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If x ≠ –1 and y ≠ ±1, then what is the value of (xy)^2 − x^2/(x + 1)(y

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Joined: 02 Sep 2009
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If x ≠ –1 and y ≠ ±1, then what is the value of (xy)^2 − x^2/(x + 1)(y [#permalink]

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28 May 2017, 01:47
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If x ≠ –1 and y ≠ ±1, then what is the value of $$\frac{(xy)^2 − x^2}{(x + 1)(y^2 − 1)}$$ ?

(1) x = 3/2
(2) y = 2

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Re: If x ≠ –1 and y ≠ ±1, then what is the value of (xy)^2 − x^2/(x + 1)(y [#permalink]

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28 May 2017, 06:00
The expression after simplication will become
$$\frac{(x^2)}{(x+1)}$$

1. Have a unique value of x will yield a unique value for the expression. Sufficient.
2. Knowing the value of y has no impact on the outcome of the expression.
We necessarily need to know the value of x. Hence, insufficient. (Option A)
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Joined: 22 Oct 2016
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Re: If x ≠ –1 and y ≠ ±1, then what is the value of (xy)^2 − x^2/(x + 1)(y [#permalink]

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19 Sep 2017, 08:30
How do you do the simplification?
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Re: If x ≠ –1 and y ≠ ±1, then what is the value of (xy)^2 − x^2/(x + 1)(y [#permalink]

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19 Sep 2017, 08:53
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Suchilou wrote:
How do you do the simplification?

Hi Suchilou

the equation can be written as $$\frac{(x^2y^2-x^2)}{(x+1)(y^2-1)}$$

Tasking $$x^2$$ as common, you get $$\frac{x^2(y^2-1)}{(x+1)(y^2-1)}$$

Solve this to get
$$\frac{x^2}{(x+1)}$$

Hence by knowing the value of $$x$$, you can solve the question
Re: If x ≠ –1 and y ≠ ±1, then what is the value of (xy)^2 − x^2/(x + 1)(y   [#permalink] 19 Sep 2017, 08:53
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If x ≠ –1 and y ≠ ±1, then what is the value of (xy)^2 − x^2/(x + 1)(y

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