extra credit problem i got :(

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• June 17th 2006, 03:41 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, 03:48 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, 04:24 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, 04:31 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, 04:36 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, 05:04 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, 05:12 PM
OReilly
Please, don't post twice same message.
• June 17th 2006, 05:28 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, 05:48 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, 07:48 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, 08:01 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).