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by pankajks2010 » Fri May 13, 2011 5:33 am
b cannot be 1. when b becomes 1, the numerator would become 0 and thus, the fraction would result into 0, which would be then different from the RHS.
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by Chaitanya_1986 » Fri May 13, 2011 5:46 am
since given a=1, 1-b/c = 1

-->> 1-b=c
-->> b=1-c

when c=0, b = 1
c=1, b=0
c=2, b=-1
c=3, b=-2

B cannot be 2
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by Anurag@Gurome » Sun May 15, 2011 7:21 pm
Chaitanya_1986 wrote:since given a=1, 1-b/c = 1

-->> 1-b=c
-->> b=1-c

when c=0, b = 1
c=1, b=0
c=2, b=-1
c=3, b=-2

B cannot be 2
What when c = -1?
Then, (1-b)/-1 = 1.
Or b = 2.
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by niazsna786 » Tue May 17, 2011 11:02 am
this is a good question .. given is a = b --- (1)
and (a - b)/c = 1 --- (2)
which can be written as (1-b) = c -- (3)
for ever value of c we get a value for b ..
but we should consider the second equation (a-b)/c = 1 is possible only if c is not zero
esle it would become infinity
so value of c should not be zero ..
for value c= 0 in eqn 3 we get b = 1 .. c = 0 is not possible as said above so b cannot have value 1 .
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