Question Number 215048 by Hanuda354 last updated on 27/Dec/24
$$\mathrm{Solve}\:\:\mathrm{for}\:\:{a},\:{b},\:{c}\:\in\:\mathbb{R} \\ $$$$\:\:\:\frac{\mathrm{1}}{{a}}\:+\:\frac{\mathrm{1}}{{b}+{c}}\:=\:\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\:\:\:\frac{\mathrm{1}}{{b}}\:+\:\frac{\mathrm{1}}{{c}+{a}}\:=\:\frac{\mathrm{1}}{\mathrm{3}} \\ $$$$\:\:\:\frac{\mathrm{1}}{{c}}\:+\:\frac{\mathrm{1}}{{a}+{b}}\:=\:\frac{\mathrm{1}}{\mathrm{4}} \\ $$
Commented by Abdullahrussell last updated on 27/Dec/24
![let, a+b+c=x we get from eqn. (1),(2),(3) respectivly ab+ac=2x, bc+ab=3x, ac+bc=4x adding all we get: ab+bc+ca=((9x)/2) ⇒ab=(x/2) , bc=((5x)/2) , ca=((3x)/2) ⇒((ac)/(bc))=(a/b)=((3x)/2)×(2/(5x))=(3/5)⇒a:b=3:5 ⇒((ab)/(ac))=(b/c)=(x/2)×(2/(3x))=(1/3)⇒b:c=1:3 ⇒a:b:c=3:5:15 [3k:5k:15k] ⇒a+b+c=23k=x ab=(x/2)⇒3k×5k=((23k)/2)⇒k=((23)/(30)) ⇒a=((23)/(10)), b=((23)/6), c=((23)/2) (Ans).](https://www.tinkutara.com/question/Q215049.png)
$$\:{let},\:{a}+{b}+{c}={x} \\ $$$$\:{we}\:{get}\:{from}\:{eqn}.\:\left(\mathrm{1}\right),\left(\mathrm{2}\right),\left(\mathrm{3}\right)\:{respectivly} \\ $$$$\:\:{ab}+{ac}=\mathrm{2}{x},\:{bc}+{ab}=\mathrm{3}{x},\:{ac}+{bc}=\mathrm{4}{x} \\ $$$$\:{adding}\:{all}\:{we}\:{get}:\:{ab}+{bc}+{ca}=\frac{\mathrm{9}{x}}{\mathrm{2}} \\ $$$$\:\Rightarrow{ab}=\frac{{x}}{\mathrm{2}}\:,\:{bc}=\frac{\mathrm{5}{x}}{\mathrm{2}}\:,\:{ca}=\frac{\mathrm{3}{x}}{\mathrm{2}} \\ $$$$\:\Rightarrow\frac{{ac}}{{bc}}=\frac{{a}}{{b}}=\frac{\mathrm{3}{x}}{\mathrm{2}}×\frac{\mathrm{2}}{\mathrm{5}{x}}=\frac{\mathrm{3}}{\mathrm{5}}\Rightarrow{a}:{b}=\mathrm{3}:\mathrm{5} \\ $$$$\:\Rightarrow\frac{{ab}}{{ac}}=\frac{{b}}{{c}}=\frac{{x}}{\mathrm{2}}×\frac{\mathrm{2}}{\mathrm{3}{x}}=\frac{\mathrm{1}}{\mathrm{3}}\Rightarrow{b}:{c}=\mathrm{1}:\mathrm{3} \\ $$$$\:\Rightarrow{a}:{b}:{c}=\mathrm{3}:\mathrm{5}:\mathrm{15}\:\:\left[\mathrm{3}{k}:\mathrm{5}{k}:\mathrm{15}{k}\right]\: \\ $$$$\:\Rightarrow{a}+{b}+{c}=\mathrm{23}{k}={x} \\ $$$$\:{ab}=\frac{{x}}{\mathrm{2}}\Rightarrow\mathrm{3}{k}×\mathrm{5}{k}=\frac{\mathrm{23}{k}}{\mathrm{2}}\Rightarrow{k}=\frac{\mathrm{23}}{\mathrm{30}} \\ $$$$\:\Rightarrow{a}=\frac{\mathrm{23}}{\mathrm{10}},\:{b}=\frac{\mathrm{23}}{\mathrm{6}},\:{c}=\frac{\mathrm{23}}{\mathrm{2}}\:\:\left({Ans}\right). \\ $$
Commented by Hanuda354 last updated on 27/Dec/24
$${Thanks} \\ $$