kuttingchai wrote:

Bunuel wrote:

If S and T are non-zero numbers and \frac{1}{S} + \frac{1}{T} = S + T, which of the following must be true?A.

ST = 1B.

S + T = 1C.

\frac{1}{S} = TD.

\frac{S}{T} = 1E. None of the above

OE:\frac{1}{S} + \frac{1}{T} = S + T -->

\frac{T+S}{ST}=S+T --> cross-multiply:

S+T=(S+T)*ST -->

(S+T)(ST-1)=0 --> either

S+T=0 or

ST=1. So, if

S+T=0 is true then none of the options must be true.

Answer: E.

P.S. Please read

carefully and follow:

rules-for-posting-please-read-this-before-posting-133935.html Please pay attention to the rule #3. Thank you.

Hello Bunuel,

How did u get S+T=(S+T)*ST --> (S+T)(ST-1)=0 ??

What did i miss in he below equation?? how come you got (S+T)(ST-1)=0 ?? Ca you please explain

1/S+1/T = S+T

(S+T) = (S+T) (ST)

divide both side by (S+T) we get

1=1(ST)

therefore ST=1

Thank you.

I think I got the answer

from your previous post I got "

Never reduce equation by variable (or expression with variable), if you are not certain that variable (or expression with variable) doesn't equal to zero. We can not divide by zero.

So, if you divide (reduce) s+t = (s+t)st by (s+t), you assume, with no ground for it, that (s+t) does not equal to zero thus exclude a possible solution (notice that both st=1 AND (s+t)=0 satisfy the equation).

"

therefore

1/S+1/T = S+T

(S+T) = (S+T) (ST)

0= (S+T) (ST) - (S+T)

0 = (S+T) (ST-1)

(S+T)=0 or ST = 1

Thank you