• Jan 6th 2006, 04:35 PM
jackiekka
ok these are the equations...
8k - 5n = 8r
3u + 2o = 189
5d + 2c = 6m
5L + 4m = 23b
3r - m = 3u
3k - 4i = 2i
3n - 3o = c
o - c = 4d
8i + 9r = 9n
10m + 7d = 3n
4d + 4e = 6b
4y - 6u = 2m
6b + 3c = 14m

now i need to find out what each variable really is...any help? :rolleyes:
• Jan 6th 2006, 04:40 PM
Jameson
Are you familiar with Matrices and RREF? You can use Matrices to simplify this process a lot.
• Jan 6th 2006, 04:41 PM
jackiekka
what?
no i havent what are those?
• Jan 6th 2006, 04:45 PM
Jameson
Alright. Well first of all, if your teacher won't let you use any method possible than this is pointless. If you can, this involves making a Matrix that has N rows, where N is the number of variables and N+1 columns. Do you have a calculator that can perform Matrix operations?
• Jan 6th 2006, 04:46 PM
jackiekka
problem
uh no i do not. o and this is not for school so i havent exactly learned it yet.
• Jan 6th 2006, 04:48 PM
Jameson
Alright, well this is going to require a ridiculous amount of substitution then. Find a way to substitute two of the three variables into one, solve and repeat. It'll take a while, but you can do it.
• Jan 6th 2006, 04:52 PM
jackiekka
meaning?
what exactly does that mean?
• Jan 6th 2006, 05:54 PM
ThePerfectHacker
That means, for example, you have:
\$\displaystyle x+3y=2\$
\$\displaystyle 2x+4y=1\$
Then solve for the first equation:
\$\displaystyle x=2-3y\$
Thus substitute that into 2nd equation:
\$\displaystyle 2(2-3y)+4y=1\$
Now solve for \$\displaystyle y\$
One you have \$\displaystyle y\$ you can solve for \$\displaystyle x\$.
• Jan 6th 2006, 11:18 PM
CaptainBlack
Quote:

Originally Posted by jackiekka
ok these are the equations...
8k - 5n = 8r
3u + 2o = 189
5d + 2c = 6m
5L + 4m = 23b
3r - m = 3u
3k - 4i = 2i
3n - 3o = c
o - c = 4d
8i + 9r = 9n
10m + 7d = 3n
4d + 4e = 6b
4y - 6u = 2m
6b + 3c = 14m

now i need to find out what each variable really is...any help? :rolleyes:

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