2007-03-10 17:16:12 · 7 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
there is only and only 4 combinations possible,i.e., a=1 b=2, a=2 b=1, a=0 b=1 and a=1 b=0. except these no combination cn satisfy d eqn.
2007-03-10 22:16:43 · answer #1 · answered by $#Romeo Boy#$ 2 · 0⤊ 0⤋
4
2007-03-11 01:22:12 · answer #2 · answered by gus_zalenski 5 · 0⤊ 0⤋
8
2007-03-11 01:34:15 · answer #3 · answered by Dr. haribhai p 1 · 0⤊ 0⤋
8
2007-03-11 01:20:03 · answer #4 · answered by Fresh 2 · 0⤊ 0⤋
a=0; b=2
a=2; b=0 are the only possibilities from the options you gave.
the ranges are:
2â¥aâ¥0
&
2â¥bâ¥0
2007-03-11 01:36:05 · answer #5 · answered by yupchagee 7 · 0⤊ 0⤋
(a-1)+(b-1) = 1
a + b -2 = 1
a+b = 3
a = 3-b
if a & b are whole or rational numbers there are infinite options
if by usig ll brackests instead of () brackests means we are considering only natural numbers (only the positive hole numbers)
a = 3 - b
b may only be 2 or 1
2 possible options.
2007-03-11 04:37:59 · answer #6 · answered by robertonereo 4 · 0⤊ 0⤋
4, if a,b belong to the set of integers
infinite, if a,b belong the set of real numbers
2007-03-11 01:32:10 · answer #7 · answered by makeitsimple 2 · 0⤊ 0⤋