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If x ≠ 0, what is the value of y? 1) xy + y/x − 3x − 3/x = 0 2) x = 3

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Russ19
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If x ≠ 0, what is the value of y? 1) xy + y/x − 3x − 3/x = 0 2) x = 3 [#permalink] If x ≠ 0, what is the value of y? 1) xy + y/x − 3x − 3/x = 0 2) x = 3   30 Apr 2021, 08:17
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If \(x ≠ 0\), what is the value of \( y\)?

1) \( xy + \frac{y}{x} − 3x − \frac{3}{x} = 0\)

2) \(x = 3\)
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Re: If x ≠ 0, what is the value of y? 1) xy + y/x − 3x − 3/x = 0 2) x = 3 [#permalink] If x ≠ 0, what is the value of y? 1) xy + y/x − 3x − 3/x = 0 2) x = 3   30 Apr 2021, 08:20
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sjuniv32
If \(x ≠ 0\), what is the value of \( y\)?

1) \( xy + \frac{y}{x} − 3x − \frac{3}{x} = 0\)

2) \(x = 3\)

Statement 1:

Combine x and 1/x terms:

\(x*(y - 3) + \frac{1}{x}*(y - 3) = 0\)

\((y - 3)*(x + \frac{1}{x}) = 0\)

The solution for \(x + \frac{1}{x} = 0\) is \(x^2 = -1\), hence no real solution. Then we must have \(y - 3 = 0\) and y = 3. Sufficient.

Statement 2:

Insufficient.

Ans: A
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Re: If x ≠ 0, what is the value of y? 1) xy + y/x − 3x − 3/x = 0 2) x = 3 [#permalink] If x ≠ 0, what is the value of y? 1) xy + y/x − 3x − 3/x = 0 2) x = 3   12 Jul 2021, 18:13
TestPrepUnlimited has mentioned avobe that \(x^2\) = - 1 is not a real solution

Can anyone explain the reason or concept?

I think you have said that a squared variable always has a +ve value, right?

Thanks for any response in advance!
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Re: If x ≠ 0, what is the value of y? 1) xy + y/x − 3x − 3/x = 0 2) x = 3 [#permalink] If x ≠ 0, what is the value of y? 1) xy + y/x − 3x − 3/x = 0 2) x = 3   12 Jul 2021, 20:21
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sjuniv32
TestPrepUnlimited has mentioned avobe that \(x^2\) = - 1 is not a real solution

Can anyone explain the reason or concept?

I think you have said that a squared variable always has a +ve value, right?

Thanks for any response in advance!


Yes that is correct.
A square can never be negative. The least value of a square is 0.
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If x ≠ 0, what is the value of y? 1) xy + y/x − 3x − 3/x = 0 2) x = 3 [#permalink] If x ≠ 0, what is the value of y? 1) xy + y/x − 3x − 3/x = 0 2) x = 3   13 Jul 2021, 09:23
Expert Reply
sjuniv32
TestPrepUnlimited has mentioned avobe that \(x^2\) = - 1 is not a real solution

Can anyone explain the reason or concept?

I think you have said that a squared variable always has a +ve value, right?

Thanks for any response in advance!

Hi sjuniv32 , a square naturally has to be at least 0. If we see \(x^2\) being a negative number, we just say there is no (real) solution to it or we cannot solve it. Since the x part has no solution, we look at the other side and attempt to find a solution. Hope that helps!
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Re: If x 0, what is the value of y? 1) xy + y/x 3x 3/x = 0 2) x = 3 [#permalink] If x ≠ 0, what is the value of y? 1) xy + y/x − 3x − 3/x = 0 2) x = 3   17 Feb 2024, 11:16
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Re: If x 0, what is the value of y? 1) xy + y/x 3x 3/x = 0 2) x = 3 [#permalink]
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