Who Understands Algebra?

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Old 05-04-2011, 08:21 AM
  #111  
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I taught my nephews to write out in words anything they could figure out first. Somehow putting each small item into words re-arranges the information so you can begin to see how the equation can be formed.

So, what answer did you get? (Might as well get your answer checked here too!) (If I'm not around, I'm sure someone else will help.)
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Old 05-04-2011, 08:28 AM
  #112  
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I am so stumped on solving it too. I cant remember if I need to undo the fraction by multiplying both sides of the equations by the reciprocal or combining the x 's first. Could you explain it?
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Old 05-04-2011, 08:29 AM
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I have not read all of the replies so sorry if this is a repeat, but I went back to school at 38 and I had already failed Algebra twice. It scared the crap out of me to think I was going to have to do it AGAIN!! Well us olduns seem to have better study habits then the younguns cause bar none the olduns were the best in the class. Every class. I wouldn't worry about it much. Just the fact that you are asking the question says you will do fine. If the teachers are anything like ours were when they see that you are serious about getting an education they will bend over backwards to help you get it.
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Old 05-04-2011, 08:38 AM
  #114  
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Originally Posted by Lneal
I am so stumped on solving it too. I cant remember if I need to undo the fraction by multiplying both sides of the equations by the reciprocal or combining the x 's first. Could you explain it?
I always found it easiest to break it down as much as possible:

x + 3/4x = 1400

x (1 + 3/4) =1400

x (7/4) = 1400

Now you multiply by the reciprocal to isolate the x

4/7 [x(7/4)] = 1400 (4/7)

x = 5600/7

x = 800

It has been a long time since I have done algebra, I used to love it and excel at it but when you don't use it you lose it so hopefully I did it right. Would wait for a confirmation from one of our algebra teachers first!
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Old 05-04-2011, 08:50 AM
  #115  
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It's required now, at least (1) course.

I finished up Intermediate Algebra this past few weeks, and the first attempt at it last Summer I failed with a 64; then I was made to re-take the same course again and passed with a 94 this past March.

If you need Books or CDs on the course, send me a PM I may be able to help you on that; all my text/lab books and CDs here.
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Old 05-04-2011, 08:50 AM
  #116  
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Sure feline fanatic, I would like to know the correct answer.
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Old 05-04-2011, 08:58 AM
  #117  
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feline fanatic solved it!
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Old 05-04-2011, 09:04 AM
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Thanks, so 800 cars should be 3/4 of 1400. And that would make the problem correct then.
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Old 05-04-2011, 09:09 AM
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Originally Posted by Prism99
X = # cars going east-west (this is the unknown)
3/4 X = # cars going north-south
1400= total number of cars

Total number of cars = # going east-west + # going north-south

1400 = X + 3/4 X

X + 3/4 X = 1400
No. Go back to the original written-out descriptions of what each item is. Feline fanatic solved for X; X = 800. What is X? X is the #cars going east-west. Plug the value of X back into the equation:
X + 3/4 X = 1400
800 + 3/4(800) = 1400
800 + 600 = 1400

Look at your statement:
Thanks, so 800 cars should be 3/4 of 1400. And that would make the problem correct then.

Look at the equation. 800 is the # cars going east-west. 600 is the # cars going north-south. 1400 is the sum of 800 and 600.

What is 3/4 of 1400? 1,050. It has no relationship to anything in the equation. The 3/4 was 3/4 of 800 (X).
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Old 05-04-2011, 09:12 AM
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Oh now I see. I wasn't finished with the problem. You guys are such a big help for an oldie like me.
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