x/y=6/5 find value of
(x^2+y^2)/(x^2-y^2)

• x/y=6/5
  while taking as proportions x=6k and y=5k
  thus (x^2+y^2)=((6k)^2+(5k)^2)
  =>(36(k^2))+(25(k^2))
  =>61(k^2)
  similarly (x^2-y^2)=((6k)^2-(5k)^2)
  =>(36(k^2))-(25(k^2))
  =>11(k^2)
  finally (x^2+y^2)/(x^2-y^2)=61(k^2)/11(k^2)=61/11
  
• 11 years agoHelpfull: Yes(4) No(1)
• using componendo and dividendo rule,
  x/y = 6/5
  (x+y)/(x-y)=(6+5)/(6-5)
  (x+y)/(x-y)=11,here (x-y)=(x+y)/11
  Now we have
  (x^2+y^2)/(x^2-y^2)=(x^2+y^2)/(x+y)(x-y)
  x^2+y^2=(x+y)^2-2xy
  now,(x+y)^2-2xy/(x+y)(x-y)
  11((x+y)^2-2xy)/(x+y)^2, here we substitute the value of (x-y)=(x+y)/11
  now take (x+y)^2 common,
  we have,
  11(1-2xy/(x+y)^2)
  11(1-2xy/(x^2+y^2+2xy))
  11(1-2y/x-2x/y-1)
  we have x/y=6/5,y/x=5/6 , now
  11(-2*5/6-2*6/5)
  (-44.73) Ans
• 11 years agoHelpfull: Yes(2) No(1)
• ans: 61/11
  solution: (x2+y2)/(x2-y2)= ((x-y)^2-2xy)/(x2-y2)
  = (x-y)^2/(x2-y2) - 2xy/(x2-y2)
  = (x-y)^2/((x+y)(x-y)) - 2xy/((x+y)(x-y))
  = (x-y)/(x+y) - 2xy/(xy(1+y/x)(x/y-1)
  = (x-y)/x+y) - 2/(1+y/x)(x/y-1) ------- eqn1
  we know x/y=6/5,,aplly componendo and dividend
  x-y/x+y=1/11
  put in eqn the value of x-y/x+y, x/y and y/x
• 11 years agoHelpfull: Yes(0) No(3)