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

Question

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

PareshGmat · Accepted Answer

Answer = B = 2

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

gillesalex07 · Answer

For y=2 the equation gives 1=0 which is impossible.

Answer (B)

DangerPenguin · Answer

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

Answer B!!

kinghyts · Answer

Answer should be B)

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

All numbers apart from y = 2 can be equated correctly.

devmillenium2k · Answer

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

Bunuel · Answer

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.
 
Answer: B.

Temurkhon · Answer

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

ZulfiquarA · Answer

Solve the algebric equation 2x-y-xy=0
2x=y+xy
If y is 2 then the equation could not be equal in any case
as 2x=2+2x so  (B)

bumpbot · Answer

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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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