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Bunuel
\(\frac{x + 1}{x^2 - 1} - \frac{x - 2}{x^2 - 4x + 4}\)

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

Answer: Option E
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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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Bunuel
\(\frac{x + 1}{x^2 - 1} - \frac{x - 2}{x^2 - 4x + 4}\)

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.

Answer is E.
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\(\frac{x + 1}{x^2 - 1} - \frac{x - 2}{x^2 - 4x + 4}\)

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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­Be careful about cancelling (x+1):

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