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Re: If S and T are non-zero numbers and [#permalink]

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18 Mar 2013, 11:06

1

This post received KUDOS

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. \(ST = 1\) B. \(S + T = 1\) C. \(\frac{1}{S} = T\) D. \(\frac{S}{T} = 1\) E. 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.

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

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. \(ST = 1\) B. \(S + T = 1\) C. \(\frac{1}{S} = T\) D. \(\frac{S}{T} = 1\) E. 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.

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

Re: If S and T are non-zero numbers and [#permalink]

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18 Mar 2013, 11:14

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. \(ST = 1\) B. \(S + T = 1\) C. \(\frac{1}{S} = T\) D. \(\frac{S}{T} = 1\) E. 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.

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). "

Re: If S and T are non-zero numbers and [#permalink]

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12 Jul 2014, 17:12

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Re: If S and T are non-zero numbers and [#permalink]

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26 Dec 2015, 12:34

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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Re: If S and T are non-zero numbers and [#permalink]

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27 Dec 2015, 08:46

1

This post received KUDOS

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. \(ST = 1\) B. \(S + T = 1\) C. \(\frac{1}{S} = T\) D. \(\frac{S}{T} = 1\) E. 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.

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. \(ST = 1\) B. \(S + T = 1\) C. \(\frac{1}{S} = T\) D. \(\frac{S}{T} = 1\) E. 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.

Check out this awesome article about Anderson on Poets Quants, http://poetsandquants.com/2015/01/02/uclas-anderson-school-morphs-into-a-friendly-tech-hub/ . Anderson is a great place! Sorry for the lack of updates recently. I...