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Re: Is x^3/y = x + x/y? [#permalink]
21 Jun 2012, 01:07

Expert's post

Is x^3/y = x + x/y?

Is \(\frac{x^3}{y} = x + \frac{x}{y}\)? --> \(\frac{x^3}{y} = \frac{x(y+1)}{y}\)?

(1) y = 8. The question becomes: is \(\frac{x^3}{8} = \frac{9x}{8}\)? --> is \(x^3=9x\)? --> is \(x(x^2-9)=0\)? --> is \(x=0\), \(x=3\) or \(x=-3\)? We don't know that. Not sufficient.

(2) y = x^2 - 1. The question becomes: is \(\frac{x^3}{x^2-1} = \frac{x(x^2-1+1)}{x^2-1}\)? --> is \(\frac{x^3}{x^2-1} = \frac{x^3}{x^2-1}\)? The answer to this question is YES. Sufficient.

Re: Is x^3/y = x + x/y? [#permalink]
21 Jun 2012, 02:03

Expert's post

shivanigs wrote:

Thanks a lot,that really helps.I cancelled the x in A,silly mistake.

Even if you could cancel \(x\) in (1) (for example if we were told that \(x\neq{0}\)) question would become: is \(x^2=9\)? Or is \(x=3\) or \(x=-3\)? And since we don't know whether that's true, then the statement still would be insufficient. _________________

Re: Is x^3/y = x + x/y? [#permalink]
08 Aug 2012, 15:56

Bunuel wrote:

Even if you could cancel \(x\) in (1) (for example if we were told that \(x\neq{0}\)) question would become: is \(x^2=9\)? Or is \(x=3\) or \(x=-3\)? And since we don't know whether that's true, then the statement still would be insufficient.

Sorry if I was unclear. I meant for statement 1. If you take this approach you get \(x^2-9=0 --> (x-3)(x+3)\). But according to the post above the potential x's should include a zero also. So this means that this approach is not allowed?

Sorry if I was unclear. I meant for statement 1. If you take this approach you get \(x^2-9=0 --> (x-3)(x+3)\). But according to the post above the potential x's should include a zero also. So this means that this approach is not allowed?

We cannot divide \(x^3=9x\) by \(x\) because \(x\) can be zero and division by zero is not allowed.

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) \(x^3=9x\) by \(x\), you assume, with no ground for it, that \(x\) does not equal to zero thus exclude a possible solution (notice that \(x=0\) satisfies \(x^3=9x\)).

Originally posted on MIT Sloan School of Management : We are busy putting the final touches on our application. We plan to have it go live by July 15...