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# If x and y are different integers and x^2 = xy, which of the

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If x and y are different integers and x^2 = xy, which of the [#permalink]

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15 Jul 2007, 07:15
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If x and y are different integers and x^2 = xy, which of the following must be true ?

I. x = 0
II. y = 0
III. x = -y

(A) I only
(B) II only
(C) III only
(D) I and III only
(E) I, II, and III
[Reveal] Spoiler: OA

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Last edited by Bunuel on 19 Jan 2014, 09:19, edited 1 time in total.
Renamed the topic, edited the question, added the OA and moved to PS forum.
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Re: different integers [#permalink]

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15 Jul 2007, 07:37
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alimad wrote:
If x and y are different integers and x^2 = xy, which of the following must be true ?

1. x = 0
2. y = 0
3. x = -y

(a). I only
(b). II only
(c). III only
(d). I and III only
(e). I, II, and III

Please provide explaination...Thanks

A for me.

Re-arrange equation: x^2 - xy = 0 => x*(x-y) = 0
Therefore, either x=0 or x=y. We know that x and y are different, so x must be equal to 0.
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15 Jul 2007, 07:41
Answer choice A is correct.

The question gives you x^2=xy
you can simplify that as x^2-xy=0
x(x-y)=0 the solution will be either x=0 or x=y, since x and y are different integers x cannot equal y, that leaves you x=0 to be the only answer.
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Re: different integers [#permalink]

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15 Jul 2007, 09:45
alimad wrote:
If x and y are different integers and x^2 = xy, which of the following must be true ?

1. x = 0
2. y = 0
3. x = -y

(a). I only
(b). II only
(c). III only
(d). I and III only
(e). I, II, and III

Please provide explaination...Thanks

x^2 - xy = 0; x=0 or x=y but it is provided that x<>y
so only A.
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16 Jul 2007, 09:00
I have a question. 4 and -4 are the same integer?
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If x and y are different integers and x^2 = xy, which of the [#permalink]

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08 Feb 2011, 04:12
If x and y are different integers and x^2 = xy, which of the following must be true ?

I. x = 0
II. y = 0
III. x = -y

(A) I only
(B) II only
(C) III only
(D) I and III only
(E) I, II, and III
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Re: problem solving [#permalink]

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08 Feb 2011, 05:33
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Lolaergasheva wrote:
if x and y ARE different integers and x^2=xy which of the following must be true?

x=0
y=0
x=-y

a. 1 only
b. 2 only
c. 3 only
d. 1 and 3
e. 1,2 and 3

$$x^2=xy$$ --> $$x(x-y)=0$$ --> either $$x=0$$ or $$x=y$$ but as given that $$x$$ and $$y$$ are different numbers than the second option is out and we have: $$x=0$$. So only I is always true (in fact because of the same reason that $$x$$ and $$y$$ are different numbers II and III are never true).

Answer: A.
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Re: different integers [#permalink]

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07 Oct 2011, 16:33
Doesn't x=-1 y=1 disprove A?

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Re: different integers [#permalink]

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07 Oct 2011, 16:48
ekang1026 wrote:
Doesn't x=-1 y=1 disprove A?
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x=-1 AND y=1 doesn't satisfy the expression x^2=xy

x^2=xy
(-1)^2=-1*1
1=-1. Not correct.

x^2=xy
x^2-xy=0
x(x-y)=0
Means;
x=y OR x=0
We know $$x \ne y$$. So, x must be equal to 0.
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Re: different integers [#permalink]

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03 Nov 2011, 14:23
Okay, something is off here for me

What I did is this

X ^ 2 = xy

I took the square root of x ^ 2 so I ended up with

x = square root of (x.y)
which means x = square root of (x) * square root of (y)

Therefore, square root of (y) will equal x/square root of (x)

If x is = 0 then it will be undefined. You can't divide by 0.

The same applies to the value of square root of (x) which will equal x/square root of (y). Therefore, y can't equal 0.

Anyone help me here?

Thanks
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Re: different integers [#permalink]

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03 Nov 2011, 22:07
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Lstadt wrote:
Okay, something is off here for me

What I did is this

X ^ 2 = xy

I took the square root of x ^ 2 so I ended up with

x = square root of (x.y)
which means x = square root of (x) * square root of (y)

Therefore, square root of (y) will equal x/square root of (x)

If x is = 0 then it will be undefined. You can't divide by 0.

The same applies to the value of square root of (x) which will equal x/square root of (y). Therefore, y can't equal 0.

Anyone help me here?

Thanks

First of all, square root of $$x^2$$ is |x|, not x. Do not take the square root until and unless you really need to.
$$x^2 = xy$$
You do not divide both sides by x here. You lose out on a solution.
What you can very safely do is $$x^2 - xy = 0$$
$$x(x-y) = 0$$
Now, either x = 0 or x = y or both
Since x and y are different, x must be 0.

When x = 0, no matter what the value of y, $$x^2$$ is equal to xy since both sides are equal to 0.
Since x = 0, you cannot re-write this as $$y = x^2/x$$
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