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# System of Equations

Registered User regular
edited September 2009
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)

SkyEye on
Steam: Autumn_Thunder - SC2: AutumnThundr.563 (NA) - Hearthstone: AutumnThundr.1383

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## Posts

• Registered User regular
edited September 2009
Well there are probably two solutions, but I found one of them for sure
Spoiler:

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• Registered User regular
edited September 2009
Well there are probably two solutions, but I found one of them for sure
Spoiler:

They're nonzero; they are values of resistors, and I'm pretty sure all the answers have to be at least 8.33.

Steam: Autumn_Thunder - SC2: AutumnThundr.563 (NA) - Hearthstone: AutumnThundr.1383

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• Registered User regular
edited September 2009
oh okay mathematica? or is this one of those obnoxious "show your work" kind of deals

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• Registered User regular
edited September 2009
SkyEye wrote: »
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)

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.

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• Registered User regular
edited September 2009
SkyEye wrote: »
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)

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.

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• When life gives you lemons... ...eat your delicious lemonsRegistered User regular
edited September 2009
If this is for circuit design, you should just pin down one of your resistors and work out the other two from that.

Homogeneous distribution of your varieties of amuse-gueule
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• Registered User regular
edited September 2009
Savant wrote: »
SkyEye wrote: »
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)

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.
Spoiler:

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• Registered User regular
edited September 2009
Savant wrote: »
SkyEye wrote: »
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)

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.
Spoiler:

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.

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