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04 May 2020, 07:37
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B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
85%
(hard)
Question Stats:
48% (01:56) correct
52%
(01:53)
wrong
based on 263
sessions
History
Date
Time
Result
Not Attempted Yet
The equation \(x + y = xy = \frac{x}{y}\) has how many real solutions?
A) 0
B) 1
C) 2
D) 3
E) More than 3
A) 0
B) 1
C) 2
D) 3
E) More than 3
04 May 2020, 15:36
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\(xy = \frac{x}{y}\)
\(xy - \frac{x}{y} = 0\)
\(x(y-\frac{1}{y}) = 0\)
\(\frac{x(y^2-1)}{y} = 0\)
Either x=0 or y = 1 or -1
Case 1- x=0
x+y=xy
y= 0
But \(\frac{x}{y}\) is undefined at y= 0
Reject this case
Case 2- y= 1
x+y=xy
x+1= x
1=0
Reject this one too
Case 3- y= -1
x+y=xy
x-1= -x
\(x=\frac{1}{2}\)
At \(x=\frac{1}{2}\) and \(y=-1\)
\(x+y = -\frac{1}{2}\); \(xy=\frac{-1}{2}\); \(\frac{x}{y}= \frac{-1}{2}\) [We got one]
B
\(xy - \frac{x}{y} = 0\)
\(x(y-\frac{1}{y}) = 0\)
\(\frac{x(y^2-1)}{y} = 0\)
Either x=0 or y = 1 or -1
Case 1- x=0
x+y=xy
y= 0
But \(\frac{x}{y}\) is undefined at y= 0
Reject this case
Case 2- y= 1
x+y=xy
x+1= x
1=0
Reject this one too
Case 3- y= -1
x+y=xy
x-1= -x
\(x=\frac{1}{2}\)
At \(x=\frac{1}{2}\) and \(y=-1\)
\(x+y = -\frac{1}{2}\); \(xy=\frac{-1}{2}\); \(\frac{x}{y}= \frac{-1}{2}\) [We got one]
B
BrentGMATPrepNow
General Discussion
Updated on: 05 May 2020, 02:18
Originally posted by Enchanting on 04 May 2020, 11:53.
Last edited by Enchanting on 05 May 2020, 02:18, edited 1 time in total.
Last edited by Enchanting on 05 May 2020, 02:18, edited 1 time in total.
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BrentGMATPrepNow
