Question was: x,y,z are positive integers if x(y+z)=63, y(z+x)=40, z(y+x)=33. what is the value of x+y+z?

• x ( y+z ) 49
  y ( z+x ) 40
  z ( y+x ) 33
  
  x 7
  y 4
  z 3
  
  x+y+z 14
  s type mistake solution is
  
  
  
• 8 years agoHelpfull: Yes(12) No(6)
• xy+xz=63
  yz+yx=40
  zy+zx=33
  
  2(xy+xz+yz)=63+33+40
  xy+yz+xz=136/2
  xy+z(y+x)=68
  xy=68-33
  xy=35=7*5
  xz=28=7*4
  x=7
  y=5
  z=4
  
  x+y+z=16
• 8 years agoHelpfull: Yes(12) No(10)
• 2(xy+yz+zx)=136
  xy+yz+zx=68
  we know xy+xz=63
  so yz=68-63=5
  like that we will get zx=28 and xy=35
  yz=5 so z=5/y
  zx=28 so x*5/y=28 so x/y=28/5
  xy=35 so putting x=28y/5 we will get (28/5)y^2=35 so y=5/2
  from this we will get x=14 and z=2
  so x+y+z=14+2.5+2=18.5
• 8 years agoHelpfull: Yes(5) No(3)
• 24 is the right ans.
• 8 years agoHelpfull: Yes(2) No(2)
• x=7
  y=4
  z=3
• 8 years agoHelpfull: Yes(1) No(1)
• x+y+z=48 by solving given equation
• 8 years agoHelpfull: Yes(0) No(1)
• sry
  x=7
  y=4
  z=3
  
  if x ( y+z ) 49
  y ( z+x ) 40
  z ( y+x ) 33
  
• 8 years agoHelpfull: Yes(0) No(3)
• ans is 2580
• 8 years agoHelpfull: Yes(0) No(1)