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Re: If xy ≠ 0 and x^2y^2 − xy = 6, which of the following could be y in te [#permalink]

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18 Oct 2015, 20:44

Bunuel wrote:

If xy ≠ 0 and x^2y^2 − xy = 6, which of the following could be y in terms of x?

I. 1/(2x) II. -2/x III. 3/x

(A) I only (B) II only (C) I and II (D) I and III (E) II and III

Kudos for a correct solution.

xy(xy-1) = 6

Possible solutions (2,3) (-2,-3) (1,6) (-1,-6) (3,2) (-3,-2) (-6,-1)

1) 1/2x = y so xy = 1/2 but we know that xy has to be integer - (see solutions above) 2) -2/x =y so xy = -2 (xy-1 = -3) Yes we can get this. 3) 3/x = y so xy =3 (yes we can get this one too)

Ans E.
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If xy ≠ 0 and x^2y^2 − xy = 6, which of the following could be y in terms of x?

I. 1/(2x) II. -2/x III. 3/x

(A) I only (B) II only (C) I and II (D) I and III (E) II and III

Kudos for a correct solution.

We are given the equation, (x^2)(y^2) – xy = 6, and, although it may not be obvious, the equation is a quadratic-format equation. Thus, our first step is to set the equation equal to zero. We then factor the left side and, finally, solve for xy.

(x^2)(y^2) – xy - 6 = 0

(xy – 3)(xy + 2) = 0

xy – 3 = 0 or xy + 2 = 0

xy = 3 or xy = -2

Since we need y in terms of x, we can isolate y in both of our equations.

y = 3/x or y = -2/x

Thus, the expressions in II and III are correct.

Answer: E
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If xy ≠ 0 and x^2y^2 − xy = 6, which of the following could be y in terms of x?

I. 1/(2x) II. -2/x III. 3/x

(A) I only (B) II only (C) I and II (D) I and III (E) II and III

Kudos for a correct solution.

Hi, its a 700 level Q, BUT is literally begging you to make it a simpler Q by using substitution.. we have an equation - \(x^2y^2 − xy = 6\) we have tto find OUT of 3, which all will be values of y in terms of x..

two ways-

1) factorize and then find answers 2) substitution, as we have our answers right in front of us and any value that fits in IS the answer..

[b]we can actually eliminate I at the first glance as it is giving us a 2 in denominator and hence LHS will become some fraction, but lets still try..[/b]

EQ - \(x^2y^2 − xy = 6\) I. 1/(2x)......\(x^2\frac{1}{2x}^2 − x\frac{1}{2x} = 6\) ..... \(\frac{1}{4}-\frac{1}{2} = 6\)... NO II. -2/x .......\(x^2\frac{-2}{x}^2 − x\frac{-2}{x} = 6\) ..... \(4-(-2) = 6\)..YES III. 3/x .......\(x^2\frac{3}{x}^2 − x\frac{3}{x} = 6\) ..... \(9-3 = 6\)..YES

If xy ≠ 0 and \(x^2y^2 − xy = 6\), which of the following could be y in terms of x?

I. 1/(2x) II. -2/x III. 3/x

(A) I only (B) II only (C) I and II (D) I and III (E) II and III

Kudos for a correct solution.

\(x^2y^2 − xy = 6\)

i.e. \(xy(xy − 1) = 6\) [Please observe that xy and (xy-1) are two consecutive number] But 3*2 = 6 also, (-2)*(-3) = 6

therefore, xy = 3 or xy = -2

i.e. , y = 3/x or y=-2/x

Answer: option E
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Re: If xy ≠ 0 and x^2y^2 − xy = 6, which of the following could be y in te [#permalink]

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16 Dec 2017, 04:32

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