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Question Number 37037 by jayanta11 last updated on 08/Jun/18

If ((a+b−c)/(a+b)) = ((b+c−a)/(b+c)) =((c+a−b)/(c+a)) and a+b+c ≠ 0 then which of the following is true 1) a=−b=c 2)−a=−b=c 3) a=b=c 4) a=b≠c

$${If}\:\frac{{a}+{b}−{c}}{{a}+{b}}\:=\:\frac{{b}+{c}−{a}}{{b}+{c}}\:=\frac{{c}+{a}−{b}}{{c}+{a}}\:{and}\: \\ $$$$\mathrm{a}+\mathrm{b}+\mathrm{c}\:\neq\:\mathrm{0}\:{then}\:{which}\:{of}\:{the}\:{following} \\ $$$$\left.{i}\left.{s}\:{true}\:\mathrm{1}\right)\:{a}=−{b}={c}\:\mathrm{2}\right)−{a}=−{b}={c}\: \\ $$$$\left.\mathrm{3}\left.\right)\:\mathrm{a}=\mathrm{b}=\mathrm{c}\:\mathrm{4}\right)\:\mathrm{a}=\mathrm{b}\neq{c} \\ $$

Answered by math1967 last updated on 08/Jun/18

3)a=b=c ((a+b)/(a+b))−(c/(a+b))=((b+c)/(b+c))−(a/(b+c))=((c+a)/(c+a))−(b/(c+a)) (c/(a+b))=(a/(b+c))=(b/(c+a)) ((a+b)/c)+1=((b+c)/a)+1=((c+a)/b)+1 ((a+b+c)/c)=((a+b+c)/a)=((a+b+c)/b) (1/c)=(1/a)=(1/b) [(a+b+c)≠0] ∴c=a=b⇒a=b=c

$$\left.\mathrm{3}\right){a}={b}={c} \\ $$$$\frac{{a}+{b}}{{a}+{b}}−\frac{{c}}{{a}+{b}}=\frac{{b}+{c}}{{b}+{c}}−\frac{{a}}{{b}+{c}}=\frac{{c}+{a}}{{c}+{a}}−\frac{{b}}{{c}+{a}} \\ $$$$\frac{{c}}{{a}+{b}}=\frac{{a}}{{b}+{c}}=\frac{{b}}{{c}+{a}} \\ $$$$\frac{{a}+{b}}{{c}}+\mathrm{1}=\frac{{b}+{c}}{{a}}+\mathrm{1}=\frac{{c}+{a}}{{b}}+\mathrm{1} \\ $$$$\frac{{a}+{b}+{c}}{{c}}=\frac{{a}+{b}+{c}}{{a}}=\frac{{a}+{b}+{c}}{{b}} \\ $$$$\frac{\mathrm{1}}{{c}}=\frac{\mathrm{1}}{{a}}=\frac{\mathrm{1}}{{b}}\:\:\left[\left({a}+{b}+{c}\right)\neq\mathrm{0}\right] \\ $$$$\therefore{c}={a}={b}\Rightarrow{a}={b}={c} \\ $$

Commented by Rasheed.Sindhi last updated on 08/Jun/18

Nice!

$$\mathcal{N}{ice}! \\ $$

Answered by Rasheed.Sindhi last updated on 08/Jun/18

((a+b−c)/(a+b)) = ((b+c−a)/(b+c)) =((c+a−b)/(c+a)) =(((a+b−c)+(b+c−a)+(c+a−b))/((a+b)+(b+c)+(c+a))) =((a+b+c)/(2(a+b+c)))=(1/2) 2(a+b−c)=a+b 2(b+c−a)=b+c 2(c+a−b)=c+a a+b−2c=0⇒b=2c−a^∗ b+c−2a=0⇒2c−a+c−2a=0⇒3c=3a^∗ c+a−2b=0⇒c+a−4c+2a=0⇒3a=3c^∗ ^∗ a=b=c

$$\frac{{a}+{b}−{c}}{{a}+{b}}\:=\:\frac{{b}+{c}−{a}}{{b}+{c}}\:=\frac{{c}+{a}−{b}}{{c}+{a}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\left({a}+{b}−{c}\right)+\left({b}+{c}−{a}\right)+\left({c}+{a}−{b}\right)}{\left({a}+{b}\right)+\left({b}+\mathrm{c}\right)+\left({c}+{a}\right)}\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{{a}+{b}+{c}}{\mathrm{2}\left({a}+{b}+\mathrm{c}\right)}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\mathrm{2}\left({a}+{b}−{c}\right)={a}+{b} \\ $$$$\mathrm{2}\left({b}+{c}−{a}\right)={b}+{c} \\ $$$$\mathrm{2}\left({c}+{a}−{b}\right)={c}+{a} \\ $$$$ \\ $$$${a}+{b}−\mathrm{2}{c}=\mathrm{0}\Rightarrow{b}=\mathrm{2}{c}−{a}^{\ast} \\ $$$${b}+{c}−\mathrm{2}{a}=\mathrm{0}\Rightarrow\mathrm{2}{c}−{a}+{c}−\mathrm{2}{a}=\mathrm{0}\Rightarrow\mathrm{3}{c}=\mathrm{3}{a}^{\ast} \\ $$$${c}+{a}−\mathrm{2}{b}=\mathrm{0}\Rightarrow{c}+{a}−\mathrm{4}{c}+\mathrm{2}{a}=\mathrm{0}\Rightarrow\mathrm{3}{a}=\mathrm{3}{c}^{\ast} \\ $$$$\:^{\ast} {a}={b}={c} \\ $$

