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Topic: Need help #2
Replies: 2   Last Post: Apr 18, 2005 11:22 PM

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Sally

Posts: 2
Registered: 4/19/05
Re: Need help #2
Posted: Apr 18, 2005 11:22 PM
  Click to see the message monospaced in plain text Plain Text   Click to reply to this topic Reply

"Vincent Johns" <vjohns@alumni.caltech.edu> wrote in message
news:7GZ8e.8666$An2.6898@newsread2.news.pas.earthlink.net...
> Sally wrote:
>> "ticbol" <ticbol@yahoo.com> wrote in message
>> news:1113850912.701699.24240@f14g2000cwb.googlegroups.com...
>>

>>>x/4 +y/3 = 1
>>>Multiply both sides by 3*4
>>>(x/4)(3*4) +(y/3)(3*4) = 1(3*4)
>>>(3*4*x)/4 +(3*4*y)/3 = 12
>>>3x +4y = 12
>>>That is how.
>>>(I hope you know how to cancel quantities that are common to both
>>>numerator and denominator in the same fraction.)
>>>

> [...]
>>>
>>>>Are you saying that to clear those two specific
>>>> fractions above, (x/4 & y/3), I should multiply
>>>> them both by 12? Product of 3 & 4?

>>>
>>>Yes.
>>>We have to multiply x/4 by 4 so that the fraction will "clear" or be
>>>"removed" or be changed to a whole number.
>>>Likewise, we have to multiply y/3 by 3 to to clear this fraction.
>>>Why 3*4?
>>>Because those are the denominators.
>>>We multiply x/3 not by 3 only. We multiply it by the product 3*4 which
>>>is 12, of course. (Later on in your studies you will find why it is
>>>better to leave it as 3*4 than 12.)

> [...]
>>>
>>>Regards,
>>>ticbol

>>
>> Okay, I now understand how to work these problems with your method of
>> multiplying both sides by the product of the denominators. Thank you once
>> again ticbol for your assistance. And I will take you up on the keep
>> asking questions when I run into future problems. Much appreciated ;)

>
> When you learned to add fractions containing only integer constants, such
> as
>
> 3/4 + 1/6
>
> you were probably told that you needed to convert these to equivalent
> fractions that shared a denominator (or divisor, or number below the
> fraction bar), which is correct.
>
> You were probably also told that the only correct denominator to use was
> the lowest common multiple (LCM) of the two denominators; in the case of 4
> and 6, this would be 12 = 4 * 3 = 6 * 2 . This is not really correct; any
> common multiple will do, it's just that the LCM keeps the arithmetic
> simpler, since you deal with smaller numbers. But you could simplify
>
> 3/4 + 1/6
>
> by rewriting it as
>
> (3/4)*(6/6) + (1/6)*(4/4) [multiply top & bottom of each fraction
> by the denominator of the other fraction]
> = (3*6)/(4*6) + (1*4)/(6*4)
> = (18)/(4*6) + (4)/(4*6) [simplify top of each fraction]
> = (22)/(4*6) [add tops (numerators or dividends)]
> = (11*2)/(4*3*2) [factor 2 out of top & bottom]
> = (11)/(4*3) [divide top & bottom by common factor 2]
> = 11/12
>
> You get the correct answer this way; it just involves the extra step of
> reducing the fraction by dividing top & bottom by 2, which you would not
> have needed to do if you'd used the LCM of 12 instead of 4*6 = 24.
>
> Many times you'll have denominators like (x + 13) which have no factors,
> so looking for an LCM that is less than the product of the denominators
> would just be a waste of time; it's often simpler to follow ticbol's
> advice and just multiply the denominators. Expressing denominators as a
> product of factors, such as (3 * 4) instead of (12), makes it easier to
> cancel common factors on top & bottom of a fraction, but this is perhaps a
> matter of personal preference.
>
> Your final answer, however, should normally be expressed in lowest terms,
> with no factor common to both top and bottom; (2/3) is preferable to
> (4/6). Mathematically, there's no difference, but fractions in lowest
> terms are usually easier to understand and work with, so expressnig them
> in lowest terms is customary. (For any purists reading this, notice that
> I said "usually"; I know there are exceptions -- you write for your
> intended audience.)
>
> -- Vincent Johns <vjohns@alumni.caltech.edu>
> Please feel free to quote anything I say here.


Hi Vincent, thank you for taking the time to explain this to me. With enough
help/pratice I have figured out how to work the solutions to the problems I
was working on. I truly appreciate your assistance.





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