If (x+y):(y+z):(z+x)=4:6:5 and (x+y+z)=15,then the value of z is

• Z=7
  
  (X+Y)=4a ---(i)
  (Y+Z)=6a ---(ii)
  (Z+X)=5a ---(iii)
  So (X+Y)+(Y+Z)+(Z+X)=2(X+Y+Z)=4a+6a+5a=15a ,Given (X+Y+Z)=15, So 2(X+Y+Z)=30=15a, a=2
  Substituting value of a=2, in (i), (ii) & (iii), we have
  X+Y=8 ---(iv)
  Y+Z=12 ---(v) &
  Z+X=10 ---(vi)
  On solving (iv), (v) & (vi), X=3, Y=5, Z=7
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• Given that (x+y):(y+z):(z+x)=4:6:5
  so let x+y=4p.............(i),where p is a ratio constant
  y+z=6p..................(ii) and
  z+x=5p..........(iii)
  now adding (i),(ii),(iii) we get,
  2(x+y+z)=15p
  2*15=15p[as,x+y+z=15]
  so,p=2
  so,x+y=4*2
  x+y=8........(iv)
  y+z=12....(v)
  z+x=10.....(vi)
  now we know, x+y+z=15
  so putting the value of eqn(iv)
  we get, 8+z=15
  so,z=7
  
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• x+y=4x,
  y+z=6x,
  z+x=5x
  after that we can write ,2(x+y+z)=4x+6x+5x
  2*15=15x,
  x=2..
  so,the value of z will be,z+2=10,
  
  Z=8....
  
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• by the rule of ratio- proportions the ans of z=7
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• (x+y)/(y+z)=4/6...solving this 6x+2y-4z=0..eq1
  
  (y+z)/(z+x)=6/5..solving this 6x-5y+z=0...eq2
  
  on solving 1 and 2 get
  7y+5z=0
  y=-5z/7
  now putting the value of y in eq2
  we get
  x=19z/21
  
  eq x+y+z=15
  put the value of x, and y
  19z/21-5z/7+z=15..solve the
  z=63/5
  y=-9
  x=57/5
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