Solving For A Variable In Multi-Variable Equations

allo

How do I find solutions for an equation like:
x(1+1/y)+((y**3)/x)-(x**2)(4/y)=(y/2)-y+(4)(y**4)-4

Another, less complicated that I also am confused about is something like:
x**3-x/y=1/y

Can one simplify all equations to an x=... form? Can these equations be simplified to x=... or some solvable form there of?

JJacquelin

Thanks to obvious elementary transformations, one can write both equations on the form of cubic equations :
a*x**3 +b*x**2 +c*x +d = 0
where a, b, c, d are functions of y (but consider them just as coefficients)
Then one can solve the cubic equations (Cardano's formula). The roots are expressed as functions of a, b, c and d. So, the roots x=... are expessed as functions of the parameter y.

allo

Thanks, JJacquelin. Cardano was the sort of thing I was looking for.


When I solve using the transformations of the first I get this form though:

(x)(C)+(1/x)*(C)-(x**2)*(C)=C

Where C is any number, not necessarily itself.

Here we have powers of 1, -1, and 2. Cardano solves cubics, cubics are polynomials, polynomials must have integer powers.

Are there polynomials`esque functions that allow... use... implement rational numbers? And has there been any work done in developing a method of solving them?

Is this a case of a rational function?
Transformations...
x**2C+C-x**3C=Cx MULTIPLY BY X
(x**2C-x**3C)=Cx-C SUBTRACT X's COEF
(x**2C-x**3C)/(Cx-C)=1 MY RATIONAL FUNCTION??
??

HallsofIvy

I have no clue what "C is not necessarily itself" could possible mean! Given the equation above, the first thing I would do is divide both sides by C to get rid of it - or did "not necessarily itself" mean that the different "C"s could represent different numbers- that's very bad notation. If that was what you meant, use different letters!

Polynomials must have positive integer powers. The equation you give has integer powers but [itex]1/x= x^{-1}[/itex] so it is NOT a polynomial equation. Multiply both sides of the equation by x to get Cx^2+ C- x^3= Cx which is a cubic equation.

I have no idea what you mean by "use... inmplement rational numbers".
Do you mean solving rational equations?

Why do that when both JJaquelin and I have shown that you can write it as a cubic polynomial? The "standard" way of solving equations involving rational functions is to multiply through by the "least common denominator" so that you have a polynomial equation.

In other words, if I were given the rational equation [itex](x^2- x^3)/(x- 1)= 1[/itex] and asked to solve it, the first thing I would do is multiply on both sides by x- 1 to get [itex]x^2- x^3= x- 1[/itex] or [itex]x^3- x^2+ x- 1= 0[/itex] and then solve that cubic equation. That's easy- I observe that x= 1 is a solution and, dividing by x- 1, [itex]x^3- x^2+ x- 1= (x- 1)(x^2+ 1)= 0[/itex] so that x= 1 is the only real number solution to [itex]x^2- x^2+ x- 1= 0[/itex] although x= i and x= -i also satify it.

And then, I would need to check back into my original equation. Although x= 1 is a solution to [itex]x^2- x^2+ x- 1= 0[/itex], it is NOT a solution to [itex](x^2- x^3)/(x- 1)= 1[/itex] since setting x= 1 would make the denominator 0. That equation has no real number solutions, but it is still true that x= i and x= -i are solutions.

allo

Hallsoflvy, Why does it appear to make you angry that I do not know these things?
I feel your answer was given with contempt, ridicule, and impatience.

I admit and understand my own ignorance. That is why I came here to ask these questions... because I do not know how to solve them myself, and the material I was looking into did not cover the questions I was asking in the manner in which I was asking them.

There was a time when you also did not know the answers to these questions. If it does not make you happy to answer questions about mathematics, I would suggest to you to perhaps not answer questions about mathematics.

I appreciate your thorough and speedy reply, but honestly I would rather wait a week for someone to answer me in a non threatening and encouraging manner than to receive another answer in such a way.



In answer to your question...
`C is any number, not necessarily itself ` was my own way of saying that each place there is a C in the equation you can find a numerical value, but though I use the same variable to represent each value, each value is not necessarily the same.

JJacquelin

Where do you see non-integer powers in your equations ?
Just multiply by x the equation and the powers of x will be 3, 2, 1 and 0. That is a polynomial equation (cubic).
Try to understand all that HallsofIvy has explained.
I especially agree with :

allo

Yeah, HallsofIvy was correct, I meant to say positive integers. This was an overlooked typo.