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I have one doubt in this question. when we say "non zero numbers", does it always mean non zero counting numbers (i.e > 0) or non zero real numbers (+ve and -ve)? as in later scenatio ST can be 1, S = +/-1 and T =+/-1 and statement holds true. and if its former scenario then "none" remains the only choice when S != T.

Can you help me make sense of inferences, as I think I am over thinking and confused with many inferences.

I have one doubt in this question. when we say "non zero numbers", does it always mean non zero counting numbers (i.e > 0) or non zero real numbers (+ve and -ve)? as in later scenatio ST can be 1, S = +/-1 and T =+/-1 and statement holds true. and if its former scenario then "none" remains the only choice when S != T.

Can you help me make sense of inferences, as I think I am over thinking and confused with many inferences.

rgds,

A nonzero number is a number which is NOT 0, so it can be any real number except 0, positive or negative.
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I have one doubt in this question. when we say "non zero numbers", does it always mean non zero counting numbers (i.e > 0) or non zero real numbers (+ve and -ve)? as in later scenatio ST can be 1, S = +/-1 and T =+/-1 and statement holds true. and if its former scenario then "none" remains the only choice when S != T.

Can you help me make sense of inferences, as I think I am over thinking and confused with many inferences.

rgds,

A nonzero number is a number which is NOT 0, so it can be any real number except 0, positive or negative.

I have one doubt in this question. when we say "non zero numbers", does it always mean non zero counting numbers (i.e > 0) or non zero real numbers (+ve and -ve)? as in later scenatio ST can be 1, S = +/-1 and T =+/-1 and statement holds true. and if its former scenario then "none" remains the only choice when S != T.

Can you help me make sense of inferences, as I think I am over thinking and confused with many inferences.

rgds,

A nonzero number is a number which is NOT 0, so it can be any real number except 0, positive or negative.

so the answer has to be ST = 1? right?

No. Consider S = 2 and T = -2. In this case none of the options are true. The correct answer is E.
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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

\(\frac{1}{S} + \frac{1}{T} = S + T\);

\(\frac{T+S}{ST}=S+T\)\(\rightarrow\);

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.

Answer: E

How can we cross multiply ST, ST can be negative as well

Notice that we have an equation, not an inequality. So, it does not matter whether ST is negative. Only for inequalities we should flip the sign when multiplying by negative value.
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I have doubts with this question.. have same problem as others here: after manipulations we get the same expression st=1 like in the answer choice, and we see such questions very often by GMAT, and the right answer is just a manipulated version of the question.
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When I did this question I noticed that if you reduce by S+T and ended up with ST = 1, you'll see that 1/S = T is the same solution. That tipped me off.

Hi Bunuel , I fell in the same pitfall as many other users did.

Simplifying S+T is basically letting us get rid of a possible solution, which is exactly what the question wants us to do to take the wrong path. My question is: how do we notice that a certain algebraic operation is going to neglect a potential solution? Can we assume as a rule-of-thumb that multiplying is always more reliable than dividing unknown values?

Great question! I'm not happy to get it wrong, but I'm happy to have learned the proper way to do it correctly in the future.

I got as far as trying to rationalize the denominator of the 1/S + 1/T. But then after that, I had no idea what to do. I didn't think that bringing them both to one side and setting the other equal to zero was the way to solve it.