SOLUTION: My question is {{{x^2+y^2+4x-14y+53=256/81}}} I know how to start it .. {{{x^2+4x+2+y^2-14y+49}}} .... but now what do I do with the 53? If I move it on the other side then do

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equations Lessons  -> Quadratic Equation Lesson -> SOLUTION: My question is {{{x^2+y^2+4x-14y+53=256/81}}} I know how to start it .. {{{x^2+4x+2+y^2-14y+49}}} .... but now what do I do with the 53? If I move it on the other side then do      Log On


   

Question 5847: My question is x%5E2%2By%5E2%2B4x-14y%2B53=256%2F81
I know how to start it ..
x%5E2%2B4x%2B2%2By%5E2-14y%2B49 .... but now what do I do with the 53?
If I move it on the other side then do I multiply it with 256%2F81 or what??
After that step I would get %28x%2B2%29%5E2%2B%28y-7%29%5E2
I did not get any further..
PLEASE HELP ME SOLVE THE REST
Thank-you!
Answer by longjonsilver(2297) About Me  (Show Source):
You can put this solution on YOUR website!
x%5E2%2By%5E2%2B4x-14y%2B53=256%2F81
x%5E2%2B4x+%2By%5E2-14y+=+256%2F81+-+53
x%5E2%2B4x%2B4+%2By%5E2-14y+=+256%2F81+-+53%2B4
x%5E2%2B4x%2B4+%2By%5E2-14y%2B49+=+256%2F81+-+53%2B4%2B49 --if you had known at the beginning that the 2 numbers you wanted - the 4 and 49 - where the 53, then we could just have used that...but we aren't mind readers, so tough lol)

x%5E2%2B4x%2B4+%2By%5E2-14y%2B49+=+256%2F81

%28x%2B2%29%5E2+%2B+%28y-7%29%5E2+=+256%2F81
%28x%2B2%29%5E2+%2B+%28y-7%29%5E2+=+%2816%2F9%29%5E2

--> standard form :-)

jon.