SOLUTION: m = (1/3)(xy)^2 I'm solving for x m = 1/3(x^2^y2) m = x^2y^2/3 3m = x^2y^2 3m/y^2 = x^2 ±√3m/y^2 = x >>>> or is it ±√3m/y = x, this is my problem ±3^1

Algebra ->  Radicals -> SOLUTION: m = (1/3)(xy)^2 I'm solving for x m = 1/3(x^2^y2) m = x^2y^2/3 3m = x^2y^2 3m/y^2 = x^2 ±√3m/y^2 = x >>>> or is it ±√3m/y = x, this is my problem ±3^1      Log On


   

Question 1157165: m = (1/3)(xy)^2

I'm solving for x
m = 1/3(x^2^y2)
m = x^2y^2/3
3m = x^2y^2
3m/y^2 = x^2
±√3m/y^2 = x >>>> or is it ±√3m/y = x, this is my problem
±3^1/2m^1/2/(y^2)^1/2 = x
±3^1/2m^1/2/y = x

±√3m/y = x
If this is the correct answer, how is the next step?
Answer by Edwin McCravy(20055) About Me  (Show Source):
You can put this solution on YOUR website!
It's usually best and easiest to clear of fractions first:

m+=+%281%2F3%29%28xy%29%5E2

Clear the fraction by multiplying both sides by 3

3m+=+%28xy%29%5E2

Use the principle of square roots:

%22%22+%2B-+sqrt%283m%29=xy

Divide both sides by y:

%22%22+%2B-+sqrt%283m%29%2Fy=x

Edwin