How can I solve 4 equations with 4 unknowns?

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Homework Help Overview

The original poster is working on a mathematical problem involving four equations with four unknowns. They have attempted substitution methods but encountered squared unknowns in their equations.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss methods to eliminate mixed terms (xy) from the equations and explore transformations to standard forms, such as using trigonometric identities. There are attempts to express the equations in terms of hyperbolic functions and questions about the validity of these transformations.

Discussion Status

Participants are actively engaging with each other's suggestions, questioning the steps taken, and exploring various mathematical approaches. Some guidance has been offered regarding the transformation of equations, but there is no explicit consensus on the best method to proceed.

Contextual Notes

There are indications of confusion regarding the manipulation of terms and the assumptions made in the equations. The original poster has expressed uncertainty about the relationships between the variables and the transformations being discussed.

malsch
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Homework Statement


First of all hi,

I was working some mathematics and ended up with 4 unknowns with 4 equations. I tried substituting equations but ended up with squared unknowns. The following is what i ended up with so far.

The Attempt at a Solution



23a902129184660.jpg


Any help on how to solve this would be greatly appreciated.

Thank you.
 
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welcome to pf!

hi malsch! welcome to pf! :smile:

essentially, you have two equations of the form Px2 + Qxy + Ry2 = S

(i don't know whether this is the quickest way :redface:, but …)

i'd make it more symmetrical by getting rid of the xy terms, giving

x2/a2 + y/b2 = 1​

so x = acosθ, y = bsinθ,

and then use standard trigonometric identities to arrive at an equation in cos2θ and sin2θ :wink:

(it it's x2/a2 minus y/b2 = C2, use cosh and sinh instead of cos and sin)
 
Last edited:
hi. thanks for your reply tiny-tim :)

i didn't understand how you can get rid of the xy terms and end up with x2/a2 + y/b2 = C2.

10x again
 
now that i come to it, i am guessing i can do the following:

Px2 + Qxy + Ry2 = S

=> (xA + yB)(xC + yD) = S

and find the values of x and y simultaneausly
 
hi malsch! :smile:

(try using the X2 icon just above the Reply box :wink:)
malsch said:
now that i come to it, i am guessing i can do the following:

Px2 + Qxy + Ry2 = S

=> (xA + yB)(xC + yD) = S

and find the values of x and y simultaneausly

i don't see how that helps :confused:
malsch said:
i didn't understand how you can get rid of the xy terms and end up with x2/a2 + y/b2 = C2.

eliminate the xy terms (in your case, the CG terms) …

what do you get? :smile:
 
yes you are right, that doesn't help :(.

You mean dividing through out by xy (CG in my case)?
 
no, you have two equations with xy in …

combine them in such a way that there's no xy

(for example, put xy on the LHS in one, then substitute the RHS for xy in the other)
 
ok now i came up with:

4.5*10-7 = -G2 + 98,652.1C2

I was wondering how you can convert this to cos and sin.

And by doing so, you would end up by an unknown angle, right?
 
malsch said:
ok now i came up with:

4.5*10-7 = -G2 + 98,652.1C2

I was wondering how you can convert this to cos and sin.

And by doing so, you would end up by an unknown angle, right?

that minus means you'll need to use cosh and sinh instead …

divide by a constant to get it into the form

C2/a2 - G2/b2 = 1​

and then tanh = b/a

(btw, i got the formula wrong in my first post :redface:, the RHS has to be 1, I've edited it now)
 
  • #10
thats ok hehe. i found the angle using tanh = b/a.

now i am working with these formulas:

C = acoshθ, G = bsinhθ,

but how did you conclude to them? i know cosθ + sinθ = 1, and coshθ - sinhθ = 1
 
  • #11
malsch said:
thats ok hehe. i found the angle using tanh = b/a.

now i am working with these formulas:

C = acoshθ, G = bsinhθ,

but how did you conclude to them? i know cosθ + sinθ = 1, and coshθ - sinhθ = 1

Those last two formulas aren't true. You're missing a few things, namely 2's.

cos^2(\theta) + sin^2(\theta)=1

cosh^2(\theta) - sinh^2(\theta) = 1
 
  • #12
yes yes you are right. ok i think i know how to continue from here.

Thanks Char. Limit and Tiny Tim especially ;)
 

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