Mathematical   Problem?
484601
The previous poster didn't see the whole problem. This is a question about modular arithmetic, not standard algebra. The equations are to be done modulo n and the solutions are to be integers.\n\nBy adding 3 times the first to the second, we need 5x=8. By subtracting twice the first from the second, we get -5y=-2, so 5y=2. So all that is required is that 5 be invertible modulo n, i.e., that 5 does not divide n. For example, if n=4, x=0 and y=2 works.
900562
I'd suggest searching for such problems here on yahoo answers.  You can find out how to do it.
695947
D\n\nEither is sufficient.  \n1) will lead to 2 equations,2 unknowns which yield a unique solution for x and y:  x=y+1, x+y=77, therefore x=39, y=38.  xy=1482 \n\n2) If the sum is 77 and x and y have the same tens digit, then there are two solutions for {x,y}, namely {38,39} and {39,38}, however the product xy is the same. xy=1482
103215
With this given data, it is not possible\n\nalthough the area is totalling to be 70 for both cases (7*10 and two pieces of 8*8 + 1*6) but with one division only, this is not possible\n\nPLEASE DO LET ME KNOW IF THERE IS SOME SOLUTION (i found it very interesting)
112971
There's a problem with equation. Go back and check it.\nAs it stands, no x would satisfy this.\nInfity is an abstraction, not a number.
951771
That's a question.\n\nThe answer is 42
60395
If this is a math question then anything is possible.
222284
duh...?  what's your question, Mam?  \n\n=0(
40850
-3 <4x\n x < -3/4\n solution set : {- infinity ,.............,-3/4}
880255
I can just give you a link to help you solve it.\n\nhttp://www.ticalc.org/pub/83plus/basic/math/geometry/
156773
when x=3\na) 9\nb) 38\nc) 23\nd) 9\ne) 4 and 2/3\n\nwhen x=2\na) 7\nb) 30\nc) 17\nd) 7\ne) 6\n\nwhen x=1\na) 5\nb) 22\nc) 11\nd) 5\ne) 10\n\nwhen x=0\n\na) 3\nb) 14\nc) 5\nd) 3\ne) ERROR, can not divide by zero
705654
i dont see anything wrong the way u solved it(mathematically).if there was any condition or state is mentioned about that thing in time or wrt time then u have to reconsider the problem.
674498
1+1=2. That's it. The answer for everything else would be simply 42
665576
x^2<4x\n\ncase 1: x<=0\nno solution\n\ncase 2: x>0\ndivide by x\nx<4\n\ntherfore the solutions are 0<x<4
329355
Try solving this one:\n\nDetermine all pairs of positive integers (a, b) such that\n\na^2/(2ab^2-b^3+1)\n\nis a positive integer.\n\nNote that in 2ab^2 just the b is squared.\n\nGood luck!
924405
Is it possible to have 2 equal signs in a problem?\nI am only in algebra but I haven't seen one like that before.\nSorry I couldn't help.
966925
There are 3 solutions for ABCDEFGHJ -\n\n2 6 7 8 3 5 1 4 9\n2 6 7 9 5 3 1 4 8\n6 7 8 4 1 3 2 5 9