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If (x + y)/(xy) = 1, then y =

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Bunuel
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If (x + y)/(xy) = 1, then y = [#permalink] New post   05 Mar 2014, 02:09
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80% (01:10) correct 20% (02:02) wrong based on 1708 sessions
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If (x + y)/(xy) = 1, then y =

(A) x/(x - 1)
(B) x/(x + 1)
(C) (x - 1)/x
(D) (x + 1)/x
(E) x
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A
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Bunuel
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Re: If (x + y)/(xy) = 1, then y = [#permalink] New post   05 Mar 2014, 02:09
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SOLUTION

If (x + y)/(xy) = 1, then y =

(A) x/(x - 1)
(B) x/(x + 1)
(C) (x - 1)/x
(D) (x + 1)/x
(E) x

\(\frac{x + y}{xy} = 1\);

\(x + y = xy\);

\(x = xy-y\);

\(x = y(x-1)\);

\(y=\frac{x}{x-1}\).

Answer: A.
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PareshGmat
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Re: If (x + y)/(xy) = 1, then y = [#permalink] New post   05 Mar 2014, 19:34
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\(Answer = A = \frac{x}{(x - 1)}\)

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

x + y = xy

xy - y = x

y(x-1) = x

\(y = \frac{x}{(x-1)}\)
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Re: If (x + y)/(xy) = 1, then y = [#permalink] New post   10 Feb 2015, 02:02
Bunuel wrote:
SOLUTION

If (x + y)/(xy) = 1, then y =

(A) x/(x - 1)
(B) x/(x + 1)
(C) (x - 1)/x
(D) (x + 1)/x
(E) x

\(\frac{x + y}{xy} = 1\);

\(x + y = xy\);

\(x = xy-y\);

\(x = y(x-1)\);

\(y=\frac{x}{x-1}\).

Answer: A.



Could you please explain the last step ? i.e. how you got from x= y(x-1) to your solution?!

Yo divided x by y and then you got x/y=(x-1) and how do you get the x over now ?
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Bunuel
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Re: If (x + y)/(xy) = 1, then y = [#permalink] New post   10 Feb 2015, 05:55
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sauberheine1 wrote:
Bunuel wrote:
SOLUTION

If (x + y)/(xy) = 1, then y =

(A) x/(x - 1)
(B) x/(x + 1)
(C) (x - 1)/x
(D) (x + 1)/x
(E) x

\(\frac{x + y}{xy} = 1\);

\(x + y = xy\);

\(x = xy-y\);

\(x = y(x-1)\);

\(y=\frac{x}{x-1}\).

Answer: A.



Could you please explain the last step ? i.e. how you got from x= y(x-1) to your solution?!

Yo divided x by y and then you got x/y=(x-1) and how do you get the x over now ?


Divide both parts of \(x = y(x-1)\) by x-1 to get \(\frac{x}{x-1}=y\).
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Re: If (x + y)/(xy) = 1, then y = [#permalink] New post   22 Nov 2015, 10:54
I had a very strange way of solving this. Might not always work.

I plugged in numbers and already wasted enough time trying to figure out which numbers to plug in that I didn't want to go back and solve algebraically.

So what numbers added to each other and divided by the product of each other equals 1? The only ones I could think of included 0 making the solution undefined. X=1 Y=0 ...... Answer A is also Undefined while all the others give some sort of solution.

Warning: I suck at math so take my advice with a grain of salt.
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Re: If (x + y)/(xy) = 1, then y = [#permalink] New post   25 Nov 2015, 15:45
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Hi DJ1986,

This question is a type of 'in terms of' question - the prompt gives you an equation and asks you to isolate one of the variables 'in terms of' the other variable(s) in the equation. These types of questions are typically best solved with Algebra (TESTing VALUES can get complicated and you might not easily find values that 'fit' the given equation.

To answer your question though, X=2, Y=2 gives us (2+2)/(2)(2) = 1. Using those values would lead us to 2 answers though (Answers A and E).

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Re: If (x + y)/(xy) = 1, then y = [#permalink] New post   22 May 2017, 18:39
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Quote:

If (x + y)/(xy) = 1, then y =

(A) x/(x - 1)
(B) x/(x + 1)
(C) (x - 1)/x
(D) (x + 1)/x
(E) x


We can simplify the given equation:

(x + y)/(xy) = 1

x + y = xy

x = xy - y

x = y(x - 1)

x/(x - 1) = y

Answer: A
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Re: If (x + y)/(xy) = 1, then y = [#permalink] New post   24 Oct 2022, 11:59
ScottTargetTestPrep wrote:
Quote:

If (x + y)/(xy) = 1, then y =

(A) x/(x - 1)
(B) x/(x + 1)
(C) (x - 1)/x
(D) (x + 1)/x
(E) x


We can simplify the given equation:

(x + y)/(xy) = 1

x + y = xy

x = xy - y

x = y(x - 1)

x/(x - 1) = y

Answer: A


Thanks ScottTargetTestPrep, to clarify at this stage x + y = xy, y = xy-x , y = x(y-1) would have been correct too right if it's in the answer choice? Thanks
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Re: If (x + y)/(xy) = 1, then y = [#permalink] New post   24 Oct 2023, 20:22
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