anonymous
  • anonymous
x^2-y^2 = 2*(x-y)^2 What is the ratio of the smaller variable to the larger variable? Please explain.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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phi
  • phi
@ven pretty close. @mri can you fix his algebra?
anonymous
  • anonymous
@Phi: Fail from my side think I did a mistake :/
anonymous
  • anonymous
I'll try to fix let me try it before you fix it please.

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More answers

anonymous
  • anonymous
yes, (x-y)^2 doesnt equal x^2 - y^2
phi
  • phi
x^2-y^2 = 2*(x-y)^2 (x-y)(x+y)= 2*(x-y)(x-y) keep going...
anonymous
  • anonymous
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?
phi
  • phi
ratio of the smaller variable to the larger variable? 1:3 Yes
anonymous
  • anonymous
Awesome. Thanks phi
anonymous
  • anonymous
I got 1:3 as well but how do you understand what is smaller and which is larger phi?
phi
  • phi
although when I got to (x+y)=2(x-y) I did: x+y = 2x - 2y 0= x - 3y 3y=x
anonymous
  • anonymous
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?
phi
  • phi
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)
anonymous
  • anonymous
Ah got it, thank you :)
phi
  • phi
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?
anonymous
  • anonymous
ambiguous?
anonymous
  • anonymous
I looked it up. Um.. It only had one answer.
anonymous
  • anonymous
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
anonymous
  • anonymous
Try plugging the negative numbers back in to the original equation.
phi
  • phi
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.
anonymous
  • anonymous
Okay, thanks, i gotta go eat dinner

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