Bike Forums > Foo > Need help with some easy math problems! Thanks.
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 Foo Off-Topic chit chat with no general subject.

 01-26-06, 09:09 PM #1 Katrogen Team BYRDS Thread Starter     Join Date: Oct 2005 Location: Boise, Idaho Bikes: Specialized Allez Elite Double Posts: 449 Mentioned: 0 Post(s) Tagged: 0 Thread(s) Quoted: 0 Post(s) 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.
 01-26-06, 09:16 PM #2 jschen riding once again     Join Date: Oct 2005 Location: San Diego, CA Bikes: '06 Cervelo R3, '05 Specialized Allez Posts: 7,359 Mentioned: 0 Post(s) Tagged: 0 Thread(s) Quoted: 0 Post(s) 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... __________________ If you notice this notice then you will notice that this notice is not worth noticing.
 01-26-06, 09:17 PM #3 Katrogen Team BYRDS Thread Starter     Join Date: Oct 2005 Location: Boise, Idaho Bikes: Specialized Allez Elite Double Posts: 449 Mentioned: 0 Post(s) Tagged: 0 Thread(s) Quoted: 0 Post(s) Thanks I really appreciate it!
 01-26-06, 09:21 PM #4 jschen riding once again     Join Date: Oct 2005 Location: San Diego, CA Bikes: '06 Cervelo R3, '05 Specialized Allez Posts: 7,359 Mentioned: 0 Post(s) Tagged: 0 Thread(s) Quoted: 0 Post(s) 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. __________________ If you notice this notice then you will notice that this notice is not worth noticing.
 01-26-06, 09:38 PM #5 jschen riding once again     Join Date: Oct 2005 Location: San Diego, CA Bikes: '06 Cervelo R3, '05 Specialized Allez Posts: 7,359 Mentioned: 0 Post(s) Tagged: 0 Thread(s) Quoted: 0 Post(s) Never mind... MERTON corrected his solution guide. Very well done. __________________ If you notice this notice then you will notice that this notice is not worth noticing. Last edited by jschen; 01-26-06 at 09:48 PM.
 01-26-06, 09:40 PM #6 slvoid 2-Cyl, 1/2 HP @ 90 RPM     Join Date: Oct 2003 Location: NYC Bikes: 04' Specialized Hardrock Sport, 03' Giant OCR2 (SOLD!), 04' Litespeed Firenze, 04' Giant OCR Touring, 07' Specialized Langster Comp Posts: 15,762 Mentioned: 0 Post(s) Tagged: 0 Thread(s) Quoted: 1 Post(s) 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.
 01-26-06, 09:43 PM #7 ChAnMaN Specialized Member     Join Date: Nov 2003 Location: I live in a small town Bikes: 2004 Specialized Allez Posts: 973 Mentioned: 0 Post(s) Tagged: 0 Thread(s) Quoted: 0 Post(s) im confused are the problems (x/x)-3 = x+(4/x) or x/(x-3) = (x+4)/x __________________ You can never be too Specialized Click here if any of the following apply to you: 1 You dont like Specialized, 2 You drive a SUV, 3 Your name is George Bush
 01-26-06, 09:50 PM #8 jschen riding once again     Join Date: Oct 2005 Location: San Diego, CA Bikes: '06 Cervelo R3, '05 Specialized Allez Posts: 7,359 Mentioned: 0 Post(s) Tagged: 0 Thread(s) Quoted: 0 Post(s) 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. __________________ If you notice this notice then you will notice that this notice is not worth noticing.
 01-26-06, 09:55 PM #9 Katrogen Team BYRDS Thread Starter     Join Date: Oct 2005 Location: Boise, Idaho Bikes: Specialized Allez Elite Double Posts: 449 Mentioned: 0 Post(s) Tagged: 0 Thread(s) Quoted: 0 Post(s) Its a proportion problem. All this info really helps guys. Thanks alot.
01-26-06, 10:24 PM   #10
slvoid
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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.

01-27-06, 05:12 PM   #11
Katrogen
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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.