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If (x^2+1)y=3, which of the following is not...

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by BTGmoderatorLU » Wed Nov 22, 2017 2:27 pm
If (x^2+1)y=3, which of the following is not a possible value for y?

A. 3/2
B. 2
C. 8/3
D. 3
E. 7/2

The OA is E.

I'm confused with this PS question. Please, can any expert assist me with it? Thanks in advanced.
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by Jay@ManhattanReview » Wed Nov 22, 2017 8:29 pm
LUANDATO wrote:If (x^2+1)y=3, which of the following is not a possible value for y?

A. 3/2
B. 2
C. 8/3
D. 3
E. 7/2

The OA is E.

I'm confused with this PS question. Please, can any expert assist me with it? Thanks in advanced.
We have y(x^2 + 1) = 3

=> x^2 + 1 = 3/y => x^2 = 3/y - 1 => x = √(3/y - 1)

x = √[(3 - y)/y]

Since x is a real number, y must be less than equal to 3, else (3 - y) would be a negative number and x would then be an imaginary number.

Thus, any value greater than 3 is not qualified for y. The correct answer is 7/2 = 3.5.

The correct answer: E

Hope this helps!

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by Scott@TargetTestPrep » Sat Oct 12, 2019 4:01 pm
BTGmoderatorLU wrote:If (x^2+1)y=3, which of the following is not a possible value for y?

A. 3/2
B. 2
C. 8/3
D. 3
E. 7/2

The OA is E.

I'm confused with this PS question. Please, can any expert assist me with it? Thanks in advanced.
We are given that (x^2 + 1)y = 3.

Since x^2 is always nonnegative, we see that x^2 + 1 must always be greater than or equal to 1. Thus, in order for the product of a number greater than or equal to 1 and y to be 3, y has to be less than or equal to 3. Thus, y cannot equal 7/2.

Answer: E

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