Question

x^2-y^2 = 2*(x-y)^2
What is the ratio of the smaller variable to the larger variable? Please explain.

Answer 1

@ven pretty close. @mri can you fix his algebra?

Answer 2

@Phi: Fail from my side think I did a mistake :/

Answer 3

I'll try to fix let me try it before you fix it please.

Answer 4

yes, (x-y)^2 doesnt equal x^2 - y^2

Answer 5

x^2-y^2 = 2*(x-y)^2
(x-y)(x+y)= 2*(x-y)(x-y)
keep going...

Answer 6

Get rid of the (x-y) on both sides to get
(x+y)=2(x-y)
(x+y)/(x-y)=2/1
Cross multiply?
2x-2y=x+y
x = 3y
Answer is 1:3?

Answer 7

ratio of the smaller variable to the larger variable? 1:3 Yes

Answer 8

Awesome. Thanks phi

Answer 9

I got 1:3 as well but how do you understand what is smaller and which is larger phi?

Answer 10

although when I got to
(x+y)=2(x-y)
I did:
x+y = 2x - 2y
0= x - 3y
3y=x

Answer 11

That wouldve been quicker, but, oh well, at least i got it.
@vengeance one of them is larger and one is smaller, not sure if that answers your question. Could you be more clear?

Answer 12

if we have x = 3y
then, to use a concrete example, let y=1 so x=3
ratio of smaller to larger is 1:3
of course, this will work for any y value (except 0)

Answer 13

Ah got it, thank you :)

Answer 14

I think there is more to this than I thought. We could have y= -1, x= -3
and the ratio of small to large is -3:-1 = 3:1
Is this one of those ambiguous questions?

Answer 15

ambiguous?

Answer 16

I looked it up. Um.. It only had one answer.

Answer 17

Maybe there is a secret rule to ratios that when converting to positive you take the reciprocal. Maybe like when dividing by a negative in inequalities

Answer 18

Try plugging the negative numbers back in to the original equation.

Answer 19

I just did a quick goole search. It looks like you can take ratios of negatives, but generally it is not done, so I would assume positive numbers unless specifically told otherwise.

yes, negative numbers work in your original equation.

Answer 20

Okay, thanks, i gotta go eat dinner