Kirchhoff's rule and multivariable algebra

AI Thread Summary
The discussion focuses on solving a set of equations related to voltage and current physics problems, specifically using Kirchhoff's rules. The participant seeks help with algebraic manipulation to isolate variables. Suggestions include solving for A in the second equation and substituting it into the third equation. The conversation emphasizes the importance of correctly substituting values to simplify the equations. The final goal is to find a solution for the variables involved using the provided equations.
FiveAlive
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I'm working on some voltage/current physics problems and I think I understand the concepts however I believe the algebra is holding me back(not my strong point). Could someone make some suggestions how I could solve for at least one of these variables by using a combination of theses three equations?

eq 1) A+B=H

eq 2) A(0.1) + 12 = B(0.01) +14

eq 3) B(0.01) + H(1.20) = 12

I believe that the first step is to expand eq 3 since H=A+B into

B(0.01) + A(1.2) + B(1.2) = 12 but beyond that I am unsure.

thanking you in advance,
Linus
 
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Solve eq 2) for A. Plug that into your last equation
 
Is this what you mean?

A(0.1) + 12 = B(0.01) +14
A= 20+ B(0.1)


Cheers, thanks so much.
 
FiveAlive said:
Is this what you mean?

A(0.1) + 12 = B(0.01) +14
A= 20+ B(0.1)


Cheers, thanks so much.

Yes.

Now put {20+ B(0.1)} into B(0.01) + A(1.2) + B(1.2) = 12 , in place of A.
 
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks

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