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(S+T)(ST−1)=0. Either S+T=0 or ST=1. Now, notice that if S+T=0 is true then none of the options must be true.

The correct answer is E

Question: I understand the way of the equation, however, what I would have done is interfere at the following step: S+T = (S+T)*ST -> (S+T)/(S+T)=ST -> ST = 1

Is there some rule which forbids me to take this step? Or is the only option to realize so, that you perform the above given equation as well and realise that "S+T = 0" negates all other options than E.... ??

Thanks in advance, best regards

P.S. Sry if the format is terrible, this is the first question I am copying out of somewhere.

If S and T are non-zero numbers and 1S+1T=S+T, which of the following must be true? A. ST=1 B. S+T=1 C. 1/S=T D. S/T=1 E. none of the above Explanation provided: 1/S + 1/T = S+T; T+S/ST = S+T→; Cross-multiply: S+T=(S+T)∗ST; (S+T)(ST−1)=0. Either S+T=0 or ST=1. Now, notice that if S+T=0 is true then none of the options must be true. The correct answer is E Question: I understand the way of the equation, however, what I would have done is interfere at the following step: S+T = (S+T)*ST -> (S+T)/(S+T)=ST -> ST = 1 Is there some rule which forbids me to take this step? Or is the only option to realize so, that you perform the above given equation as well and realise that "S+T = 0" negates all other options than E.... ?? Thanks in advance, best regards P.S. Sry if the format is terrible, this is the first question I am copying out of somewhere.

Yes, this is not correct way of cancelling. I'll show you one example. 5* 0 = 3*0 if we cancel 0 on both side, we get 5=3. is it correct? No.

The crux (and the rule) is, we can cancel out a term only when we know its not 0. So the way it is done in explanation is absolutely correct and the right method.

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.

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.

Answer: E.

Bunuel, in your solution I need to ask one thing. Inequalities, there is a rule that if you dont know the sign of denominator, then dont cross multiply. In your solution, how can you be so sure of the sign of ST. Please let me know if i am missing something

We are concerned with the sign when we cross-multiply an inequality because this operation might affect (flip) its sign (> to <, for example) but it's always safe to cross-multiply an equation.
_________________

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

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18 Mar 2013, 10:06

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

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

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27 Dec 2015, 07:46

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 1S+1T=S+T, which of the following must be true?

A. ST=1 B. S+T=1 C. 1/S=T D. S/T=1 E. none of the above

Explanation provided:

1/S + 1/T = S+T;

T+S/ST = S+T→;

Cross-multiply: S+T=(S+T)∗ST;

(S+T)(ST−1)=0. Either S+T=0 or ST=1. Now, notice that if S+T=0 is true then none of the options must be true.

The correct answer is E

Question: I understand the way of the equation, however, what I would have done is interfere at the following step: S+T = (S+T)*ST -> (S+T)/(S+T)=ST -> ST = 1

Is there some rule which forbids me to take this step? Or is the only option to realize so, that you perform the above given equation as well and realise that "S+T = 0" negates all other options than E.... ??

Thanks in advance, best regards

P.S. Sry if the format is terrible, this is the first question I am copying out of somewhere.

There was this fun derivation that my math teacher showed us in school. Just to demonstrate how cancelling of 0 could yield wrong results. He claimed that he could prove that 1=2 and hence all numbers are equal.

It goes as below:

Let, \(a=b\)

Multiplying both sides by a, We get

\(a^2 = ab\)

Subtracting \(b^2\) from both sides

\(a^2 - b^2 = ab - b^2\)

\((a+b)(a-b) = b(a-b)\)

Cancelling \((a-b)\) on both sides,

\(a+b = b\)

Since \(a=b\)

\(a+a = a\)

\(2a = a\)

\(2=1\)
_________________

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Yes, this is not correct way of cancelling. I'll show you one example. 5* 0 = 3*0 if we cancel 0 on both side, we get 5=3. is it correct? No.

The crux (and the rule) is, we can cancel out a term only when we know its not 0. So the way it is done in explanation is absolutely correct and the right method.

Hope it helps

Thanks - lol I am an idiot - ...! I even thought of the zero number things, but mistakenly memorized the prompt as telling me that the respective equation could not be "0", though it only said that each number alone is non-zero....

Thanks guys!
_________________

Exhaust your body, proceed your mind, cultivate your soul.

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

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07 Nov 2012, 00:19

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.

Answer: E.

Bunuel, in your solution I need to ask one thing. Inequalities, there is a rule that if you dont know the sign of denominator, then dont cross multiply. In your solution, how can you be so sure of the sign of ST. Please let me know if i am missing something
_________________

In this step: s+t = (s+t)st can't we just cancel (s+t) and get ---> st =1?

thanks, -K

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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10 Feb 2013, 05:03

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.

Bunuel, Since S+T=0 OR ST=1 and the question asks whatmust be true, the answer is E ? Another way to answer the question. . Is my reasoning right?
_________________

hope is a good thing, maybe the best of things. And no good thing ever dies.

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.

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

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10 Feb 2013, 06:04

I understand what you are trying to say bunuel.. my question is that since the equation results in 2 soln and we have a OR .. . that is soln 1 OR soln 2 and the question asks for MUST be true..

So based on this reasoning, can we say the answer is E..?

For something must be true , we cannot have soln 1 OR soln 2.. we need to have 1 soln / soln1 AND soln2..

Hope you are getting what I am trying to ask..
_________________

hope is a good thing, maybe the best of things. And no good thing ever dies.

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

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10 Feb 2013, 13:55

Sachin9 wrote:

I understand what you are trying to say bunuel.. my question is that since the equation results in 2 soln and we have a OR .. . that is soln 1 OR soln 2 and the question asks for MUST be true..

So based on this reasoning, can we say the answer is E..?

For something must be true , we cannot have soln 1 OR soln 2.. we need to have 1 soln / soln1 AND soln2..

Hope you are getting what I am trying to ask..

I would say that don't generalize this point. You know that because there are two solutions, any of the given options need not be a MUST. But if u really have an option that says (s+t)(st-1)=0 then that MUST be true.
_________________

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

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10 Feb 2013, 18:17

Vips0000 wrote:

Sachin9 wrote:

I understand what you are trying to say bunuel.. my question is that since the equation results in 2 soln and we have a OR .. . that is soln 1 OR soln 2 and the question asks for MUST be true..

So based on this reasoning, can we say the answer is E..?

For something must be true , we cannot have soln 1 OR soln 2.. we need to have 1 soln / soln1 AND soln2..

Hope you are getting what I am trying to ask..

I would say that don't generalize this point. You know that because there are two solutions, any of the given options need not be a MUST. But if u really have an option that says (s+t)(st-1)=0 then that MUST be true.

But if u really have an option that says (s+t)(st-1)=0 then that MUST be true.

Ididn;t understand this.. if (s+t)(st-1)=0 then either s+t=0 or st-1=0.. we still have a OR here
_________________

hope is a good thing, maybe the best of things. And no good thing ever dies.

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