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algebra ?


1123581321

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I was wondering, is it that you can only add and subtract like terms... but what about multiplication and division..

 

and with powers or exponents like, 6^8, can you call them power coefficients... like the coefficient of terms such as 56t or 4ab etc...

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You need to be a little clearer in what you are asking. You can multiply two terms to give a product term

 

[math] 4x . 3y = 12xy [/math]

 

With exponents (ie the 2 in [math] x^2 [/math]) you can only only really work on them if the base (ie the x in [math] x^2 [/math]) are the same and you are multiplying (dividing is the same with negative exponent)

 

ie [math] x^2 . x^3 = x^5[/math] but [math] x^2 . y^3 \neq xy^5 \neq (xy)^5 [/math]

 

whenever you are unsure of stuff like this sub in simple numbers and check it still works (or fails)

 

[math] 2^2 . 2^3 = x^5 [/math]

[math] 4 * 8 = 32 [/math]

 

but

[math] 2^2 . 3^3 \neq 6^5 [/math]

[math]4 * 27 = 108\[/math] but [math]6^5=7776[/math]

Edited by imatfaal
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You need to be a little clearer in what you are asking. You can multiply two terms to give a product term

 

[math] 4x . 3y = 12xy [/math]

 

With exponents (ie the 2 in [math] x^2 [/math]) you can only only really work on them if the base (ie the x in [math] x^2 [/math]) are the same and you are multiplying (dividing is the same with negative exponent)

 

ie [math] x^2 . x^3 = x^5[/math] but [math] x^2 . y^3 \neq xy^5 [/math]

 

whenever you are unsure of stuff like this sub in simple numbers and check it still works (or fails)

 

[math] 2^2 . 2^3 = x^5 [/math]

[math] 4 * 8 = 32 [/math]

 

but

[math] 2^2 * 3^3 \neq 6^5 [/math]

[math]4 * 27 = 108\[/math] but [math]6^5=7776[/math]

 

Or simplify the exponentiation to multiplication

 

[math] x^2 * y^3 = x * x * y * y * y [/math]

[math] xy^5 = xy * xy * xy * xy * xy * xy [/math]

Edited by baric
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Or simplify the exponentiation to multiplication

 

[math] xy^5 = xy * xy * xy * xy * xy * xy [/math]

 

 

ridiculous....

 

[math]xy = x*y[/math]

 

so [math] xy^5 = x*y*y*y*y*y [/math]

 

and, bracket always important.

 

[math] xy * xy * xy * xy * xy * xy = (xy)^5 = x^5y^5[/math]

 

actually, the exponent simplify multiplication... i dun think it's happen the other way :)

Edited by Vastor
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ridiculous....

 

I thought it was obvious that I was following the notation of the person I quoted who used [math] xy^5 [/math] as a visual shorthand for [math] (xy)^5 [/math]

 

My point about converting the exponentiation to multiplication still holds.

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ok, that makes sense, but what about misusing, such as: X^2 - x

 

or such a one as B^5c - 3abc...

 

What i meant before is that, can you take 3abc away from B^5c?. in another question, do you treat the 3 as the quantities of the abc letters or do you associate it just as itself (number 3)? or both..

 

so would you end up with '3ab^4' ?

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I thought it was obvious that I was following the notation of the person I quoted who used [math] xy^5 [/math] as a visual shorthand for [math] (xy)^5 [/math]

 

My point about converting the exponentiation to multiplication still holds.

 

No - I was showing why it didn't work, I will edit my initial to make it clearer.

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sorry, also what i meant was, can the exponents of terms, ie. X^3 be regarded as power/exponent coefficients ?, therefore when its said that like terms only differ in their coefficients, they mean the powers as well...

Edited by 1123581321
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ok, that makes sense, but what about misusing, such as: X^2 - x

 

or such a one as B^5c - 3abc...

 

What i meant before is that, can you take 3abc away from B^5c?. in another question, do you treat the 3 as the quantities of the abc letters or do you associate it just as itself (number 3)? or both..

 

so would you end up with '3ab^4' ?

 

You can only add and subtract like terms. You can also factor out a term from an addition/subtraction to make it easy to work with. Again try it with real numbers

 

EDIT

 

OK I see your follow-up. No you cannot think of things being like terms that can added if their exponents are different. However you can factorise and take out most of the x term and leave something that looks like this x^2(x+3) which might help.

Edited by imatfaal
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I thought it was obvious that I was following the notation of the person I quoted who used [math] xy^5 [/math] as a visual shorthand for [math] (xy)^5 [/math]

 

My point about converting the exponentiation to multiplication still holds.

 

 

let's me explain you about the use of bracket and others that u should know...

 

[math] xy^5 = x^1y^5 = (x)^1(y)^5[/math]

 

based on the equation, u can see that the '1' actually not shown becoz everything would result the same

 

 

[math] () = * [/math] in different way, let's see

 

[math] a = xy [/math]

 

[math] a^5 = (xy)^5 = (x)^5*(y)^5 \neq xy^5 = (x)^1(y)^5 = x^1*y^5 [/math]

 

() usually used to substitute an unknown while having the same effect of * which is multiply

Edited by Vastor
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let's me explain you about the use of bracket and others that u should know...

 

[math] xy^5 = x^1y^5 = (x)^1(y)^5[/math]

 

 

Please, I understand all of that. That is below basic algebra.

 

I was simply trying to follow the same visual format of the person I was quoting for clarify rather than going back and nitpicking his post over a minor formatting error.

 

And if you had read the text of the post you just quoted, I wouldn't have to repeat that.

Edited by baric
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