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

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Is x^3/y = x + x/y? [#permalink]  21 Jun 2012, 00:55
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Is x^3/y = x + x/y?

(1) y = 8
(2) y = x^2 - 1
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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.

Hope it's clear.
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Re: Is x^3/y = x + x/y? [#permalink]  21 Jun 2012, 01:57
Thanks a lot,that really helps.I cancelled the x in A,silly mistake.
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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.
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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.

Why isn't this allowed?

$$x^3 = xy + x$$

$$\frac{x^3-x}{x}=y$$

$$x^2-1=y$$

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Re: Is x^3/y = x + x/y? [#permalink]  08 Aug 2012, 16:07
superpus07 wrote:

$$x^3 = xy + x$$

$$\frac{x^3-x}{x}=y$$

$$x^2-1=y$$

$$x^3 = xy + x$$
$$x^3=x(y+1)$$
$$x^2= y+1$$ = Statement 2 (Answer B)
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Re: Is x^3/y = x + x/y? [#permalink]  08 Aug 2012, 16:40
SOURH7WK wrote:
superpus07 wrote:

$$x^3 = xy + x$$

$$\frac{x^3-x}{x}=y$$

$$x^2-1=y$$

$$x^3 = xy + x$$
$$x^3=x(y+1)$$
$$x^2= y+1$$ = Statement 2 (Answer B)

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?
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Re: Is x^3/y = x + x/y? [#permalink]  09 Aug 2012, 02:01
Expert's post
superpus07 wrote:
SOURH7WK wrote:
superpus07 wrote:

$$x^3 = xy + x$$

$$\frac{x^3-x}{x}=y$$

$$x^2-1=y$$

$$x^3 = xy + x$$
$$x^3=x(y+1)$$
$$x^2= y+1$$ = Statement 2 (Answer B)

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 $$x^3=9x$$).

Hope it's clear.
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Re: Is x^3/y = x + x/y?   [#permalink] 09 Aug 2012, 02:01
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