Solve-for-a-b-and-c-1-a-1-b-c-1-2-1-b-1-c-a-1-3-1-c-1-a-b-1-4-

Question Number 202400 by MATHEMATICSAM last updated on 26/Dec/23

Solve for a, b and c

\frac{1}{a} + \frac{1}{b + c} = \frac{1}{2}

\frac{1}{b} + \frac{1}{c + a} = \frac{1}{3}

\frac{1}{c} + \frac{1}{a + b} = \frac{1}{4}

Commented by mr W last updated on 11/Apr/24

s o l u t i o n s e e Q 206294

Answered by Rasheed.Sindhi last updated on 27/Dec/23

{\begin{cases} \frac{1}{a} + \frac{1}{b + c} = \frac{1}{2} \\ \frac{1}{b} + \frac{1}{c + a} = \frac{1}{3} \\ \frac{1}{c} + \frac{1}{a + b} = \frac{1}{4} \end{cases}

{\begin{cases} \frac{a + b + c}{a (b + c)} = \frac{1}{2} \Rightarrow \frac{a (b + c)}{a + b + c} = 2 \\ \frac{a + b + c}{b (c + a)} = \frac{1}{3} \Rightarrow \frac{b (c + a)}{a + b + c} = 3 \\ \frac{a + b + c}{c (a + b)} = \frac{1}{4} \Rightarrow \frac{c (a + b)}{a + b + c} = 4 \end{cases}

{\begin{cases} a (b + c) = 2 (a + b + c) \dots (i) \\ b (c + a) = 3 (a + b + c) \dots (i i) \\ c (a + b) = 4 (a + b + c) \dots (i i i) \end{cases}

{\begin{cases} 6 a (b + c) = 12 (a + b + c) \dots (i) \\ 4 b (c + a) = 12 (a + b + c) \dots (i i) \\ 3 c (a + b) = 12 (a + b + c) \dots (i i i) \end{cases}

12 (a + b + c)

= 6 a (b + c)

= 4 b (c + a)

= 3 c (a + b)

(i i) - (i) : b (c + a) - a (b + c) = a + b + c

(i i i) - (i i) : c (a + b) - b (c + a) = a + b + c

a b + b c - a b - a c = a c + b c - b c - a b

b c - a c = a c - a b

a b + b c - 2 a c = 0

\frac{1}{a} - \frac{2}{b} + \frac{1}{c} = 0

\frac{6 a (b + c)}{a + b + c} = \frac{4 b (c + a)}{a + b + c} = \frac{3 c (a + b)}{a + b + c} = 12

6 a (b + c) = 4 b (c + a) = 3 c (a + b) = 12 (a + b + c)

b + c = \frac{2 (a + b + c)}{a}

c + a = \frac{3 (a + b + c)}{b}

a + b = \frac{4 (a + b + c)}{c}

2 (a + b + c) = \frac{2 (a + b + c)}{a} + \frac{3 (a + b + c)}{b} + \frac{4 (a + b + c)}{c}

2 (a + b + c) - \frac{2 (a + b + c)}{a} - \frac{3 (a + b + c)}{b} - \frac{4 (a + b + c)}{c} = 0

(a + b + c) (2 - \frac{2}{a} - \frac{3}{b} - \frac{4}{c}) = 0

a + b + c = 0 \lor 2 - \frac{2}{a} - \frac{3}{b} - \frac{4}{c} = 0

\frac{2}{a} + \frac{3}{b} + \frac{4}{c} = 2

2 a b c - 2 b c - 3 a c - 4 a b = 0

a + b + c = \frac{a (b + c)}{2} = \frac{b (c + a)}{3} = \frac{c (a + b)}{4}

s = \frac{a (s - a)}{2} = \frac{b (s - b)}{3} = \frac{c (s - c)}{4}

s^{3} = \frac{a (s - a)}{2} \cdot \frac{b (s - b)}{3} \cdot \frac{c (s - c)}{4}

= \frac{a b c}{24} (s - a) (s - b) (s - c)

\dots .

Answered by Frix last updated on 26/Dec/23

Simply solve the 1^{st} for a, insert and solve

the 2^{nd} for b, insert and solve the 3^{rd} for c

a = \frac{23}{10} b = \frac{23}{6} c = \frac{23}{2}

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