Method of multipliers
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(1) I am assuming this is not a calculus question where dx, dy and dz have a different meaning.

Variable d can be removed as a common factor:

ax/((b-c)yz)=by/((c-a)zx)=cz/((a-b)xy)

Take the first pair and remove common factor z:

ax/((b-c)y)=by/((c-a)x); ax^2(c-a)=by^2(b-c); ax^2=by^2(b-c)/(c-a); by^2=ax^2(c-a)/(b-c)

The second pair have common factor x:

by^2(a-b)=cz^2(c-a); by^2=cz^2(c-a)/(a-b); cz^2=by^2(a-b)/(c-a)

First and last have common factor y:

ax^2(a-b)=cz^2(b-c); ax^2=cz^2(b-c)/(a-b); cz^2=ax^2(a-b)/(b-c)

We have a pair of equations for each of the three quantities ax^2, by^2, cz^2.

So,

ax^2=by^2(b-c)/(c-a)=cz^2(b-c)/(a-b), by^2/(c-a)=cz^2/(a-b) or

(A) by^2/(cz^2)=(c-a)/(a-b)

by^2=ax^2(c-a)/(b-c)=cz^2(c-a)/(a-b), ax^2/(b-c)=cz^2/(a-b) or

(B) ax^2/(cz^2)=(b-c)/(a-b)

cz^2=by^2(a-b)/(c-a)=ax^2(a-b)/(b-c), by^2/(c-a)=ax^2/(b-c) or

(C) by^2/(ax^2)=(c-a)/(b-c)

The relative sizes of a, b and c can be:

(1) a<b<c (2) a<c<b (3) b<a<c (4) b<c<a (5) c<a<b (6) c<b<a

The quotient of two squares is always positive, so we can see the relative signs of the quotients of the other variables.

Apply each of these to (A), (B) and (C), looking for consistency:

(1A)<0 implies bc<0 and b<0, c>0 because b<c (1B)>0 implies ac>0 and a>0 because c>0 (1C)<0 implies ab<0 and b<0 so b<a, but b>a: INCONSISTENT

(2A)<0 implies bc<0 and b>0, c<0 because b>c (2B)<0 implies ac<0 and a>0 because c<0 (2C)>0 implies ab>0 and b>0 but c<0 and a>0 making c<a, but c>a: INCONSISTENT

(3A)>0 implies bc>0 (3B)<0 implies ac<0 and a<0, c>0 and b>0 because a<c (3C)<0 implies ab<0 and b>0 and so b>a, but b<a: INCONSISTENT

(4A)<0 implies bc<0 and b<0, c>0 because b<c (4B)<0 implies ac<0 and a<0 because c>0 (4C)>0 implies ab>0, true, because a and b are both negative, but c<a making c<0 when c>0: INCONSISTENT

(5A)>0 implies bc>0 (5B)<0 implies ac<0 and c<0, b<0, a>0 because a>c (5C)<0 implies ab<0, but b>a which cannot be if b<0: INCONSISTENT

(6A)<0 implies bc<0 and c<0, b>0 because b>c (6B)>0 implies ac>0 and a<0, but b<a and b>0 which cannot be: INCONSISTENT

So there are no solutions because there is inconsistency throughout.

(2) I am assuming this is a calculus problem.

Assume a, b and c are constants and x, y and z are variables.

adx/((b-c)y)=bdy/((c-a)x); b(b-c)ydy=a(c-a)xdx

Integrating: b(b-c)y^2/2=a(c-a)x^2/2 (constant to be added later)

a(c-a)x^2-b(b-c)y^2=k a constant.

A similar result follows for the other two pairs of variables (x,z and y,z) by separation of variables and using the other two pairs of equations:

a(a-b)x^2-c(b-c)z^2=p and b(a-b)y^2-c(c-a)z^2=q

 

by Top Rated User (1.1m points)

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