ab+a+b=250 bc+b+c=300 ca+c+a=216
find a+b+c

• ab+a+b=250
  10a+b+a+b=250
  11a+2b=250----(1)
  like wise
  11b+2c=300----(2)
  11c+2a=216----(3)
  adding 1,2,3
  we get 13(a+b+c)=766
  a+b+c=766/13=58.92----ans
• 10 years agoHelpfull: Yes(10) No(4)
• ab+ a+ b = 250---1
  bc+b+c = 300---2
  ca+ c+ a= 216----3
  Adding 1 2 3
  2(a+b+c) + (ab+bc+ca) = 766
  2(a+b+C) = 766 - ab - bc- ca
  2(a+ b+ c) = 766 -(a + b - 250) - (b+c -300) -(c+ a -216)
  2(a+b+c) = 766 + 250+300+216 -2(a +b+c)
  4(a+b+c) = 766+766
  a+b+c = 383
• 10 years agoHelpfull: Yes(9) No(5)
• given,
  ab + a + b = 250
  a(1+b) + b = 250
  
  a = (250−b)/(1+b)--------(1)
  
  ac + a + c = 300
  c(1+a) + a = 300
  c = (300−a)/(1+a)
  substitute eqn(1) in d above eqn, we get
  
  c=(50+301b)/251 -------(2)
  
  given bc+b+c=216
  substitute eqn(2) in d above eqn, we get
  43b^2 +86b -7738=0
  and we get
  b=12.45 or -14.45 ( we have [-b +or- sqrt(b^2 - 4ac)]/2a )
  substitue b=12.45 in eqn(1), we get
  a=17.66
  substitue b=12.45 in eqn(2), we get
  c=15.13
  so,
  a=17.66
  b=12.45
  c=15.13
  
  so
  a+b+c= 17.66+12.45+15.13 = 45.24
• 10 years agoHelpfull: Yes(5) No(0)
• 
  ab + a + b = 250
  a(1+b) + b = 250
  
  a = (250−b)(1+b)
  
  ac + a + c = 300
  c(1+a) + a = 300
  
  c = (300−a)(1+a)
  
  =300−(250−b)(1+b)1+(250−b)(1+b)=300(1+b)−(250−b)1+b+250−b
  
  =300(1+b)−(250−b)251=(50+301b)251=(50+301b)251
  
  bc + b + c = 216
  
  (50+301b)b251+b+(50+301b)251=216
  
  (50 + 301b)b + 251b + 50 + 301b = 54216
  
  301b2 +602b−54216 =0
  
  b=12.46
  
  a =(250−b)(1+b)=(250−12.46)(1+12.46) = 17.65
  c = (300−a)(1+a)=(300−17.65)(1+17.65) = 15.14
  
  These values of a,b and c statisfies the given conditions,as given below
  ab + a + b = 17.65 *12.46 + 17.65 + 12.46 ≈ 250
  ac + a + c = 17.65 *15.14 + 17.65 + 15.14 ≈ 300.21
  bc + b + c = 12.46 *15.14 + 12.46 + 15.14 ≈ 216
  
  Hence,
  
• 10 years agoHelpfull: Yes(0) No(8)