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# For any numbers x and y, x<>y = 2x - y - xy. If x<>y = 0, then which

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Math Expert
Joined: 02 Sep 2009
Posts: 62637
For any numbers x and y, x<>y = 2x - y - xy. If x<>y = 0, then which  [#permalink]

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13 Nov 2014, 08:32
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Question Stats:

65% (01:58) correct 35% (02:18) wrong based on 212 sessions

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Tough and Tricky questions: Algebra.

For any numbers $$x$$ and $$y$$, $$x \diamond y = 2x - y - xy$$. If $$x \diamond y = 0$$, then which of the following CANNOT be $$y$$?

A. 3
B. 2
C. 0
D. $$-\frac{4}{3}$$
E. -2

Kudos for a correct solution.

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Concentration: General Management, Technology
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Re: For any numbers x and y, x<>y = 2x - y - xy. If x<>y = 0, then which  [#permalink]

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13 Nov 2014, 19:25
3
1

Given that 2x - y - xy = 0

Bring "x" in terms of "y"

2x - xy = y

$$x = \frac{y}{2-y}$$

Denominator cannot be 0

y cannot be 2
Intern
Joined: 29 Sep 2014
Posts: 15
Re: For any numbers x and y, x<>y = 2x - y - xy. If x<>y = 0, then which  [#permalink]

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13 Nov 2014, 09:09
2
For y=2 the equation gives 1=0 which is impossible.

Manager
Joined: 22 Sep 2012
Posts: 120
Concentration: Strategy, Technology
WE: Information Technology (Computer Software)
Re: For any numbers x and y, x<>y = 2x - y - xy. If x<>y = 0, then which  [#permalink]

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13 Nov 2014, 19:33
1

We can reduce 2x - y - xy = 0 to y = 2x/1+x

All numbers apart from y = 2 can be equated correctly.
Intern
Joined: 20 Jan 2013
Posts: 33
Re: For any numbers x and y, x<>y = 2x - y - xy. If x<>y = 0, then which  [#permalink]

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14 Nov 2014, 04:34
1
Given $$x \diamond y$$ = 2x - y - xy.
And $$x \diamond y$$ = 0

Let's Put this value in first equation. So, 2x-xy=y => x= y/2-y
Now we can check the given options one by one from the above equation.

a) y=3 => x = -3 Possible
b) y=2 => x= infinite, Y can not be 2.

hence Answer is B --> 2
Math Expert
Joined: 02 Sep 2009
Posts: 62637
Re: For any numbers x and y, x<>y = 2x - y - xy. If x<>y = 0, then which  [#permalink]

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14 Nov 2014, 08:41
1
2
Bunuel wrote:

Tough and Tricky questions: Algebra.

For any numbers $$x$$ and $$y$$, $$x \diamond y = 2x - y - xy$$. If $$x \diamond y = 0$$, then which of the following CANNOT be $$y$$?

A. 3
B. 2
C. 0
D. $$-\frac{4}{3}$$
E. -2

Kudos for a correct solution.

Official Solution:

For any numbers $$x$$ and $$y$$, $$x \diamond y = 2x - y - xy$$. If $$x \diamond y = 0$$, then which of the following CANNOT be $$y$$?

A. 3
B. 2
C. 0
D. $$-\frac{4}{3}$$
E. -2

We must determine which of the answer choices cannot be a value of $$y$$.

If $$x \diamond y = 2x - y - xy$$ and $$x \diamond y = 0$$, we can combine equations to get: $$2x - y - xy = 0$$. Because the problem tells us that one value for $$y$$ is impossible, one answer choice will make the equation $$2x - y - xy = 0$$ false.

Plug in each answer choice and see which makes the equation false.

A. If $$y = 3$$, then $$2x - y - xy = 0$$ becomes $$2x - 3 - x(3) = 0$$. Combine like terms: $$-x - 3 = 0$$, so $$x = -3$$. Because $$x$$ is a variable, it can be equal to any quantity. Therefore, this equation is not false.

B. If $$y = 2$$, then $$2x - 2 - x(2) = 0$$. The $$x$$ terms cancel, leaving $$-2 = 0$$. This is false, so B is correct. Double-check by making sure that no other answer yields a false equation.

C. If $$y = 0$$, then $$2x - 0 - x(0) = 0$$. This simplifies to $$2x = 0$$, which is not false.

D. If $$y = -\frac{4}{3}$$ then $$2x - (-\frac{4}{3}) - x(-\frac{4}{3}) = 0$$. Combine like terms: $$\frac{10}{3}x + \frac{4}{3} = 0$$. Isolate the $$x$$ term: $$\frac{10}{3}x = -\frac{4}{3}$$, so $$x = -\frac{2}{5}$$. This is not false.

E. If $$y = -2$$, then $$2x - (-2) - x(-2) = 0$$. Combine like terms: $$4x +2 = 0$$. Isolate the $$x$$ term: $$4x = -2$$, so $$x = -\frac{1}{2}$$. This is not false, so only choice B yields a false equation.

Choice B is the correct answer.

Alternatively, we can solve this problem using algebra. Recall that a value can be impossible for a variable in two ways: if the value of the variable causes a division by 0 or if the value forces the equation to look for the square root of a negative number. Since there are no exponents or roots in this equation, the answer will likely involve dividing by zero. If plugging in a certain value for $$y$$ causes a division by zero, we should look for $$y$$ in the denominator of a fraction. Solving the equation for $$x$$ may give us a fraction in terms of $$y$$.

First factor out the $$x$$ to get $$x(2 - y) - y = 0$$.

Add $$y$$ to both sides: $$x(2 - y) = y$$.

Divide both sides by $$2 - y$$ to get $$x = \frac{y}{2-y}$$.

If $$y = 2$$, the denominator becomes $$2 - 2 = 0$$. Any value that makes a denominator equal to 0 cannot be a valid solution to an equation because dividing by 0 is undefined. Therefore, $$y$$ cannot be equal to 2.

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Re: For any numbers x and y, x<>y = 2x - y - xy. If x<>y = 0, then which  [#permalink]

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13 Nov 2014, 10:52
A. plug in 3 for y to get 2x-3x-3=0 or -x=3. This could work
B. plug in 2 for y to get 2x-2-2x=0 or 0=2 and this is not true

Director
Joined: 23 Jan 2013
Posts: 511
Schools: Cambridge'16
For any numbers x and y, x<>y = 2x - y - xy. If x<>y = 0, then which  [#permalink]

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14 Nov 2014, 22:49
Means that

2x-y-xy=0
x(2-y)-y=0
x(2-y)=y
so, test every option

A) x(2-3)=3 is possible when x=-3
B) x(2-2)=2 is not possible in any x
C) x(2-0)=0 is possible when x=0
D) x(2--4/3)=-4/3 is possible when x=-4/3:10/3
E) x(2--2)=-2 is possible when x=-1/2

B
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Re: For any numbers x and y, x<>y = 2x - y - xy. If x<>y = 0, then which  [#permalink]

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23 Nov 2019, 00:26
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Re: For any numbers x and y, x<>y = 2x - y - xy. If x<>y = 0, then which   [#permalink] 23 Nov 2019, 00:26
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