Algebraic Manipulation: Solving Complex Equations

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The discussion focuses on manipulating two equations that equal VCC/3 to derive a new expression for RBB as a function of RE. The user initially struggled with the complexity of the equations but realized that equating the left sides of the original equations simplifies the problem. By substituting back into the original equations, they successfully arrived at the desired expression involving VBB, Vf, and VCC. The user overcame previous confusion caused by an extra factor in their calculations. This approach highlights the importance of careful algebraic manipulation in solving complex equations.
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Homework Statement



Somehow the upper two statements are manipulated to get the lower one but i can't work out how... (see attatched image)

Homework Equations



N/A

The Attempt at a Solution



Oh you don't want to see mypages of scribble. I can't get close to that expression!
 

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In the first two equations you have expressions that are both equal to VCC/3, so you can equate the two expressions on the left sides.

Doing that, you can solve for RBB as a constant multiple of RE.

Then, substitute in either of your first two equations to get the desired equation in VBB, Vf, and VCC.
 
ahh, finally got it... the last time i tried to do it that way i got and extra factor somehow and that's what was messing me up.

thanks!
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
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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