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.