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# If 2*y^(-2) + 3*y^(-1) - 14 = 0, which of the following could be the

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If 2*y^(-2) + 3*y^(-1) - 14 = 0, which of the following could be the  [#permalink]

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08 May 2018, 01:51
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65% (01:58) correct 35% (02:10) wrong based on 124 sessions

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If $$2*y^{-2} + 3*y^{-1} - 14 = 0$$, which of the following could be the value of y?

A. 2
B. 1/2
C. 2/7
D. -1/2
E. -7

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Re: If 2*y^(-2) + 3*y^(-1) - 14 = 0, which of the following could be the  [#permalink]

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08 May 2018, 02:21
2
The given equation can be changed to
$$14y^2-3y-2=0$$

$$14y^2 - 7y + 4y -2 =0$$

7y (2y-1) + 2 (2y-1) =0

(7y+2) (2y -1) =0

y can be either -2/7 or 1/2.

Ans :- B
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Re: If 2*y^(-2) + 3*y^(-1) - 14 = 0, which of the following could be the  [#permalink]

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08 May 2018, 03:59

Solution

Given:
• 2∗$$y^{−2}+3∗y^{−1}$$−14=0

To find:
• The possible value of y

Approach and Working:
• If we assume the value of $$y^{-1}$$ to be x, we can rewrite the equation as
o 2$$x^2$$ + 3x – 14 = 0
Or, 2$$x^2$$ + 7x – 4x – 14 = 0
Or, (x – 2) (2x + 7) = 0
Or, x = 2 , $$\frac{-7}{2}$$

• Replacing the value of x with y^-1, we can write:
o $$\frac{1}{y}$$ = 2
Or, y = $$\frac{1}{2}$$
o Similarly, $$\frac{1}{y} = \frac{-7}{2}$$
Or, y =$$\frac{-2}{7}$$

As per the options, the correct answer is option B.

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Re: If 2*y^(-2) + 3*y^(-1) - 14 = 0, which of the following could be the  [#permalink]

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08 May 2018, 04:04
Bunuel wrote:
If $$2*y^{-2} + 3*y^{-1} - 14 = 0$$, which of the following could be the value of y?

A. 2
B. 1/2
C. 2/7
D. -1/2
E. -7

I think back solving will be the easiest way to deal with this question. First simplify the equation the plug in values. B is the correct answer. Here it is clear that fraction will be the correct answer and no negative value won't work. Thus u can't A and E immediately.
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Re: If 2*y^(-2) + 3*y^(-1) - 14 = 0, which of the following could be the  [#permalink]

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09 May 2018, 01:38
Bunuel wrote:
If $$2*y^{-2} + 3*y^{-1} - 14 = 0$$, which of the following could be the value of y?

A. 2
B. 1/2
C. 2/7
D. -1/2
E. -7

For such questions, back solving is the best and the fastest way.

First simplify the given equation and then try each answer choice.

B is the correct choice.

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Re: If 2*y^(-2) + 3*y^(-1) - 14 = 0, which of the following could be the  [#permalink]

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09 May 2018, 16:45
Bunuel wrote:
If $$2*y^{-2} + 3*y^{-1} - 14 = 0$$, which of the following could be the value of y?

A. 2
B. 1/2
C. 2/7
D. -1/2
E. -7

Simplifying we have:

2/y^2 + 3/y - 14 = 0

Multiplying by y^2 we have:

2 + 3y - 14y^2 = 0

14y^2 - 3y - 2 = 0

(7y + 2)(2y - 1) = 0

7y + 2 = 0 → y = -2/7

2y - 1 = 0 → y = 1/2

Since only 1/2 is given as one of the choices. Choice B is the correct answer.

Alternate solution:

If we let x = y^-1, then the equation can be rewritten as 2x^2 + 3x - 14 = 0. Let’s solve it:

(2x + 7)(x - 2) = 0

2x + 7 = 0 → x = -7/2

x - 2 = 0 → x = 2

Since x = y^-1 = 1/y, y = 1/x. Therefore, y is either 1/(-7/2) = -2/7 or 1/2. Since only 1/2 is given as one of the choices. Choice B is the correct answer.

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Re: If 2*y^(-2) + 3*y^(-1) - 14 = 0, which of the following could be the  [#permalink]

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16 Mar 2019, 19:48
ScottTargetTestPrep wrote:
Bunuel wrote:
If $$2*y^{-2} + 3*y^{-1} - 14 = 0$$, which of the following could be the value of y?

A. 2
B. 1/2
C. 2/7
D. -1/2
E. -7

Simplifying we have:

2/y^2 + 3/y - 14 = 0

Multiplying by y^2 we have:

2 + 3y - 14y^2 = 0

14y^2 - 3y - 2 = 0

(7y + 2)(2y - 1) = 0

7y + 2 = 0 → y = -2/7

2y - 1 = 0 → y = 1/2

Since only 1/2 is given as one of the choices. Choice B is the correct answer.

Alternate solution:

If we let x = y^-1, then the equation can be rewritten as 2x^2 + 3x - 14 = 0. Let’s solve it:

(2x + 7)(x - 2) = 0

2x + 7 = 0 → x = -7/2

x - 2 = 0 → x = 2

Since x = y^-1 = 1/y, y = 1/x. Therefore, y is either 1/(-7/2) = -2/7 or 1/2. Since only 1/2 is given as one of the choices. Choice B is the correct answer.

Hello ScottTargetTestPrep!!

Whys is it like the follwoing?

14y^2 - 3y - 2 = 0

(7y + 2)(2y - 1) = 0

Shouldn't the sum of the roots be -3?

Regards!
Re: If 2*y^(-2) + 3*y^(-1) - 14 = 0, which of the following could be the   [#permalink] 16 Mar 2019, 19:48
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