ab-a-b-5-bc-b-c-14-ac-a-c-9-find-a-b-c-

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

{\begin{cases} a b + a + b = 5 \\ b c + b + c = 14 \\ a c + a + c = 9 \end{cases}

find a + b + c =___

Answered by bramlex last updated on 04/Jul/20

(1) a = \frac{5 - b}{b + 1} = \frac{9 - c}{c + 1}

(2) b = \frac{14 - c}{c + 1}

(3) \frac{5 - (\frac{14 - c}{c + 1})}{1 + (\frac{14 - c}{c + 1})} = \frac{9 - c}{c + 1}

\frac{5 c + 5 - 14 + c}{c + 1 + 14 - c} = \frac{9 - c}{c + 1}

\frac{6 c - 9}{15} = \frac{9 - c}{c + 1} \Leftrightarrow \frac{2 c - 3}{5} = \frac{9 - c}{c + 1}

(2 c - 3) (c + 1) = 45 - 5 c

2 c^{2} - c - 3 + 5 c - 45 = 0

2 c^{2} + 4 c - 48 = 0; c^{2} + 2 c - 24 = 0

(c + 6) (c - 4) = 0

{\begin{cases} c = 4 \land a = 1 \land b = 2 \Rightarrow a + b + c = 7 \\ c = - 6 \land a = - 3 \land b = - 4 \Rightarrow a + b + c = - 13 \end{cases}

(B ramlex \oplus)

Answered by nimnim last updated on 04/Jul/20

ab + a + b + 1 = 6 \Rightarrow (a + 1) (b + 1) = 6 \dots \dots \dots (1)

bc + b + c + 1 = 15 \Rightarrow (b + 1) (c + 1) = 15 \dots \dots . (2)

ac + a + c + 1 = 10 \Rightarrow (a + 1) (c + 1) = 10 \dots \dots . . (3)

(1) \times (2) \times (3)

{(a + 1)}^{2} {(b + 1)}^{2} {(c + 1)}^{2} = 900

(a + 1) (b + 1) (c + 1) = \pm 30 \dots \dots (4)

Solving (1), (2), (3) and (4) we get

a + 1 = \pm 5 \Rightarrow a = 4, - 6

b + 1 = \pm 2 \Rightarrow b = 1, - 3

c + 1 = \pm 3 \Rightarrow c = 2, - 4

∴ (a + b + c) = 7, - 13

Commented by Rasheed.Sindhi last updated on 04/Jul/20

C o o l!

Answered by 1549442205 last updated on 04/Jul/20

(1) \Leftrightarrow (a + 1) (b + 1) = 6

(2) \Leftrightarrow (b + 1) (c + 1) = 15

(3) \Leftrightarrow (c + 1) (a + 1) = 10

{[(a + 1) (b + 1) (c + 1)]}^{2} = 900 \Leftrightarrow (a + 1) (b + 1) (c + 1) = \pm 30

\Rightarrow {\begin{cases} a + 1 = \pm 2 \\ c + 1 = \pm 5 \\ b + 1 = \pm 3 \end{cases} \Leftrightarrow (a, b, c) \in {(2, 3, 5); (- 2, - 3, - 5)}

a + b + c \in {7; - 13}