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Hello everyone.
For years, everyone's been talking about plugging in numbers for algebra problems. You pick a number for the variable, plug it in, get an answer, and then find that answer in the answer choices.
I think this is dangerous, because it only works sometimes, and I think that algebra or some other trick is always important as a backup. Plugging in should be done by people who understand it and when not to use it.
Sorry for waxing on about that, but I ran across a problem today that would be very difficult to plug numbers into, but is also very time consuming using straightforward algebra. You might expect it to work out well with substitution, but you'll see that it doesn't. So I thought I'd post it here and see what everyone thought about it, and if there's some clever thing that I'm not seeing.
--------------
If 2x + 3y = 1, what is (x/2) + (y/3) in terms of y?
A) y/5
B) (1 - 3y)/2
C) (1 - 3y)/4
D) (3y + 4)/15
E) (3 - 5y)/12
For years, everyone's been talking about plugging in numbers for algebra problems. You pick a number for the variable, plug it in, get an answer, and then find that answer in the answer choices.
I think this is dangerous, because it only works sometimes, and I think that algebra or some other trick is always important as a backup. Plugging in should be done by people who understand it and when not to use it.
Sorry for waxing on about that, but I ran across a problem today that would be very difficult to plug numbers into, but is also very time consuming using straightforward algebra. You might expect it to work out well with substitution, but you'll see that it doesn't. So I thought I'd post it here and see what everyone thought about it, and if there's some clever thing that I'm not seeing.
--------------
If 2x + 3y = 1, what is (x/2) + (y/3) in terms of y?
A) y/5
B) (1 - 3y)/2
C) (1 - 3y)/4
D) (3y + 4)/15
E) (3 - 5y)/12
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25 Aug 2004, 18:49
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I guess I am losing my mind. But for (x/2) + (y/2), I am getting (1-y)/4. This is not an option listed.
