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System of Equations

SkyEyeSkyEye Registered User regular
edited September 2009 in Help / Advice Forum
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

Posts

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

  • SkyEyeSkyEye 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

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

  • Dunadan019Dunadan019 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.

  • SavantSavant Registered User regular
    edited September 2009
    Dunadan019 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.

  • Mojo_JojoMojo_Jojo What's beef A mouth full of goldRegistered 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
  • Dunadan019Dunadan019 Registered User regular
    edited September 2009
    Savant wrote: »
    Dunadan019 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:

  • SavantSavant Registered User regular
    edited September 2009
    Dunadan019 wrote: »
    Savant wrote: »
    Dunadan019 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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