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)


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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
	${ (((1/a) + (1/(b + c)) = (1/2))),(((1/b) + (1/(c + a)) = (1/3) )),(((1/c) + (1/(a + b)) = (1/4))) :} { (( ((a+b+c)/(a(b + c))) = (1/2)⇒((a(b + c))/(a+b+c))=2)),(( ((a+b+c)/(b(c + a))) = (1/3)⇒((b(c+a))/(a+b+c))=3 )),(( ((a+b+c)/(c(a + b))) = (1/4)⇒((c(a+b))/(a+b+c))=4)) :} { (( a(b+c)=2(a+b+c)...(i))),(( b(c+a)=3(a+b+c)...(ii) )),(( c(a+b)=4(a+b+c)...(iii))) :} { (( 6a(b+c)=12(a+b+c)...(i))),(( 4b(c+a)=12(a+b+c)...(ii) )),(( 3c(a+b)=12(a+b+c)...(iii))) :} 12(a+b+c) = 6a(b+c) = 4b(c+a) =3c(a+b) (ii)−(i): b(c+a)−a(b + c)=a+b+c (iii)−(ii): c(a+b)−b(c+a)=a+b+c ab+bc−ab−ac=ac+bc−bc−ab bc−ac=ac−ab ab+bc−2ac=0 (1/a)−(2/b)+(1/c)=0 ((6a(b + c))/(a+b+c))=((4b(c+a))/(a+b+c))=((3c(a+b))/(a+b+c))=12 6a(b + c)=4b(c+a)=3c(a+b)=12(a+b+c) b+c=((2(a+b+c))/a) c+a=((3(a+b+c))/b) a+b=((4(a+b+c))/c) 2(a+b+c)=((2(a+b+c))/a)+((3(a+b+c))/b)+((4(a+b+c))/c) 2(a+b+c)−((2(a+b+c))/a)−((3(a+b+c))/b)−((4(a+b+c))/c)=0 (a+b+c)(2−(2/a)−(3/b)−(4/c))=0 a+b+c=0 ∨ 2−(2/a)−(3/b)−(4/c)=0 (2/a)+(3/b)+(4/c)=2 2abc−2bc−3ac−4ab=0 a+b+c=((a(b + c))/2)=((b(c + a))/3)=((c(a + b))/4) s=((a(s−a))/2)=((b(s−b))/3)=((c(s−c))/4) s^3 =((a(s−a))/2)∙((b(s−b))/3)∙((c(s−c))/4) =((abc)/(24))(s−a)(s−b)(s−c) ....$
	${\begin{cases} \frac{1}{a} + \frac{1}{b + c} = \frac{1}{2} \\ \frac{1}{b} + \frac{1}{c + a} = \frac{1}{3} \\ \frac{1}{c} + \frac{1}{a + b} = \frac{1}{4} \end{cases}$ ${\begin{cases} \frac{a + b + c}{a (b + c)} = \frac{1}{2} \Rightarrow \frac{a (b + c)}{a + b + c} = 2 \\ \frac{a + b + c}{b (c + a)} = \frac{1}{3} \Rightarrow \frac{b (c + a)}{a + b + c} = 3 \\ \frac{a + b + c}{c (a + b)} = \frac{1}{4} \Rightarrow \frac{c (a + b)}{a + b + c} = 4 \end{cases}$ ${\begin{cases} a (b + c) = 2 (a + b + c) . . . (i) \\ b (c + a) = 3 (a + b + c) . . . (i i) \\ c (a + b) = 4 (a + b + c) . . . (i i i) \end{cases}$ ${\begin{cases} 6 a (b + c) = 12 (a + b + c) . . . (i) \\ 4 b (c + a) = 12 (a + b + c) . . . (i i) \\ 3 c (a + b) = 12 (a + b + c) . . . (i i i) \end{cases}$ $12 (a + b + c)$ $= 6 a (b + c)$ $= 4 b (c + a)$ $= 3 c (a + b)$ $(i i) - (i) : b (c + a) - a (b + c) = a + b + c$ $(i i i) - (i i) : c (a + b) - b (c + a) = a + b + c$ $a b + b c - a b - a c = a c + b c - b c - a b$ $b c - a c = a c - a b$ $a b + b c - 2 a c = 0$ $\frac{1}{a} - \frac{2}{b} + \frac{1}{c} = 0$ $\frac{6 a (b + c)}{a + b + c} = \frac{4 b (c + a)}{a + b + c} = \frac{3 c (a + b)}{a + b + c} = 12$ $6 a (b + c) = 4 b (c + a) = 3 c (a + b) = 12 (a + b + c)$ $b + c = \frac{2 (a + b + c)}{a}$ $c + a = \frac{3 (a + b + c)}{b}$ $a + b = \frac{4 (a + b + c)}{c}$ $2 (a + b + c) = \frac{2 (a + b + c)}{a} + \frac{3 (a + b + c)}{b} + \frac{4 (a + b + c)}{c}$ $2 (a + b + c) - \frac{2 (a + b + c)}{a} - \frac{3 (a + b + c)}{b} - \frac{4 (a + b + c)}{c} = 0$ $(a + b + c) (2 - \frac{2}{a} - \frac{3}{b} - \frac{4}{c}) = 0$ $a + b + c = 0 \lor 2 - \frac{2}{a} - \frac{3}{b} - \frac{4}{c} = 0$ $\frac{2}{a} + \frac{3}{b} + \frac{4}{c} = 2$ $2 a b c - 2 b c - 3 a c - 4 a b = 0$ $a + b + c = \frac{a (b + c)}{2} = \frac{b (c + a)}{3} = \frac{c (a + b)}{4}$ $s = \frac{a (s - a)}{2} = \frac{b (s - b)}{3} = \frac{c (s - c)}{4}$ $s^{3} = \frac{a (s - a)}{2} \cdot \frac{b (s - b)}{3} \cdot \frac{c (s - c)}{4}$ $= \frac{a b c}{24} (s - a) (s - b) (s - c)$ $. . . .$
	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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