \\
    \end{array}
    $

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

    $\displaystyle \frac{{16480 - 8240 + 49440 + 41200}}{6} = 16480$
    thank you *bows*
    does this show that y = z?
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Senior Member OReilly's Avatar
    Joined
    Mar 2006
    Posts
    340
    Quote Originally Posted by spiderz
    thank you *bows*
    does this show that y = z?
    No,
    $\displaystyle
    y = - \frac{{8240}}{6}
    $
    and
    $\displaystyle
    z = 8240
    $
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Super Member malaygoel's Avatar
    Joined
    May 2006
    From
    India
    Posts
    648
    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.
    Follow Math Help Forum on Facebook and Google+

  11. #11
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    10
    (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).
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Extra Credit Problem I dont understand at all!
    Posted in the Trigonometry Forum
    Replies: 1
    Last Post: Dec 9th 2011, 02:49 PM
  2. could use some help on an extra credit problem
    Posted in the Calculus Forum
    Replies: 1
    Last Post: May 10th 2011, 03:48 PM
  3. Calc 2 extra credit problem
    Posted in the Calculus Forum
    Replies: 4
    Last Post: Mar 17th 2009, 04:38 AM
  4. Extra-credit problem
    Posted in the Pre-Calculus Forum
    Replies: 5
    Last Post: Mar 17th 2008, 08:25 PM
  5. Extra credit word problem
    Posted in the Geometry Forum
    Replies: 1
    Last Post: Oct 27th 2006, 01:21 AM

Search Tags


/mathhelpforum @mathhelpforum