q-4000 / q +4000

how do you slove it?

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Request for Question Clarification by answerguru-ga on 04 Dec 2005 16:29 PST

Hi there,

The only way to solve for a variable is if it is part of an equation.
Is there something missing in your original question?

Answerguru is correct - but you can simplify this expression as follows.

The key is reproducing what you have on the bottom in the top.

(q-4000) / (q+4000)
 = (q+4000-8000) / (q+4000)
 = [(q+4000) / (q+4000)] - [8000 / (q+4000)]
 = 1 - [8000 / (q+4000)]

Note that in the above, the square brackets [ ] are unnecessary under
standard rules but have been added for clarity.

Maybe this is not an equation but a function? E.g. F(q) = (q-4000) / (q+4000).

If so F can be calculated depending on the value of q.
E.g.:
F(5000) = (5000-4000)/(5000+4000) = 1000/9000 = 1/9.
F(6000) = 2/10 = 1/5

For completeness we should show how to solve for q, given a specific value of F(q):

Let (q-4000) / (q+4000) = c. Then
q - 4000 = c (q + 4000)
=> q - 4000 = cq + 4000c
=> q - cq = 4000 c + 4000
=> q (1-c) = 4000 (c+1)
=> q = 4000 (c+1)/(1-c), where c is not equal to 1.

Obviously c = 1 is not allowed since this implies q - 4000 = q + 4000,
which can never be true.

there's another trick,which in this particular question won't get you
anywhere,but still its a good tricks on a difential and integeral
maths (on the Limits' field) :

(q-4000)/(q+4000) =
=(q-4000)*(q+4000)/(q+4000)*(q+4000)=

At this case ,its not that make any differences,but if the case was
the opposites,which 'q-4000' may be negetive,and could gives a zero
which undifiend,then this trick would help.

hope I get you and others a proper answered