Question Number 202540 by MATHEMATICSAM last updated on 29/Dec/23
![If (b/(a + b)) = ((3a − b − c)/(2b + c − a)) = ((3c − a)/(2a − b + 3c)) [a + b + c ≠ 0] then show that a = b = c.](https://www.tinkutara.com/question/Q202540.png)
$$\mathrm{If}\:\frac{{b}}{{a}\:+\:{b}}\:=\:\frac{\mathrm{3}{a}\:−\:{b}\:−\:{c}}{\mathrm{2}{b}\:+\:{c}\:−\:{a}}\:=\:\frac{\mathrm{3}{c}\:−\:{a}}{\mathrm{2}{a}\:−\:{b}\:+\:\mathrm{3}{c}}\: \\ $$$$\left[{a}\:+\:{b}\:+\:{c}\:\neq\:\mathrm{0}\right]\:\mathrm{then}\:\mathrm{show}\:\mathrm{that}\:{a}\:=\:{b}\:=\:{c}. \\ $$
Answered by som(math1967) last updated on 29/Dec/23
![(b/(a+b))=((3a−b−c)/(2b+c−a))=((3c−a)/(2a−b+3c)) ((3b)/(3a+3b))=((3a−b−c)/(2b+c−a))=((3c−a)/(2a−b+3c)) ∴ each ratio =((3b+3a−b−c+3c−a)/(3a+3b+2b+c−a+2a−b+3c)) =((2(a+b+c))/(4(a+b+c)))=(1/2) [(a+b+c)≠0] ∴(b/(a+b))=(1/2)⇒2b=a+b∴a=b again ((3c−a)/(2a−b+3c))=(1/2) ⇒6c−2a=2a−a+3c [∵ a=b] ⇒3c=3a ∴c=a ∴a=b=c](https://www.tinkutara.com/question/Q202550.png)
$$\:\frac{{b}}{{a}+{b}}=\frac{\mathrm{3}{a}−{b}−{c}}{\mathrm{2}{b}+{c}−{a}}=\frac{\mathrm{3}{c}−{a}}{\mathrm{2}{a}−{b}+\mathrm{3}{c}} \\ $$$$\frac{\mathrm{3}{b}}{\mathrm{3}{a}+\mathrm{3}{b}}=\frac{\mathrm{3}{a}−{b}−{c}}{\mathrm{2}{b}+{c}−{a}}=\frac{\mathrm{3}{c}−{a}}{\mathrm{2}{a}−{b}+\mathrm{3}{c}} \\ $$$$\therefore\:{each}\:{ratio} \\ $$$$=\frac{\mathrm{3}{b}+\mathrm{3}{a}−{b}−{c}+\mathrm{3}{c}−{a}}{\mathrm{3}{a}+\mathrm{3}{b}+\mathrm{2}{b}+{c}−{a}+\mathrm{2}{a}−{b}+\mathrm{3}{c}} \\ $$$$=\frac{\mathrm{2}\left({a}+{b}+{c}\right)}{\mathrm{4}\left({a}+{b}+{c}\right)}=\frac{\mathrm{1}}{\mathrm{2}}\:\:\:\left[\left({a}+{b}+{c}\right)\neq\mathrm{0}\right] \\ $$$$\therefore\frac{{b}}{{a}+{b}}=\frac{\mathrm{1}}{\mathrm{2}}\Rightarrow\mathrm{2}{b}={a}+{b}\therefore{a}={b} \\ $$$${again}\:\frac{\mathrm{3}{c}−{a}}{\mathrm{2}{a}−{b}+\mathrm{3}{c}}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{6}{c}−\mathrm{2}{a}=\mathrm{2}{a}−{a}+\mathrm{3}{c}\:\:\left[\because\:{a}={b}\right] \\ $$$$\Rightarrow\mathrm{3}{c}=\mathrm{3}{a}\:\therefore{c}={a} \\ $$$$\:\:\:\therefore{a}={b}={c}\: \\ $$