Question Number 101742 by bemath last updated on 04/Jul/20

 { ((ab+a+b = 5)),((bc + b+c = 14)),((ac + a+c = 9)) :}  find a+b+c = ___

Answered by bramlex last updated on 04/Jul/20

(1) a = ((5−b)/(b+1)) = ((9−c)/(c+1))  (2) b = ((14−c)/(c+1))   (3)  ((5−(((14−c)/(c+1))))/(1+(((14−c)/(c+1))))) = ((9−c)/(c+1))  ((5c+5−14+c)/(c+1+14−c)) = ((9−c)/(c+1))  ((6c−9)/(15)) = ((9−c)/(c+1)) ⇔ ((2c−3)/5) = ((9−c)/(c+1))  (2c−3)(c+1) = 45−5c  2c^2 −c−3+5c−45 = 0  2c^2 +4c−48=0; c^2 +2c−24=0  (c+6)(c−4)=0   { ((c=4 ∧a = 1 ∧ b=2 ⇒ a+b+c = 7)),((c=−6∧a=−3∧b= −4⇒a+b+c =−13)) :}  (Bramlex ⊕)

Answered by nimnim last updated on 04/Jul/20

ab+a+b+1=6⇒(a+1)(b+1)=6.........(1)  bc+b+c+1=15⇒(b+1)(c+1)=15.......(2)  ac+a+c+1=10⇒(a+1)(c+1)=10........(3)  (1)×(2)×(3)  (a+1)^2 (b+1)^2 (c+1)^2 =900  (a+1)(b+1)(c+1)=±30......(4)  Solving (1),(2),(3) and (4) we get  a+1=±5⇒a=4,−6  b+1=±2⇒b=1,−3  c+1=±3⇒c=2,−4  ∴(a+b+c)=7,−13

Commented byRasheed.Sindhi last updated on 04/Jul/20

Cool!

Answered by 1549442205 last updated on 04/Jul/20

(1)⇔(a+1)(b+1)=6  (2)⇔(b+1)(c+1)=15  (3)⇔(c+1)(a+1)=10  [(a+1)(b+1)(c+1)]^2 =900⇔(a+1)(b+1)(c+1)=±30  ⇒ { ((a+1=±2)),((c+1=±5)),((b+1=±3)) :} ⇔(a,b,c)∈{(2,3,5);(−2,−3,−5)}  a+b+c∈{7;−13}