At a certain time, Janice notices that her digital watch reads (a) minutes after two o'clock. Fifteen minutes later, it reads (b) minutes after three o'clock. She is amused to note that (a) is six times greater than (b). What time was it when she looked at her watch for the second time?

LOONDSI
i tried this one. kinda.
i don't think this is right but my equation was this:
6a+15=b
-15 -15
6a=b-15
-b -b
-15=6a-b
-1(-15=6a-b)
15=(-6)a+b

and then i guessed and checked...

(-2,3) worked but that didn't help. now i'm stuck too.

nooni

maybe it's
6a=b? and they are 15 min apart?

Okay... I think the thing we need to do is find another equation then try the elimination method.

But what would be another equation? Hmm... :-)

Would it (possibly) be a - 15 = b, because 15 mins later, it was (b) minutes after three o'clock?

yeah...

i don't know about that equation though.
i don't think 15 fits into it....

I see no one has answered so let me give you my thoughts. First, I think you can logically deduce the answer. If a =6*b and the difference between a and b is 15 minutes,(and since adding 15 minutes to the 2:00 + a time moves PAST the hour to give 3:00+ b time) then b must be sometime between 46 minutes and the hour; therefore, anything more than 10 minutes for a won't work because that will be over 60 minutes. 9 minutes for b would make 9*6=54 and 2:54 + 15 minutes =2:69 or 3:09. Works perfectly.
8 minutes after the hour would fit into the 15 minutes before the previous hour but 2:48 + :15 = 3:03 doesn't get 3:08. And anything less than 8 (1 min through 7 minutes) doesn't give a number which will add to over the hour in 15 minutes. So 8 and 9 minutes are the only possibilities and 8 doesn't work.
I tried a + 15 = 60+b and a=6*b for the equations.
That gives me a=45+b and 6b=45+b; then
5b = 45 and b = 9 so 3:00 + b minutes = 3:09, the time Jane looked at her watch the second time.