If 2x = 3y = 6-z, then (1/x + 1/y + 1/z)
is equal to:

• Solution:
  2x = 6 -z Þ 2 = 6-z/x -------------(i)
  3y = 6 -z Þ 3 = 6-z/y -------------(ii)
  Multiplying (i)and(ii)
  2x3 = 6-z/x x 6-z/y
  6 = 6(1-z/x) *6(1-z/y)
  1 = (-z/x) +(-z/y)
  1= -z(1/x +1/y)
  1/-z = 1/x +1/y
  1/x + 1/y + 1/z = 0
  
• 10 years agoHelpfull: Yes(7) No(6)
• 0
  2x = 6 -z then 2 = 6-z/x -------------(i)
  3y = 6 -z then 3 = 6-z/y -------------(ii)
  Multiplying (i)and(ii)
  2x3 = 6-z/x x 6-z/y
  61 = 6-z/x +-z/y
  1 = (-z/x) +(-z/y)
  1= -z(1/x +1/y)
  1/-z = 1/x +1/y
  1/x + 1/y + 1/z = 0
• 10 years agoHelpfull: Yes(4) No(4)
• let 2^x = 3^y = 6^z=k
  
  so , 2 = k^(1/x)
  3 = k^(1/y)
  6 = k^(-1/z)
  
  since , 2 * 3 = 6
  
  so, 2 * 3 = k ^ (1/x) * k ^ (1/y) = k ^ [ (1/x) + (1/y) ]
  
  also , 6=k ^ ( -1/z )
  
  therefore , k ^ [ (1/x) + (1/y) ]=k ^ ( -1/z )
  
  so [ (1/x) + (1/y) ] = ( -1/z )
  
  ==> (1/x) + (1/y) -( -1/z ) = 0
  
  so (1/x) + (1/y) + (1/z ) = 0 ans.
• 7 years agoHelpfull: Yes(3) No(0)