# Help Me Express This Algebraically (sp?) **URL:** \ **Category:** Miscellaneous and Personal Stuff I Must Share\ **Created:** [November 7, 2005, 11:23pm UTC](https://boards.straightdope.com/t/help-me-express-this-algebraically-sp/329978 "2005-11-07T23:23:23Z")\ **Posts on this page:** 6\ **Page:** 1
**Author:** ![eleanorigby](https://avatars.discourse-cdn.com/v4/letter/e/ee7513/32.png) [@eleanorigby](https://boards.straightdope.com/u/eleanorigby)\ **Post date:** [November 7, 2005, 11:23pm UTC](https://boards.straightdope.com/t/help-me-express-this-algebraically-sp/329978/1 "2005-11-07T23:23:23Z")
I am not so good with math–I liked algebra, hated geometry and trigonometry. I always had trouble with word problems. I can solve most algebraic equations, I just can’t get the stories into equation form. I have wanted to figure this out for some time (I really need a life), so if anyone can help me, it would be most appreciated. I go to Curves, which has 26 stations. You are supposed to work at each station for 30 seconds, and then move on to the next when prompted to do so. A complete body workout is accomplished when you have completed the circuit twice. I find this boring–to do it all over again–and so have taken to doing each station twice in a row. Here is my question. If I start and there is someone 6(say) stations away from me–how fast will she catch up to me? I would like to figure this out so that I can finish my workout, uni-interrupted. Unlike some at Curves, I don’t think that hogging equipment and slowing someone else down just because I workout differently is a polite thing to do. I realize that I am going twice as slowly as everyone else, but I cannot figure out how long it would take for someone to catch up to my station. If I could figure it out, I would start my workout that many stations apart, if you follow me. Does any of this make sense or should I look into medication? Thanks in advance. I put this here because it is so mundane–and didn’t seem to fit GQ. ---
**Author:** ![photopat](https://avatars.discourse-cdn.com/v4/letter/p/e95f7d/32.png) [@photopat](https://boards.straightdope.com/u/photopat)\ **Post date:** [November 7, 2005, 11:37pm UTC](https://boards.straightdope.com/t/help-me-express-this-algebraically-sp/329978/2 "2005-11-07T23:37:56Z")
> [@eleanorigby](#): > > I am not so good with math–I liked algebra, hated geometry and trigonometry. > > I always had trouble with word problems. I can solve most algebraic equations, I just can’t get the stories into equation form. > > I have wanted to figure this out for some time (I really need a life), so if anyone can help me, it would be most appreciated. > > I go to Curves, which has 26 stations. You are supposed to work at each station for 30 seconds, and then move on to the next when prompted to do so. A complete body workout is accomplished when you have completed the circuit twice. > > I find this boring–to do it all over again–and so have taken to doing each station twice in a row. > > Here is my question. If I start and there is someone 6(say) stations away from me–how fast will she catch up to me? I would like to figure this out so that I can finish my workout, uni-interrupted. Unlike some at Curves, I don’t think that hogging equipment and slowing someone else down just because I workout differently is a polite thing to do. > > I realize that I am going twice as slowly as everyone else, but I cannot figure out how long it would take for someone to catch up to my station. If I could figure it out, I would start my workout that many stations apart, if you follow me. > > Does any of this make sense or should I look into medication? > Thanks in advance. I put this here because it is so mundane–and didn’t seem to fit GQ. Well, I’ll give it a shot. Since you (X) are doing two workouts at each station at once and the other person (Y) is doing 1 at a time she is moving from station to station twice as fast as you, so for every 1 station you move she’ll move 2. So the movement can be expressed as Y = 2X. In that case, to prevent somebody from catching up at the same station you would need to be more than 13 stations ahead, because Y = 2X, where X = 13, Y = 26. Meaning you would meet at the last station. If you start out 14 stations ahead of anybody else they won’t catch up with you. ---
**Author:** ![TJdude825](https://avatars.discourse-cdn.com/v4/letter/t/dec6dc/32.png) [@TJdude825](https://boards.straightdope.com/u/TJdude825)\ **Post date:** [November 8, 2005, 12:01am UTC](https://boards.straightdope.com/t/help-me-express-this-algebraically-sp/329978/3 "2005-11-08T00:01:47Z")
Put it another way, you’re moving at a rate of 1 station per minute. She’s going 1 station per half-minute or 2 stations per minute. If you know how far apart you are (measured in number of stations) then it becomes a distance/rate/time problem. ---
**Author:** ![eleanorigby](https://avatars.discourse-cdn.com/v4/letter/e/ee7513/32.png) [@eleanorigby](https://boards.straightdope.com/u/eleanorigby)\ **Post date:** [November 8, 2005, 12:20am UTC](https://boards.straightdope.com/t/help-me-express-this-algebraically-sp/329978/4 "2005-11-08T00:20:22Z")
This is very odd. I went downstairs and put laundry in the dryer and the answer popped into my head. Duh-you would need to be 13 stations away from her…and that is indeed what is here. but now I will throw a monkey wrench into the works. About every 3 minutes (or 6 stations), you are supposed to check your heart rate–so there is one 30 second period every 3 minutes. I do not do this–I continue on, but most people do stop at their station and do this, and then move on. Will you help me again, please? Danke! ---
**Author:** ![TJdude825](https://avatars.discourse-cdn.com/v4/letter/t/dec6dc/32.png) [@TJdude825](https://boards.straightdope.com/u/TJdude825)\ **Post date:** [November 8, 2005, 12:52am UTC](https://boards.straightdope.com/t/help-me-express-this-algebraically-sp/329978/5 "2005-11-08T00:52:54Z")
> [@eleanorigby](#): > > This is very odd. I went downstairs and put laundry in the dryer and the answer popped into my head. Duh-you would need to be 13 stations away from her…and that is indeed what is here. > > but now I will throw a monkey wrench into the works. About every 3 minutes (or 6 stations), you are supposed to check your heart rate–so there is one 30 second period every 3 minutes. I do not do this–I continue on, but most people do stop at their station and do this, and then move on. > > Will you help me again, please? Danke! Okay, so instead of getting through 6 stations in 3 minutes, you’ll get through 6 stations in 3 and a half minutes, right? 6/3.5 = 1.714285… stations per minute vs. your rate which is still 1 station per minute. Again, rate = distance / time, and so on. ---
**Author:** ![eleanorigby](https://avatars.discourse-cdn.com/v4/letter/e/ee7513/32.png) [@eleanorigby](https://boards.straightdope.com/u/eleanorigby)\ **Post date:** [November 8, 2005, 1:03am UTC](https://boards.straightdope.com/t/help-me-express-this-algebraically-sp/329978/6 "2005-11-08T01:03:58Z")
I will just sit here and admire the skills that others have and that I lack. I could have looked at that for a 1000 years and never come up with it. Bless you–I feel like an itch that I couldn’t reach was finally scratched.