Algebra equivalence help


by tranceical
Tags: algebra, equivalence
tranceical
tranceical is offline
#1
Feb10-14, 12:56 PM
P: 20
Hi guys,

please could someone tell me how this is equivalent and/or what the algebraic rule is?

how is this: a/as + 1

is equivalent to this: 1/s+1/a

Thanks a lot for your time and help
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ME_student
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#2
Feb10-14, 01:23 PM
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P: 61
So a/as=1/s do you agree?
Mark44
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#3
Feb10-14, 02:13 PM
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Quote Quote by tranceical View Post
Hi guys,

please could someone tell me how this is equivalent and/or what the algebraic rule is?

how is this: a/as + 1

is equivalent to this: 1/s+1/a
First off, what you wrote is ambiguous. Taken literally, what you wrote is ##\frac{a}{a}s + 1 = s + 1##, if a ≠ 0.

Assuming that's not what you meant, it could be either
##\frac{a}{as} + 1##
or ##\frac{a}{as + 1}##

Starting with 1/s + 1/a, the rule for adding fractions says that we need a common denominator, so
1/s + 1/a = a/(as) + s/(as) = (a + s)/(as). This doesn't match any interpretations of what you wrote, so I don't see that what you started with is equal to 1/a + 1/s.

tranceical
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#4
Feb10-14, 05:02 PM
P: 20

Algebra equivalence help


Thanks for the replies. Sorry for the ambiguity i should have used parentheses.

Mark44 - What i meant: how is a/(as+1) equivalent to 1/(s+(1/a))

Me_student - i understand a/as=1/s but i dont understand how the other
terms equal? i.e. how does the +1 term from a/(as+1) become 1/a?

many thanks
Mark44
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#5
Feb10-14, 05:20 PM
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P: 21,071
Quote Quote by tranceical View Post
Thanks for the replies. Sorry for the ambiguity i should have used parentheses.

Mark44 - What i meant: how is a/(as+1) equivalent to 1/(s+(1/a))
a/(as + 1) = a/[a(s + 1/a)]
Can you finish it and show that the last expression is equal to 1/(s + 1/a)?
What I did was factor a from both terms in the denominator.
Quote Quote by tranceical View Post

Me_student - i understand a/as=1/s but i dont understand how the other
terms equal? i.e. how does the +1 term from a/(as+1) become 1/a?
I explained that above.
tranceical
tranceical is offline
#6
Feb11-14, 01:31 AM
P: 20
Thanks a lot Mark44 you've made that perfectly clear to me, i can see how the expressions equal now. Much appreciated :)


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