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The numbers in the pair do not have to match.

2006-11-29 16:21:13 · 11 answers · asked by Es Nina Chica 1 in Science & Mathematics Mathematics

11 answers

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 · answer #1 · answered by barker67860 2 · 2 2

0 x 0 = 0 + 0

2006-11-29 16:27:22 · answer #2 · answered by Kevin W 2 · 0 0

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 · answer #3 · answered by Anonymous · 0 0

how about 0 and 0?

2006-11-29 16:31:28 · answer #4 · answered by usasoccer 2 · 0 0

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 · answer #5 · answered by Anonymous · 0 0

(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 · answer #6 · answered by Michael M 6 · 0 0

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 · answer #7 · answered by supensa 6 · 0 0

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 · answer #8 · answered by pzratnog 3 · 3 1

i agree with POLARBEAR because he gives the correct answer..........

2006-11-29 17:05:46 · answer #9 · answered by csp 2 · 1 0

zeroworks pretty well

2006-11-29 16:24:11 · answer #10 · answered by elmo o 4 · 0 0

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