Inverting a system of equations technique?

[Mugged]

I'm not entirely sure if this belongs in the linear algebra section, but here goes:

suppose I have two equations:

b = d + e

c = e*d

I have described b and c in terms of e and d. Is it possible to solve for e and d in terms of b and c? Can anyone provide some possible techniques?

Thank you

 

[tiny-tim]

Hi Mugged!

Use the first equation to get e = b - d,

then substitute for e into the second equation, to give you a quadratic equation in d

 

[Mugged]

oh, yeah...forgot to mention. I can't use the quadratic equation directly.

 

[tiny-tim]

I can't use the quadratic equation directly.

what do you mean?

 

[Mugged]

suppose I don't know the quadratic equation, how would i solve it then.

you see, the equations i put up come directly from a generic quadratic polynomial:

suppose I have an equation: x^2 + b*x + c = 0 and I try to put it in the form (x + e)*(x + d) = 0

the result is the equations above. So basically I'm trying to derive the quadratic formula without using it. Of course knowing the form of the quadratic formula helps you derive the forms for e and d, but I am just wondering is there a way to do it without knowing anything beforehand.

 

[HallsofIvy]

Typically one derives the quadratic formula using "completing the square".

 

[tiny-tim]

suppose I have an equation: x^2 + b*x + c = 0 …

you could write it x² + bx + 1/4 b² = -c - 1/4 b²,

which is the same as (x + b/2)² = -c - 1/4 b²

(that's completing the square)

 

[Mugged]

oh come on...

 

[chiro]

It seems like you have a nail to put in as well as a hammer, but you don't for some reason want to use the hammer. Is there any reasoning behind this?

 

[Mugged]

Well, what I am really looking for is a systematic method for solving, so what I am really after is a nail gun.

If i change the equations in some manner, id like to still be able to solve them. Do you see what i mean? Its like if you have 2nd order ODE and solve it by finding the roots...that same root finding idea can be applied to a 3rd order ODE without radically changing the method.

 

[chiro]

Well, what I am really looking for is a systematic method for solving, so what I am really after is a nail gun.

If i change the equations in some manner, id like to still be able to solve them. Do you see what i mean? Its like if you have 2nd order ODE and solve it by finding the roots...that same root finding idea can be applied to a 3rd order ODE without radically changing the method.

Could you then express the generality in terms of the mathematical relationships.

As an example instead of it being a quadratic is it a polynomial? Maybe it's a transcendental function?

Instead of e being a constant could e = r(x,y,z)?

If you give us those things, then you will get a more specific answer to your question.

Also remember that techniques for general solutions like for example solving a general polynomial analytically are not at the very least unknown with the techniques we currently have in use.

So if you gave me say an arbitrary 10th degree polynomial and I had to find it's roots, I would probably in all likelihood have to use a computational numeric root finder to get an approximate answer, as opposed to using something that will spit out an analytic method.

 

[Mugged]

the type of equation i have used to come up with the equations in my first post is irrelevant. The letters b,c,d,e are all constants, possibly complex.

Im just wondering is there some inversion technique that can be used that i haven't considered?

and of course you'd have to use numerical for 10th degree. the abel-ruffini thm says no solution in radicals exists for degrees 5 or higher.

 

[Dickfore]

The Jacobian:
$$\frac{\partial(b, c)}{\partial(d, e)} = \begin{vmatrix} 1 & 1 \\ e & d \end{vmatrix} = d - e$$
is non-zero if and only if $d \neq e$. This is a necessary and sufficient condition for the existence of the inverse transformation.