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Retired Moderator
Joined: 10 Apr 2015
Posts: 6218
Given Kudos: 136
The equation x + y = xy = x/y has how many
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05 May 2020, 10:03
05 May 2020, 10:03
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Question Stats:
40% (01:40) correct
60% (01:33) wrong based on 145 sessions
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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
Retired Moderator
Joined: 10 Apr 2015
Posts: 6218
Given Kudos: 136
Re: The equation x + y = xy = x/y has how many
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05 May 2020, 16:42
05 May 2020, 16:42
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GreenlightTestPrep wrote:
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
To begin, if \(xy = \frac{x}{y}\), we can conclude that \(y\neq0\), otherwise \(\frac{x}{y}\) is undefined.
Take: \(xy = \frac{x}{y}\)
Multiply both sides of the equation by \(y\) to get: \(xy^2 = x\)
Subtract \(x\) from both sides to get: \(xy^2 - x = 0\)
Factor to get: \(x(y^2 - 1) = 0\)
Factor again to get: \(x(y - 1)(y + 1) = 0\)
There are 3 POSSIBLE solutions to the above equation: \(x = 0\), \(y = 1\) and \(y = -1\)
Let's examine each possible case:
case i: \(x = 0\)
Substitute this into the original equation to get: \(0 + y = (0)y = \frac{0}{y}\)
For this equation to hold true, we need \(y = 0\), but we already showed that \(y\) cannot equal \(0\)
So, it can't be the case that \(x = 0\)
case ii: \(y = 1\)
Substitute this into the original equation to get: \(x + 1 = x(1) = \frac{x}{1}\)
Simplify: \(x + 1 = x = x\)
Since there are no solutions to the equation \(x + 1 = x\), it can't be the case that \(y = 1\)
case iii: \(y = -1\)
Substitute this into the original equation to get: \(x + (-1) = x(-1) = \frac{x}{-1}\)
Simplify: \(x - 1 = -x = -x\)
Now take: \(x - 1 = -x\)
Add \(x\) to both sides: \(2x - 1 = 0\)
Add \(1\) to both sides: \(2x = 1\)
Solve: \(x = 0.5\)
So, \(x = 0.5\) and \(y = -1\) is the ONLY possible solution to the given equation.
Answer: B
Cheers,
Brent
General Discussion
Re: The equation x + y = xy = x/y has how many
[#permalink]
12 May 2020, 06:33
12 May 2020, 06:33
1
GreenlightTestPrep wrote:
GreenlightTestPrep wrote:
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
To begin, if \(xy = \frac{x}{y}\), we can conclude that \(y\neq0\), otherwise \(\frac{x}{y}\) is undefined.
Take: \(xy = \frac{x}{y}\)
Multiply both sides of the equation by \(y\) to get: \(xy^2 = x\)
Subtract \(x\) from both sides to get: \(xy^2 - x = 0\)
Factor to get: \(x(y^2 - 1) = 0\)
Factor again to get: \(x(y - 1)(y + 1) = 0\)
There are 3 POSSIBLE solutions to the above equation: \(x = 0\), \(y = 1\) and \(y = -1\)
Let's examine each possible case:
case i: \(x = 0\)
Substitute this into the original equation to get: \(0 + y = (0)y = \frac{0}{y}\)
For this equation to hold true, we need \(y = 0\), but we already showed that \(y\) cannot equal \(0\)
So, it can't be the case that \(x = 0\)
case ii: \(y = 1\)
Substitute this into the original equation to get: \(x + 1 = x(1) = \frac{x}{1}\)
Simplify: \(x + 1 = x = x\)
Since there are no solutions to the equation \(x + 1 = x\), it can't be the case that \(y = 1\)
case iii: \(y = -1\)
Substitute this into the original equation to get: \(x + (-1) = x(-1) = \frac{x}{-1}\)
Simplify: \(x - 1 = -x = -x\)
Now take: \(x - 1 = -x\)
Add \(x\) to both sides: \(2x - 1 = 0\)
Add \(1\) to both sides: \(2x = 1\)
Solve: \(x = 0.5\)
So, \(x = 0.5\) and \(y = -1\) is the ONLY possible solution to the given equation.
Answer: B
Cheers,
Brent
why x = 0 and y = 0, not a solution??
