Solve an equation as a fraqtion. Is it possible?

[blevita]

Hi

I would like to solve a simple equation to two variables.
This is obvious a very simple task for the attached example but as you could imagine, i would like to do the same with much more complex equations.

So my question is, is it possible to solve such an equation to a fraqtion which consists of two variables?
I have tried with solve(...,[x/b]) but i got only an empty result.


If the prime is not able to do so, is there any othe tool out there who can do such a algebraic operation?

Thanks

[ndzied1]

The CAS is where you would do this



   

[Aries]

(11-02-2018 08:42 AM)blevita Wrote:  Hi

I would like to solve a simple equation to two variables.
This is obvious a very simple task for the attached example but as you could imagine, i would like to do the same with much more complex equations.

So my question is, is it possible to solve such an equation to a fraqtion which consists of two variables?
I have tried with solve(...,[x/b]) but i got only an empty result.


If the prime is not able to do so, is there any othe tool out there who can do such a algebraic operation?

Thanks

Hi,
you have to use the CAS ambient in order to do that
Best,

Aries

[blevita]

Thanks for your answers.

But its not what i have searched for ^^.
The problem is, that in my case, the "a"-part is not that simple.

I think i must give you the real example.

This is my equation:
((r1*r3*u2+r2*r3*u1)/(c2*p*r1*r2*r3+r1*r2+r1*r3+r2*r3)) = (-u2*p*c1*r3)

I want to know what u2/u1 is.
Therefore i have tried to do this:

solve(((r1*r3*u2+r2*r3*u1)/(c2*p*r1*r2*r3+r1*r2+r1*r3+r2*r3)) = (-u2*p*c1*r3),[u2/u1])

but it does not work

[Aries]

(11-02-2018 04:24 PM)blevita Wrote:  Thanks for your answers.

But its not what i have searched for ^^.
The problem is, that in my case, the "a"-part is not that simple.

I think i must give you the real example.

This is my equation:
((r1*r3*u2+r2*r3*u1)/(c2*p*r1*r2*r3+r1*r2+r1*r3+r2*r3)) = (-u2*p*c1*r3)

I want to know what u2/u1 is.
Therefore i have tried to do this:

solve(((r1*r3*u2+r2*r3*u1)/(c2*p*r1*r2*r3+r1*r2+r1*r3+r2*r3)) = (-u2*p*c1*r3),[u2/u1])

but it does not work

I got it … I think the argument has to be a variable name …
Best,

Aries

[rushfan]

(11-02-2018 04:24 PM)blevita Wrote:  Thanks for your answers.

But its not what i have searched for ^^.
The problem is, that in my case, the "a"-part is not that simple.

I think i must give you the real example.

This is my equation:
((r1*r3*u2+r2*r3*u1)/(c2*p*r1*r2*r3+r1*r2+r1*r3+r2*r3)) = (-u2*p*c1*r3)

I want to know what u2/u1 is.
Therefore i have tried to do this:

solve(((r1*r3*u2+r2*r3*u1)/(c2*p*r1*r2*r3+r1*r2+r1*r3+r2*r3)) = (-u2*p*c1*r3),[u2/u1])

but it does not work

Try:

Code:

solve(((r1*r3*u2+r2*r3*u1)/(c2*p*r1*r2*r3+r1*r2+r1*r3+r2*r3)) = (-u2*p*c1*r3), u2)
Ans/u1
simplify(Ans)

[ijabbott]

(11-02-2018 04:24 PM)blevita Wrote:  Thanks for your answers.

But its not what i have searched for ^^.
The problem is, that in my case, the "a"-part is not that simple.

I think i must give you the real example.

This is my equation:
((r1*r3*u2+r2*r3*u1)/(c2*p*r1*r2*r3+r1*r2+r1*r3+r2*r3)) = (-u2*p*c1*r3)

I want to know what u2/u1 is.
Therefore i have tried to do this:

solve(((r1*r3*u2+r2*r3*u1)/(c2*p*r1*r2*r3+r1*r2+r1*r3+r2*r3)) = (-u2*p*c1*r3),[u2/u1])

but it does not work

Replace u2 with u1*x and solve for x.

[blevita]

Thank you very much!
I think i got it

I love the HP-Forum.
And i love my HP-Calc.
He is much better than my old TI-NSpire. ^^