Interested to see how others do simple algebraic manipulations

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etotheipi
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Main Question or Discussion Point

I've discussed this with a few people and I've found it pretty interesting (ok, maybe not that interesting, but still...) how there's quite a bit of difference in internal thought process for performing the bog-standard algebraic manipulations/rearrangements that come up all of the time.

For instance, I'll tend to think of an equation in the form of two fractions (ignoring the 1's on the denominators, though I've put them in for clarity), and will just "slide" the quantities across either of the diagonals, since I've thought this is a pretty quick method:

##(c^{2}) = (a^{2} + b^{2} - 2ab\cos{\theta})##

##\frac{(c^{2} - a^{2} - b^{2})}{1} = \frac{-2ab\cos{\theta}}{1} \implies \frac{1}{-2ab} = \frac{\cos{\theta}}{(c^{2} - a^{2} - b^{2})}## etc.

I was wondering if there were any methods that anyone else tends to use that are more efficient? Sorry for the rather broad question!
 
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  • #2
Math_QED
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Yes, but is this really something you "imagine"? What you are doing is basically how we define fractions, not much of a trick or something like that is going on here.

Perhaps I miss the point of the question?
 
  • #3
etotheipi
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Yes, but is this really something you "imagine"? What you are doing is basically how we define fractions, not much of a trick or something like that is going on here.

Perhaps I miss the point of the question?
I've just re-read my own question and have realised that there isn't actually any substance to it. I can't actually remember what I was thinking when I wrote it... please ignore this thread from now on!
 
  • #4
I've just re-read my own question and have realised that there isn't actually any substance to it. I can't actually remember what I was thinking when I wrote it... please ignore this thread from now on!
Hi etotheipi,
Were you maybe thinking about whether we divide through to eliminate terms initially, or manipulate the expression in terms of what we want to find before doing so?

Don't discourage yourself; now that I think about it I'm sure people do it in different ways!

CS :)
 
  • #5
HallsofIvy
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The first thing I would think about before doing any algebraic manipulations is what my purpose was- what I wanted to end with. You appear to have gone from [tex]c^2= a^2+ b^2- 2abcos(\theta)[/tex] to [tex]\frac{1}{-2ab}= \frac{cos(\theta)}{c^2- a^2- b^2}[/tex]. To what purpose? You say "etc." so you intend to go further. What is your purpose?

For example, if I wanted to solve [tex]c^2= a^2+ b^2- 2abcos(\theta)[/tex] for [itex]\theta[/theta[/itex], in terms of a, b, and c, I would start by subtracting [itex]a^2+ b^2[/itex] from each side to get [itex]c^2- a^2- b^2= 2ab cos(\theta)[/itex] and then divide both sides by [itex]2ab[/itex] to get [itex]cos(\theta)= \frac{c^2- a^2- b^2}{2ab}[/itex].

If I wanted to solve for a, in terms of b, c, and [itex]\theta[/itex] I would do quite different algebra. I would write [tex]c^2= a^2+ b^2- 2ab cos(\theta)[/tex] as [tex]a^2- 2bcos(\theta) a+ b^2- c^2= 0[/tex] and solve that quadratic equation using the "quadratic formula": [tex]a= \frac{2b cos(\theta)\pm\sqrt{4b^2cos^2(\theta)- 4(b^2- c^2)}}{2}= 2bcos(\theta)\pm\sqrt{b^2cos^2(\theta)- b^2+ c^2}[/tex].
 
  • #6
etotheipi
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Thank you @HallsofIvy, I'd say I use the same approach. The context behind this is being able to do the manipulations mentally; although we could write down all the steps, I've found that the technique of shifting the different terms i.e. along the diagonals makes it much easier to do the rearrangement without paper. Yet I'm sure other techniques are available.
 

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