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Old 01-26-06, 09:09 PM   #1
Katrogen
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Need help with some easy math problems! Thanks.

I can't find any similar examples of these proportion problems.... I've found the same principle but much simpler so it doesn't help. I would totally appreciate it if one of you could walk me through one of these problems. My Geometry book doesn't even have any examples. Thanks!

x/x-3 = x+4/x

and

x+1/6 = x-1/x

Sorry if these are dumb. I just have no idea how to do it.
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Old 01-26-06, 09:16 PM   #2
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Don't worry about geometry. Think simply about algebra. Whenever you want to solve something like a/b = c/d, realize that it is equivalent to any of the following, some of which might be easier to solve than others:

a*d = c*b (a common strategy: multiply by the two denominators)
b/a = d/c (multiply by the two denominators, divide by the two numerators, aka flip equations)
a/c = d/b (multiply by one denominator, divide by one numerator)

Also note that you must go back and check that your solution does not result in a zero in the denominator (not possible in these simple cases, but possible in more complex cases).

A sample problem following...
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Old 01-26-06, 09:17 PM   #3
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Thanks I really appreciate it!
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Old 01-26-06, 09:21 PM   #4
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Suppose (x+1)/(x-1) = (x+2)/(x-2) is our problem.

Upon inspection, there's no obvious tricks, so we'll take the most common strategy (and the only one in this case that is particularly helpful) of multiplying by the two denominators.

Multiply both sides by (x-1) * (x-2), and you get (x+1)*(x-2) = (x+2)*(x-1), or
x^2 - x - 2 = x^2 + x - 2.

In this trivial case, you get -x = x, or x = 0. We double check our solution, and indeed, x = 0 satisfies the original equation. In a more general case, once you get to something like the above, it's a simple problem in algebra (in the general case to be solved by making one side equal to zero, then either factoring or falling back on the quadratic equation).

Hope that helps. Feel free to post or PM me with an attempt at one of the problems you mentioned if you want to see if you're doing everything right.
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Old 01-26-06, 09:38 PM   #5
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Never mind... MERTON corrected his solution guide. Very well done.
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Old 01-26-06, 09:40 PM   #6
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I'd just try to get rid of the damn fractions first.

x/x-3 = x+4/x then becomes x-3x=x^2+4

Now you end up with x^2+2x+4=0
Which means you typed the original equation in wrong, unless you want to end up with imaginary numbers.
Or you meant to say x/x+3, which gives you x=2. Now I'm beginning to think you meant x/(x+3) in which case you get some weird ass number 3.781 and so on forth...

Now, your 2nd one, same thing.

x+1/6 = x-1/x becomes 6x+1=6x-6/x which becomes 6x^2+x=6x^2-6 which is now idiot proof since x=-6.

Last edited by slvoid; 01-26-06 at 09:46 PM.
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Old 01-26-06, 09:43 PM   #7
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im confused are the problems
(x/x)-3 = x+(4/x)

or

x/(x-3) = (x+4)/x
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Old 01-26-06, 09:50 PM   #8
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I assume the problem is x/(x-3) = (x+4)/x) since it's supposedly a problem about proportions.

If not, see slvoid's answer key instead of MERTON's.
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Old 01-26-06, 09:55 PM   #9
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Its a proportion problem. All this info really helps guys. Thanks alot.
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Old 01-26-06, 10:24 PM   #10
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You don't need math either.

Quote:
Originally Posted by MERTON
the only way you end up with imaginary numbers is if you use that quadratic equation (if i correctly remember what imaginary numbers are)

you don't need that thing for this.
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Old 01-27-06, 05:12 PM   #11
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Quote:
Originally Posted by MERTON
the only way you end up with imaginary numbers is if you use that quadratic equation (if i correctly remember what imaginary numbers are)

you don't need that thing for this.

For the 2nd problem you use the quadratic formula if you can't factor. In this case you can, its faster.
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