# Solving For A Variable In Multi-Variable Equations

1. Sep 12, 2011

### 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?

2. Sep 13, 2011

### 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.

3. Sep 13, 2011

### allo

Thanks

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 polynomialsesque 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 http://en.wikipedia.org/wiki/Rational_function" [Broken]?
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??
??

Last edited by a moderator: May 5, 2017
4. Sep 13, 2011

### HallsofIvy

Re: Thanks

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 $1/x= x^{-1}$ 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 $(x^2- x^3)/(x- 1)= 1$ and asked to solve it, the first thing I would do is multiply on both sides by x- 1 to get $x^2- x^3= x- 1$ or $x^3- x^2+ x- 1= 0$ and then solve that cubic equation. That's easy- I observe that x= 1 is a solution and, dividing by x- 1, $x^3- x^2+ x- 1= (x- 1)(x^2+ 1)= 0$ so that x= 1 is the only real number solution to $x^2- x^2+ x- 1= 0$ 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 $x^2- x^2+ x- 1= 0$, it is NOT a solution to $(x^2- x^3)/(x- 1)= 1$ 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.

Last edited by a moderator: May 5, 2017
5. Sep 13, 2011

### allo

Poison Ivy

Hallsoflvy, Why does it appear to make you angry that I do not know these things?

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.

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.

Last edited: Sep 13, 2011
6. Sep 13, 2011

### 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 :

7. Sep 13, 2011

### allo

Typo

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