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Re: Is │x│=│y│? (1) x - y = 6 (2) x + y = 0 [#permalink]
12 Jun 2013, 11:43

I should have worded that better.

I guess that solving the problem they way I did, I would say A is correct.

I see that for x-y=6 there could be many values for x and y, however, it seems like many similar problems are solved by manipulating statements (i.e. x-y=6) and plugging into x=y or x=-y.

I understand how this problem was solved, but I want to understand WHY it was solved the way it was.

As always, thank you for for help.

Zarrolou wrote:

WholeLottaLove wrote:

My first instinct was to manipulate |x|=|y|

x=y OR x=-y

1.) says that x-y=6 which means we can get values for x and y

x=6+y y=x-6

So, for x=y

6+y=y 6=0 (Invalid)

x=x-6 0=6 (Invalid)

So, for x=-y

6+y=-y y=-3

x=-x+6 x=3

Why wouldn't we use that methodology on this problem?

Are you saying that A is sufficient?

In the case x=3 and y=-3 => |x|=|y|.

But if x=90 and y=84 then x - y = 6 but |x|\neq{|y|}.

The question asks you if x=y OR x=-y, you cannot assume that it's true in your solution to find the values of x,y for which it holds ture.

Re: Is │x│=│y│? (1) x - y = 6 (2) x + y = 0 [#permalink]
12 Jun 2013, 11:48

1

This post received KUDOS

Expert's post

WholeLottaLove wrote:

My first instinct was to manipulate |x|=|y|

x=y OR x=-y

1.) says that x-y=6 which means we can get values for x and y

x=6+y y=x-6

So, for x=y

6+y=y 6=0 (Invalid)

x=x-6 0=6 (Invalid)

So, for x=-y

6+y=-y y=-3

x=-x+6 x=3

Why wouldn't we use that methodology on this problem?

When you assume that x=y and solve the equation [x-y = 6], you WILL get invalid solutions as you got because you have anyways assumed that x=y--> x-y = 0;which contradicts the given fact.

However, when you assume that x=-y-->x+y=0, you have inherently assumed that |x| IS equal to |y| and now you are just solving for the values of x and y. Thus, the equation [x-y=6] would really not make any difference for this method.( x-y) could equal anything and you would still get |x| = |y|.

Re: Is │x│=│y│? (1) x - y = 6 (2) x + y = 0 [#permalink]
12 Jun 2013, 11:52

Ahhh! That makes sense! I am assuming |x| = |y| when I am trying to verify if it does or does not. Thanks!

vinaymimani wrote:

WholeLottaLove wrote:

My first instinct was to manipulate |x|=|y|

x=y OR x=-y

1.) says that x-y=6 which means we can get values for x and y

x=6+y y=x-6

So, for x=y

6+y=y 6=0 (Invalid)

x=x-6 0=6 (Invalid)

So, for x=-y

6+y=-y y=-3

x=-x+6 x=3

Why wouldn't we use that methodology on this problem?

When you assume that x=y and solve the equation [x-y = 6], you WILL get invalid solutions as you got because you have anyways assumed that x=y--> x-y = 0;which contradicts the given fact.

However, when you assume that x=-y-->x+y=0, you have inherently assumed that |x| IS equal to |y| and now you are just solving for the values of x and y. Thus, the equation [x-y=6] would really not make any difference for this method.( x-y) could equal anything and you would still get |x| = |y|.