If ab + b + a = 135 bc + b + c = 47

ca + a + c = 101

What is the value of a + b + c?

• ab+a+b+1 = 135 + 1
  a(b+1)+1(b+1) =136
  (a+1)(b+1) =136 —-1
  Same for further 2
  (b+1)(c+1)=48 ——2
  (a+1)(c+1)=102 ——-3
  Now make factors
  For eq1 , factors 17*8
  For 2, factors 8*6
  For 3 , factors 6*17
  So by factors
  a+1 = 17
  a=16
  b+1=8
  b=7
  c+1=6
  c=5
  a+b+c=28
• 6 years agoHelpfull: Yes(18) No(1)
• 2(ab+bc+ca)+2(a+b+c)=283
  2(135-a-b+47-b-c+101-a-c)+2(a+b+c)=283
  566-(a+b+c)=283
  2(a+b+c)=283
  a+b+c=141.5
• 6 years agoHelpfull: Yes(2) No(24)
• ab+a+b=135 -----(1)
  bc+b+c=47 -----(2)
  ca+b+c=101 -----(3)
  
  1st step:
  take equation 1 and add 1 on both sides
  ab+a+b+1=135+1
  a(b+1)+b+1=136
  b+1(a+1)=136 -------(4)
  
  take equation (2) and add 1 on both sides
  bc+b+c +1=47+1
  b(c+1)+c+1=48
  c+1(b+1)=48 ---------(5)
  
  take equation (3) & add 1 on both sides
  ca+a+c+1=101+1
  a(c+1)+c+1=102
  c+1(a+1)=102 ---------(6)
  
  solve using this 4*5/6
  ((b+1)(a+1))*((c+1)(b+1))/((c+1)(a+1))=138*48/(102)
  eliminate and solve
  (b+1)^2=6528/102
  =64
  b+1=root(64)
  b+1=8
  b=7 --------(7)
  use b=7 in (4)
  (7+1)(a+1)=136
  8a=136-8
  8a=128
  a=128/8=16
  a=16 ----------------------(8)
  sub b=7 in(5)
  (c+1)(7+1)=48
  8c=48-8
  c=40/8=5
  c=5 ------------(9)
  therefore (7)+(8)+(9)=16+7+5=28
  ANSWER: a+b+c=28
• 3 years agoHelpfull: Yes(0) No(0)