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# The values of x for which the expression above is NOT defined are

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Re: The values of x for which the expression above is NOT defined are [#permalink]
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Bunuel wrote:
$$\frac{x + 1}{x^2 - 1} - \frac{x - 2}{x^2 - 4x + 4}$$

The values of x for which the expression above is NOT defined are

A. 1 and 2 only
B. –1 and 2 only
C. –1 and –2 only
D. –1, 1, and –2
E. –1, 1, and 2

PS21172

CONCEPT: an expression is NOT DEFINED when denominator = 0

i.e. in the expression $$\frac{x + 1}{x^2 - 1} - \frac{x - 2}{x^2 - 4x + 4}$$

$$x^2 - 1 = 0$$ i.e. x = +1 or -1
and
$$x^2 - 4x + 4 = 0$$ i.e $$(x -2)^2 = 0$$ i.e. x = 2

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The values of x for which the expression above is NOT defined are [#permalink]
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GMATinsight

$$\frac{x + 1}{x^2 - 1} - \frac{x - 2}{x^2 - 4x + 4}$$

if i solve this further i get

$$\frac{x+1}{(x+1)(x-1)} - \frac{(x-2)}{(x-2)^2}$$

cancelling like terms

$$\frac{1}{x-1} - \frac{1}{x-2}$$

now the only values that give denominator 0 are 1 and 2
i know im missing -1 here, but could you help with where I went wrong? which term did I cancel that I wasn't supposed to cancel and why?
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Re: The values of x for which the expression above is NOT defined are [#permalink]
GMATinsight wrote:
Kritisood wrote:
GMATinsight

$$\frac{x + 1}{x^2 - 1} - \frac{x - 2}{x^2 - 4x + 4}$$

if i solve this further i get

$$\frac{x+1}{(x+1)(x-1)} - \frac{(x-2)}{(x-2)^2}$$

cancelling like terms

$$\frac{1}{x-1} - \frac{1}{x-2}$$

now the only values that give denominator 0 are 1 and 2
i know im missing -1 here, but could you help with where I went wrong? which term did I cancel that I wasn't supposed to cancel and why?

Kritisood

Canceling Like terms is acceptable ONLY

- If the term to be canceled is NON-ZERO

You are canceling the term (x+1) without knowing whether it's NON-ZERO - $$This is the mistake$$

Since that term also may be 0 bring the situation of $$\frac{0}{0}$$ therefore it will be WRONG to cancel it.

SImilar example
$$x^2-2x = 0$$ does NOT mean $$x*x = 2x$$ and $$x = 2$$
$$x^2-2x = 0$$ means $$x*(x-2) = 0$$ i.e. x = 0 or 2

I hope this help!

This helps a lot, thank you so much GMATinsight
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Re: The values of x for which the expression above is NOT defined are [#permalink]
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Bunuel wrote:
$$\frac{x + 1}{x^2 - 1} - \frac{x - 2}{x^2 - 4x + 4}$$

The values of x for which the expression above is NOT defined are

A. 1 and 2 only
B. –1 and 2 only
C. –1 and –2 only
D. –1, 1, and –2
E. –1, 1, and 2

PS21172

$$\frac{x + 1}{x^2 - 1} - \frac{x - 2}{x^2 - 4x + 4}$$

$$\frac{x+1}{(x+1)(x-1)} - \frac{x-2 }{ (x-2)(x-2)}$$

An expression is undefined if the denominator will equal 0.

From the above, the expression will equal 0 is $$x = -1, 1, or 2$$. Answer is E.

Notice we CAN'T cancel out $$x+1$$ on the left hand side because we could be canceling out a root.

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Re: The values of x for which the expression above is NOT defined are [#permalink]
$$\frac{x + 1}{x^2 - 1} - \frac{x - 2}{x^2 - 4x + 4}$$

The values of x for which the expression above is NOT defined are

Since denominators contain terms e.g. (x-1), (x+1) & (x-2)
The values of x for which the expression above is NOT defined are 1, -1 & 2

IMO E
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Re: The values of x for which the expression above is NOT defined are [#permalink]