Interested to see how others do simple algebraic manipulations

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In summary, the conversation discusses different methods of performing algebraic manipulations and rearrangements. One person tends to think of an equation in terms of fractions and "slides" the quantities across diagonals, while another person considers their purpose and uses different techniques accordingly. However, the original question is deemed to not have any substance and is disregarded.
  • #1
etotheipi
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
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
Math_QED said:
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 realized 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
etotheipi said:
I've just re-read my own question and have realized 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 :)
 
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  • #5
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].
 
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  • #6
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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1. How do you solve equations with variables on both sides?

To solve equations with variables on both sides, you need to first combine like terms on each side of the equation. Then, isolate the variable by performing inverse operations (such as addition/subtraction or multiplication/division) on both sides until the variable is left on one side and the constant on the other side. Finally, solve for the variable by performing any remaining operations.

2. What is the order of operations in algebraic expressions?

The order of operations in algebraic expressions is PEMDAS, which stands for Parentheses, Exponents, Multiplication/Division (from left to right), and Addition/Subtraction (from left to right). This means that you should first simplify any expressions within parentheses, then evaluate any exponents, and finally perform multiplication/division and addition/subtraction in the given order.

3. What is the difference between an equation and an expression?

An equation is a statement that shows the equality between two expressions, while an expression is a mathematical phrase that can contain variables, numbers, and operations. In other words, an equation is a complete sentence, while an expression is a fragment of a sentence.

4. How do you solve for a specific variable in a multi-variable equation?

To solve for a specific variable in a multi-variable equation, you need to first isolate the variable by performing inverse operations on both sides of the equation. Then, substitute any known values for the remaining variables and solve for the desired variable.

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Some common mistakes to avoid when simplifying algebraic expressions include not following the correct order of operations, forgetting to distribute a negative sign, and making errors when combining like terms. It is also important to pay attention to the signs of the numbers and variables, as well as to double check your work for any arithmetic errors.

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