Answered by ajfour last updated on 08/Jun/18

((a+b−c)/(a+b))=((b+c−a)/(b+c))=((c+a−b)/(c+a)) ⇒ (c/(a+b))=(a/(b+c))=(b/(c+a)) ⇒ ((a+b+c)/c)=((a+b+c)/a)=((a+b+c)/b) ⇒ a=b=c .

$$\frac{{a}+{b}−{c}}{{a}+{b}}=\frac{{b}+{c}−{a}}{{b}+{c}}=\frac{{c}+{a}−{b}}{{c}+{a}} \\ $$$$\Rightarrow\:\:\frac{{c}}{{a}+{b}}=\frac{{a}}{{b}+{c}}=\frac{{b}}{{c}+{a}} \\ $$$$\Rightarrow\:\:\frac{{a}+{b}+{c}}{{c}}=\frac{{a}+{b}+{c}}{{a}}=\frac{{a}+{b}+{c}}{{b}} \\ $$$$\Rightarrow\:\:\:\:{a}={b}={c}\:. \\ $$

Answered by tanmay.chaudhury50@gmail.com last updated on 08/Jun/18

((a+b−c)/(a+b))=((b+c−a)/(b+c))=((c+a−b)/(c+a)) 1−(c/(a+b))=1−(a/(b+c))=1−(b/(c+a)) (c/(a+b))=(a/(b+c))=(b/(c+a)) bc+c^2 =a^2 +ab bc−ab+c^2 −a^2 =0 b(c−a)+(c+a)(c−a)=0 (c−a)(b+c+a)=0 since a+b+c not equales to zero so c=a similarly a=b so a=b=c value of each ratio i ((a+b−c)/(a+b)) =((a+a−a)/(a+a)) =(a/(2a))=(1/2) value of each ratio when a+b+c=0 ((a+b−c)/(a+b)) ((−c−c)/(−c)) ((−2c)/(−c)) =2

$$\frac{{a}+{b}−{c}}{{a}+{b}}=\frac{{b}+{c}−{a}}{{b}+{c}}=\frac{{c}+{a}−{b}}{{c}+{a}} \\ $$$$\mathrm{1}−\frac{{c}}{{a}+{b}}=\mathrm{1}−\frac{{a}}{{b}+{c}}=\mathrm{1}−\frac{{b}}{{c}+{a}} \\ $$$$\frac{{c}}{{a}+{b}}=\frac{{a}}{{b}+{c}}=\frac{{b}}{{c}+{a}} \\ $$$${bc}+{c}^{\mathrm{2}} ={a}^{\mathrm{2}} +{ab} \\ $$$${bc}−{ab}+{c}^{\mathrm{2}} −{a}^{\mathrm{2}} =\mathrm{0} \\ $$$${b}\left({c}−{a}\right)+\left({c}+{a}\right)\left({c}−{a}\right)=\mathrm{0} \\ $$$$\left({c}−{a}\right)\left({b}+{c}+{a}\right)=\mathrm{0} \\ $$$${since}\:{a}+{b}+{c}\:{not}\:{equales}\:{to}\:{zero} \\ $$$${so}\:{c}={a} \\ $$$${similarly}\:{a}={b} \\ $$$${so}\:{a}={b}={c} \\ $$$${value}\:{of}\:{each}\:{ratio}\:{i} \\ $$$$\frac{{a}+{b}−{c}}{{a}+{b}} \\ $$$$=\frac{{a}+{a}−{a}}{{a}+{a}} \\ $$$$=\frac{{a}}{\mathrm{2}{a}}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${value}\:{of}\:{each}\:{ratio}\:\:{when}\:{a}+{b}+{c}=\mathrm{0} \\ $$$$\frac{{a}+{b}−{c}}{{a}+{b}} \\ $$$$\frac{−{c}−{c}}{−{c}} \\ $$$$\frac{−\mathrm{2}{c}}{−{c}} \\ $$$$=\mathrm{2} \\ $$$$ \\ $$

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