a,b,c are positive numbers such that a+b+ab= 8, b+c+bc=15 and c+a+ca = 35
What is the value of a+b+c+abc?

• add 1 on both sides to all the equations and take common elements
  now the equations become
  (a+1)(b+1)=9
  (b+1)(c+1)=16
  (c+1)(a+1)=36
  solve the equations for a , b, c
  so we get
  a=3.5 b=1 and c=7
  so the value of a+b+c+abc= 36
• 6 years agoHelpfull: Yes(19) No(9)
• GIVEN:-
  
  
  a + b + ab = 8
  
  
  b + a(1 + b) = 8
  
  
  a = {8 - b}/{1 + b}}-------( 1 )
  
  
  b + c + bc = 15
  
  
  b + c(1 + b) = 15
  
  c = {15 - b}/{1 + b}} ------( 2 )
  
  
  a + c + ac = 35
  
  {8 - b}/{1 + b} + {15 - b}/{1 + b} +{8 - b}/{1 + b}*{15 - b}/{1 + b} = 35
  
  
  From----------( 1 ) &-----------( 2 )
  
  
  {8 - b + 15 - b}/{1 + b} + {120 - 8b - 15b + b^2}/{(1 + b)^2} = 35
  
  
  {(23 - 2b)(1 + b) + (120 - 23b + b^2) = 35(1 + b)^2}
  
  
  {23 + 23b - 2b - 2b^2 + 120 - 23b + b^2 = 35 + 35b^2 + 70b}
  
  
  {143 - 35 - 2b - b^2 = 35b^2 + 70b}
  
  
  {108 - 72b - 36b^2 = 0}
  
  
  {b^2 + 2b - 3 = 0}
  
  
  {b^2 + 3b - b - 3 = 0}
  
  
  {b(b + 3) - 1(b + 3)}
  
  
  {(b + 3)(b - 1)}
  
  
  so, b = 1 , b = -3[neglect]
  
  
  because a , b , c ,are positive,
  
  
  now,
  
  
  from----------( 1 )
  
  
  a = (8 - b)(1 + b)
  
  
  a = (8 - 1)(1 + 1)
  
  
  a = 7/2
  
  
  from-----------( 2 )
  
  
  c = (15 - b)(1 + b)
  
  
  c = (15 - 1)(1 + 1)
  
  
  c = 14/2
  
  
  c = 7
  
  
  thus, the value of "a + b + c + abc" is
  
  
  7/2 + 1 + 7 + (7 * 7/2 * 1)
  
  
  {(7 + 2 + 14 )/2 + 49/2}
  
  
  (23/2 + 49/2)
  
  
  (23 + 49)/2
  
  
  72/2
  
  
  36
  
  
  HENCE, the value of (a + b + c + abc) = 36
• 6 years agoHelpfull: Yes(6) No(4)
• a-b=25.9
  a=6b
  Sub value of a in 1st eqn
  b=5.18
  a=30.08
  a+b=38.26
• 5 years agoHelpfull: Yes(2) No(2)