Our new Indie Games subforum is now open for business in G&T. Go and check it out, you might land a code for a free game. If you're developing an indie game and want to post about it, follow these directions. If you don't, he'll break your legs! Hahaha! Seriously though.
Our rules have been updated and given their own forum. Go and look at them! They are nice, and there may be new ones that you didn't know about! Hooray for rules! Hooray for The System! Hooray for Conforming!
System of Equations
SkyEye
Registered User regular
Registered User regular
I have been working on this thing for the last ten hours, no joke. It seemed so simple. Please help.
x(y+z) = 15(x+y+z)
y(x+z) = 8.33(x+y+z)
z(x+y) = 13.3(x+y+z)
x(y+z) = 15(x+y+z)
y(x+z) = 8.33(x+y+z)
z(x+y) = 13.3(x+y+z)
SkyEye on
Posts
They're nonzero; they are values of resistors, and I'm pretty sure all the answers have to be at least 8.33.
you are missing 3 equations that can be found probably by using the current relationships.
then you can either do it with substitution or matrices.
think of another relationship between the resistors.
I don't remember the rules of resistors offhand, but there is a way to solve that system to find the nonzero answers just using algebra, though it isn't a linear system. It gets a little messy though.
I'll give you a hint as to the way I figured it out. Temporarily treat (x+y+z) as another variable, say u = x+y+z, then reduce the equations such that the variables are in terms of each other and u as cleanly as you can get. If you get to the right place you can get the variables in terms of each other in such a way that you can simplify u and use it to solve for one of the original variables. You may have to do some tricks along the way so it is easy to get lost.
Also, it looks like there is a bit of rounding in the statement of the problem, which will make the answer a bit messier too. If you treat 8.33 as 25/3 and 13.3 as 40/3 then the answer should be really nice, but a slight bit different from the problem as given.
Edit: Just for posterity, you are going to do be doing stuff where you would need to keep in mind the potential troubles of division by zero if you weren't certain that you were dealing with nonzero variables. It doesn't matter for this problem given that, but don't get careless about division by zero in general.
theres an easier way to do it with simple equation manipulation now that I had time to work at it:
in a system of equations you can add subtract multiply or divide equations together so long as you add/subract/multiply/divide both sides at the same time.
which is the same as x(y+z) - y(x+z) = 15(x+y+z) -8.33(x+y+z)
you can expand the first part to xy + xz - yx + yz and since xy=yx it simplifies to xz-yz while the other side is just 6.67(x+y+z).
now if you notice, that looks alot like the third equation which you can then add to the previous equation to get: 2xz = 20(x+y+z) or xz = 10(x+y+z)
now sincy xz+yz = 13.33, you also know that yz = 3.33(x+y+z) and since yz=3.33(x+y+z), you know that xy = 5(x+y+z) simply from substitution.
so now your equations are:
xz = 10(x+y+z)
yz = 3.33(x+y+z)
xy = 5(x+y+z)
xz/xy = 10(x+y+z)/5(x+y+z)
xz/xy is z/y and 10(x+y+z)/5(x+y+z) is just 2. so z/y=2 or z=2y
now do #1 and 2
xz/yz = 10(x+y+z)/3.33(x+y+z)
xz/yz is x/y and 10(x+y+z)/3.33(x+y+z) is ~3. so x/y=3 or x=3y
now pick any equation (like #1) and substitute in those relationships
x(y+z) = 15(x+y+z) turns into (3y)(y+(2y)) = 15((3y)+y+(2y))
that simplifies to 9y^2 = 90y which you can then just solve as y=0 or y=10
now simply use your relationships to find that x=30 and z=20
z(x+y) = 13.33(x+y+z)
and substitute all those values in:
20(30+10) = 13.33(10+20+30)
800 = 13.33*60 and if you use the value of 40/3 for 13.3 you see that 40/3*60 is 800
That is essentially how I did it, but I used u = x+y+z to clean it up a little bit and not create the distraction of trying to take it apart and figure it out too early. It makes it a bit easier to see when you get to the point where you can divide it out. For me at least.