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x ≠ –1 and y ≠ ±1
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16 Aug 2018, 09:28
16 Aug 2018, 09:28
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Question Stats:
80% (01:01) correct
19% (01:16) wrong based on 26 sessions
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\(x \neq -1\) and \(y \neq \pm1\)
A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.
Quantity A |
Quantity B |
\(\frac{(xy)^2 - x^2}{(x+1)(y^2 - 1)}\) |
\(\frac{x^2}{(x+1)}\) |
A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.
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x ≠ –1 and y ≠ ±1
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15 Jun 2021, 12:16
15 Jun 2021, 12:16
1
amorphous wrote:
\(x \neq -1\) and \(y \neq \pm1\)
A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.
Quantity A |
Quantity B |
\(\frac{(xy)^2 - x^2}{(x+1 |
y^2 - 1)}\))(\(\frac{x^2}{(x+1)}\) |
A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.
Given: QUANTITY A: \(\frac{(xy)^2 - x^2}{(x+1)(y^2 - 1)}\)
Apply the power of a product law to quantity A's numerator:
QUANTITY A: \(\frac{x^2y^2 - x^2}{(x+1)(y^2 - 1)}\)
Factor the numerator:
QUANTITY A: \(\frac{x^2(y^2 - 1)}{(x+1)(y^2 - 1)}\)
Since we're told \(y \neq \pm1\), we can be certain that: \(y^2 - 1\) does not equal 0, which means we can safely divide numerator and denominator by \(y^2 - 1\) to get:
QUANTITY A: \(\frac{x^2}{x+1}\)
At this point we can see the two quantities are equal.
Answer: C
