How Do You Simplify Complex Algebraic Fractions?

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Homework Help Overview

The discussion revolves around simplifying a complex algebraic fraction involving terms like \(\frac{1}{2700}\), \(\frac{1}{3930n^2}\), and \(10^{-5}\). Participants are examining the correct approach to combine these fractions and questioning the treatment of \(10^{-5}\) in the context of fraction addition.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the correct method for combining the fractions and express confusion regarding the presence of \(10^{-5}\) in the simplification process. Some suggest verifying the expressions by substituting values for \(n\).

Discussion Status

There is ongoing exploration of the algebraic manipulation involved in the problem. Some participants have provided guidance on how to treat \(10^{-5}\) as a fraction, while others have pointed out potential misunderstandings in the original poster's approach. Multiple interpretations of the simplification process are being examined.

Contextual Notes

Participants note that the original poster's expression may not be correctly formatted as an equation, highlighting the importance of clarity in mathematical statements.

Jason-Li
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Homework Statement
Working through some fractions in loop-gain of an oscillator and stuck when comparing my answer to the learning materials...
Relevant Equations
algebra & fractions
So my final equation is:

##\frac {1} {2700} + \frac {1} {3930n^2} + 10^{-5}##

I need to boil this down, the learning materials has the following working, but I can't seem to get it
$$\frac {1} {2700} + \frac {1} {3930n^2} + 10^{-5}$$

$$\frac {3930n^2+2700+2700*3930n^2*10^{-5}} {(2700*3930n^2)}$$

But I have the following:

$$\frac {(3930n^2+2700)*10^{-5}+2700*3930n^2*10^{-5}} {(2700*3930n^2)}$$

Not sure why I have the extra 10^{-5} or how to get rid of it?

Unless the following makes mathematical sense? by making 10^{-5} = 1/ 10^{5}

$$\frac {(3930n^2+2700)*10^{5}+2700*3930n^2} {(2700*3930n^2)*10^5}$$
$$\frac {(3930n^2+2700)+2700*3930n^2*10^{-5}} {(2700*3930n^2)}$$

But the problem is $$\frac {(3930n^2+2700)*10^{5}+2700*3930n^2} {(2700*3930n^2)*10^5} ≠ \frac {(3930n^2+2700)+2700*3930n^2*10^{-5}} {(2700*3930n^2)}$$
 
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In general $$\frac 1 a + \frac 1 b + c = \frac{b + a + abc}{ab}$$
So, the book is correct.

I think you are confusing ##10^{-5}## with ##\frac 1 {10^5}## in terms of how you treat it as a fraction.
 
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PS you can always set ##n = 1## and put each expression into a calculator or spreadsheet. You'll see that your expression is incorrect.
 
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PeroK said:
In general $$\frac 1 a + \frac 1 b + c = \frac{b + a + abc}{ab}$$
So, the book is correct.

I think you are confusing ##10^{-5}## with ##\frac 1 {10^5}## in terms of how you treat it as a fraction.

PeroK,

I should've simplified it down first like you have e.g. abc...

Thanks again!
 
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PeroK said:
I think you are confusing ##10^{-5}## with ##\frac 1 {10^5}## in terms of how you treat it as a fraction.
That would be fine, since they are equal, as long as it was done correctly.
 
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Jason-Li said:
So my final equation is: ##\frac {1} {2700} + \frac {1} {3930n^2} + 10^{-5}##
Nit: That's not an equation. An equation is a statement about the equality of two expressions. An equation will always include at least one = symbol.
 
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Mark44 said:
Nit: That's not an equation. An equation is a statement about the equality of two expressions. An equation will always include at least one = symbol.
And a good way of remembering is looking at the first 4 letters of the word : Edit :equa(l)(ity)
 
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WWGD said:
And a good way of remembering is lo9king at the first 4 letters of the word : aqua(l)(ity)
aqua?
 
Mark44 said:
aqua?
Auto (in)correct strikes again. Let me edit.
 

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