Help Solve a Group of Equations

[Proton Head]

I'm trying to think over one theory, but I need to solve this equation group and I've come to a full stop.

 

(a*b^2)/c = h

 

b^3/(a*c^2) = G

 

(a*b^3)/(c^2*d^2) = k

 

(a*b)/d^3 = M

 

So trying to solve a, b ,c and d.

Any help would be appreciated.

 

Thanks!

[maverick88]

You have 4 equation with more that 4 parameters

aren't they?

[Proton Head]

No?

a, b, c and d are the unknowns.

 

h, G, k and M are constants.

[MandrakeRoot]

Have you ever thought that maybe there is not a unique solution to your system ?

But here goes some general advice :

Solve equation 1 after c, that will give you something like c= h/(a*b^2)

Plug this in everywhere you see c and then simplify => Solve equation 2 after b

Equation b after plugging in c has the form G = (a^2 *b^5)/(a*h^2), which gives you something like b = [ (G*h^2)/a ]^(1/5),

Plug this in in equation 4 and solve after d or a, and then finally put all in the last equation and solve this after the last unknown (a or d) if you like. That will give you an expression of a in terms of your constants, then you can plug this in everywhere you see a, and obtain the solution.

 

Mandrake

[bloodhound]

i had a go, u get horrible indices, so i couldnt be bothered.

[Primarygun]

Solve in terms of h, G, k and M?

[premjan]

b = (G^2 * h^2) ^(1/7)

a = G

c = (Gh)^(3/7)

d = (h/M) * (G*h)^(5/7)

 

I didn't plug it back into everything to check. It is actually a fairly simple system to solve.

[Proton Head]

Thanks for trying to help premjan, but I'm afraid that's incorrect. I checked it...

 

Mandrake Root,

I'm not good with maths so what do you mean by unique solution? Does it mean that I could get alternative solutions for each of the unknowns?

I'll try and solve it like you said, let's see if I can get it through.

[premjan]

it is all in ratios so it is bound to be very easy to solve (if you count all the exponents correctly).

[Proton Head]

Solving Equation 1 after c:

(a*b^2)/c = h

c = (a*b^2)/h

 

Plugging this in Equation 2:

b^3/(a*c^2) = G

(b^3*h^2)/(a^3*b^4) = G

 

And plugging it in Equation 3:

(a*b^3)/(c^2*d^2) = k

h^2/(a*b*d^2) = k

 

Solving Equation 2 after b:

b^3/(a*c^2) = G

b = (G*a*c^2)^(1/3)

 

Plugging in expression for c:

b = (G*a*c^2)^(1/3)

b = h^2/(G*a^3)

 

Plugging this in Equation 4, and solving for d:

(a*b)/d^3 = M

(a*h^2)/(d^3*G*a^2) = M

d = (h^2/(G*M*a^2))^(1/3)

 

One additional step for calculating c as expression of a:

c = (a*b^2)/h

c = h^3/(a^5*G^2)

 

Solving a from Equation 3 for a:

(a*b^3)/(c^2*d^2) = k

a^(19/3) = k/(G^(5/3)*h^(-5/3)*M^(2/3))

 

a = 19th Root of (k^3*h^4)/(G^5*M^2)

[Proton Head]

And so as

b = h^2/(G*a^3)

 

b = h^2/(G*19thRootOf ((k^9*h^12)/(G^15*M^6)))

 

and

c = h^3/(a^5*G^2)

 

c = h^3 / (G^2*19thRootOf ((k^15*h^20)/(G^25*M^10)))

 

and

d = (h^2/(G*M*a^2))^(1/3)

 

d = (h^2/(G*M*19thRootOf ((k^6*h^8)/(G^10*M^4))))^(1/3)

 

 

Messy...I hope I got it right.

[Proton Head]

Yes! It's right.

 

Thanks Mandrake, I never know where to start with more than 2 equations...

[MandrakeRoot]

Yes! It's right.

 

Thanks Mandrake' date=' I never know where to start with more than 2 equations...[/quote']

 

Great you got it. It is true that several non-linear equations can be quite annoying to solve;

With a unique solution i mean that there is only one couple (a,b,c,d) that satisfies your system of equations. Taking for instance :

a*b = 10, surely has more then one solution as a system.

As does a^2 -1 = 0, though it is one equation with only one unknown. I dont know where you got this system, but if it is from some physical reasoning, you might know intiutively that there is only one solution.

 

Mandrake

[Dave]

Another way of solving systems of linear equations (if you're interested), is to write them in terms of a matrix, and then perform Gaussian elimination on the matrix to get your result.

[violetendncy]

Hey Dave,

 

That's what I was thinking too....but I've only ever used matrix determinants to solve systems of linear equations with variables to the first power.....something like:

 

[ [Ax + By + Cz = K1],

[Dx + Ey + Fz = K2],

[Gx + Hy + Iz = K3] ].

 

it happens oodles of times in physics crud where you're pluggin and chuggin for systems where you've assumed conservation of linear momentum and kinetic energy (like on good pool tables) to get velocities and isolate unknown vectors....but of course the energy stuff has a Vsquared term so i usually have to jetison my matrix strategy due to lack of understanding.

 

for simple inelastic collision problems, it works great though

 

so you get to take the reduced row echelon form and stuff.....what is this Gaussian elimination (I assume it's not when Gauss took a poop)? I'd love to know anything you feel like tellin' me!

 

Thanks

[bloodhound]

its basically that violet. its usually accompanied by backsubstitution. solving the x_k th variable in the kth row and substituting the value in the (k-1)th row to solve for the x_(k-1) th variable. and then substituting those thow values into the k-2 th row to get the x_(k-2)th variable and so on, until u u solve the system.