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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.

Re: problem solving [#permalink]
08 Feb 2011, 05:33

2

This post received KUDOS

Expert's post

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).

Re: different integers [#permalink]
03 Nov 2011, 22:07

2

This post received KUDOS

Expert's post

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 \(y = x^2/x\) _________________

Re: If x and y are different integers and x^2 = xy, which of the [#permalink]
19 Jan 2014, 00:07

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