okay, I never tried this before so here goes nothing...

Tomorrow and Friday I have a test on Pre-Calculus...I've been studying forever, I usually do well, but for this chapter, I am COMPLETELY stumped on thie chapter (and most say it's easy, I guess hard things are easy for me and easy things are hard...)

okay, since a good amount of you have already graduated high school then perhaps some Sophomore level Pre-Cal should be no problem?? heheheh...

anyways, I need people to spew out everything they know about the following

Solving a system of three equations with 3 unknowns
Creating a 3D graph using a system of equations with 3 unknowns
Multivariable Linear Systems
Linear Programming

that's all i need for now...the first one is the one i need the most

heheh, thanks ^^

*bump-ness, cause it's slightly urgent-ish*

:P

Erm . . . all those are the same type of problems. Do you have an example problem you could offer?

Also, do you use matrices?

Thank you. ^-^; My mother is a math teacher . . . and I asked for her to help give an explanation. She wanted to clarify some points.

meh, it's kinda late now...dont worry though, i already took and i we got results...i aced it on my own...meh...so much for this...ill guess we can either let the thread die or convert it into a universal math thread-ness thingy....

...

yeahz...

In a system of equations, you have to solve for different variables in 2 of them, then put em back in the main and solve for the one thats left.
This is just one way to do it, the other is to set up curley brackets, and multiply by constants to get the vars to cancel. The other is to use Cramer's rule. (http://mathcs.wilkes.edu/~hatchek/Cramer.html , the number of variables is passed via the html page, if you want more than 4 vars, you can just save the page to your computer and change the tag. Its right now set up to handle 3 variables. You type in each box the co-efficients of each variable. In this case variables are represented by x[0], x[1], x[2]. {basically its x, y, or z})
Example(and its possible this isn't solvable):
f() is some number that I dont feel like inventing...
f()=x+y+z
f()=-2x-y
f()=x-y+3z

I want to put all equations in terms of a variable in equation 1, so lets solve for them.
f()=x + y + z
-->y = x + y + z
f()=-2x -y
-->2x = -y
-->-2x = y
f()= x -y + 3z
-->x - y = -3z
-->-x/3 + y/3 = z
--> ^ slap the other y=equation here
-->x/3 + (-2x)/3 = z
now put the y= and z= eq's back in the original.
f()=x + (-2x) + (x/3 + (-2x)/3)
now just solve for x.
f()=x-2x+x/3+x/3-2x/3
(combine all like terms, which is what they all are!)
f() = -x
At this point you would take the constants that would have been represented by f() {like zero or 1,2, 80, .5 etc...} and put the x into each equation in place of 'x', solving for one of the variables.

Ive never graphed 3 dimentionally, sorry no help, but hey how hard can it be..And wtf is linear programming?

edit: just read the last post.... >.< lol)