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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)

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

SkyEye on

## Posts

• Registered User regular
edited September 2009
Well there are probably two solutions, but I found one of them for sure
x=0, y=0, z=0

scrivenerjones on
• Registered User regular
edited September 2009
Well there are probably two solutions, but I found one of them for sure
x=0, y=0, z=0

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

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

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

scrivenerjones on
• 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.

• 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.

Savant on
• A lemon squeezed in the salty fist of Poseidon Registered 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.

Mojo_Jojo on
Homogeneous distribution of your varieties of amuse-gueule
• 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.
so what you do is subtract equation 2 from equation 1
x(y+z) = 15(x+y+z)
-
y(x+z) = 8.33(x+y+z)

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.

xz = 10(x+y+z)
yz = 3.33(x+y+z)
xy = 5(x+y+z)
take any 2 and divide them by each other. for example #1 and #3.

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
simple check: go to equation 3

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

• 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.
so what you do is subtract equation 2 from equation 1
x(y+z) = 15(x+y+z)
-
y(x+z) = 8.33(x+y+z)

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.

xz = 10(x+y+z)
yz = 3.33(x+y+z)
xy = 5(x+y+z)
take any 2 and divide them by each other. for example #1 and #3.

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
simple check: go to equation 3

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.

Savant on