Beat I am mathematically challenged today

[derosnec]

2 equations, 4 unknowns. do i just substitute? when doing so things get really cumbersome.

$25 transfer on stars (when i get home tonight) if someone wants to tackle it (probably easy to you geeks). first person gets $$.

6w + 6x + 4y + 2z = 5

w + x + y + z = 1

[danzasmack]

you want a solution? you know you can't have a numeric one, right?

[derosnec]

yea, so, we do inequalities right? that's fine if someone wants to do that

[rwesty]

4w + 4x + 2y = 3

rwesty on Stars.

[derosnec]

that doesn't help. oh well, maybe i'm missing something.

what if there is also a third equation:

12w+8x+8y+12z=10

does that make the problem any more approachable?

[Kimbell175113]

I'm doing it, will be done in a few minutes.

[Kimbell175113]

(this is using all three equations)

multiply the first one times 2, subtract the third one, get x=2z

first one minus 6 times the second one, get y = -2z

second one times 12, minus third, get x = .5 - y

I'm still going, this is long and there are programs that could do it, but I'm assuming you'd have to show your work or whatever.

[Alexisonfyre]

With the first 2 equations there is an infinite number of answers.

w = 0 ; x = 0 ; y = 1.5 ; z = -0.5 works

[rwesty]

Do you happen to have a 4th equation?

[danzasmack]

edit: thanks for this thread. now i'm def going back to school