extra credit problem i got :(

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June 17th 2006, 02:41 PM
spiderz

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.
June 17th 2006, 02:48 PM
spiderz

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.
June 17th 2006, 03:24 PM
OReilly

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:
$ x = y + \frac{1}{{2z}} $
or
$ x = \frac{{y + 1}}{{2z}} $
June 17th 2006, 03:31 PM
spiderz

Quote:

Originally Posted by OReilly

Can you please verify is x:
$ x = y + \frac{1}{{2z}} $
or
$ x = \frac{{y + 1}}{{2z}} $

i meant one half times Z or .5z

so it should say Y plus half of Z
June 17th 2006, 03:36 PM
OReilly

Quote:

Originally Posted by spiderz

i meant one half times Z or .5z

so it should say Y plus half of Z

You mean
$ x = y + \frac{z}{2} $
?
June 17th 2006, 04:04 PM
OReilly

If
$ x = y + \frac{z}{2} $
is correct than this is solution:

$\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} $

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

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

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

$\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:
$\frac{{8240}}{3} - \frac{{8240}}{6} + 8240 + \frac{{20600}}{3} = 16480$

$\frac{{16480 - 8240 + 49440 + 41200}}{6} = 16480$
June 17th 2006, 04:12 PM
OReilly

Please, don't post twice same message.
June 17th 2006, 04:28 PM
spiderz

Quote:

Originally Posted by OReilly

If
$ x = y + \frac{z}{2} $
is correct than this is solution:

$\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} $

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

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

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

$\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:
$\frac{{8240}}{3} - \frac{{8240}}{6} + 8240 + \frac{{20600}}{3} = 16480$

$\frac{{16480 - 8240 + 49440 + 41200}}{6} = 16480$

thank you *bows*
does this show that y = z?
June 17th 2006, 04:48 PM
OReilly

Quote:

Originally Posted by spiderz

thank you *bows*
does this show that y = z?

No,
$ y = - \frac{{8240}}{6} $
and
$ z = 8240 $
June 17th 2006, 06:48 PM
malaygoel

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. :)
June 17th 2006, 07:01 PM
ThePerfectHacker

(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).