1.5 +3 = 1.5 * 3
7 + 1.1666667 = 7 * 1.1666667
12 + (12/11) = 12 * (12/11)
x + (x/(x-1)) = x * (x/(x-1)) for all real x's where -1 >= x or x >= 2
2006-11-29 16:27:32
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answer #1
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answered by barker67860 2
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0 x 0 = 0 + 0
2006-11-29 16:27:22
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answer #2
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answered by Kevin W 2
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None. The product is always higher when you go above 2 and always lower when you go below. It just works with 2 because you said "pair". If you'd said "Find 4 numbers" then only 4 would work.
2016-03-29 16:44:47
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answer #3
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answered by Anonymous
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how about 0 and 0?
2006-11-29 16:31:28
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answer #4
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answered by usasoccer 2
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n^2 = n+n
so n^2 -2n =0
n(n-2) =0 so there are only 2 possibilities
n=0 or n=2
NO other solutions!!!!!!!!!!!!!!!!!!!!!!
if there can be a pair of numbers:
xy=x+y
xy -x -y =0
x(y-1) -y =0
x= y/(y-1)
if y=0, then x=0
if y=2, then x=2
but for y>2, the quotient y/(y-1) will not be an integer... in case you were only interested in integer solutions!
y=4,
then x =4/3
etcetcetc
2006-11-29 16:25:58
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answer #5
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answered by Anonymous
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(x)(x) = x + x
x^2 = 2x
x^2 - 2x = 0
x(x - 2)
x = 0 or 2
0x0 = 0+0
2006-11-29 16:26:39
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answer #6
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answered by Michael M 6
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If you mean xy = x + y, it's a simple graph, but unfortunately, there is only one point of intersection with x = y, and it's at (2,2).
2006-11-29 16:52:04
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answer #7
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answered by supensa 6
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xy=x+y
so xy-x=y
x(y-1)=y
x=y/(y-1)
any of those would work eg x=3/2 y=3
2006-11-29 16:25:26
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answer #8
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answered by pzratnog 3
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i agree with POLARBEAR because he gives the correct answer..........
2006-11-29 17:05:46
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answer #9
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answered by csp 2
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zeroworks pretty well
2006-11-29 16:24:11
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answer #10
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answered by elmo o 4
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