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Question-211509




Question Number 211509 by RojaTaniya last updated on 11/Sep/24
Answered by A5T last updated on 11/Sep/24
((6a+b)/(6c+d))=k=((5a+b)/(5c+d))  ⇒6a+b=6ck+dk...(i);5a+b=5ck+dk...(ii)  (i)−(ii)⇒a=ck;6(ii)−5(i)⇒b=dk  7a+b=8(7c+d)⇒7ck+dk=8(7c+d)  ⇒k=((8(7c+d))/(7c+d))=8  ((9a+b)/(9c+d))=((9ck+dk)/(9c+d))=((k(9c+d))/(9c+d))=k=8
$$\frac{\mathrm{6}{a}+{b}}{\mathrm{6}{c}+{d}}={k}=\frac{\mathrm{5}{a}+{b}}{\mathrm{5}{c}+{d}} \\ $$$$\Rightarrow\mathrm{6}{a}+{b}=\mathrm{6}{ck}+{dk}…\left({i}\right);\mathrm{5}{a}+{b}=\mathrm{5}{ck}+{dk}…\left({ii}\right) \\ $$$$\left({i}\right)−\left({ii}\right)\Rightarrow{a}={ck};\mathrm{6}\left({ii}\right)−\mathrm{5}\left({i}\right)\Rightarrow{b}={dk} \\ $$$$\mathrm{7}{a}+{b}=\mathrm{8}\left(\mathrm{7}{c}+{d}\right)\Rightarrow\mathrm{7}{ck}+{dk}=\mathrm{8}\left(\mathrm{7}{c}+{d}\right) \\ $$$$\Rightarrow{k}=\frac{\mathrm{8}\left(\mathrm{7}{c}+{d}\right)}{\mathrm{7}{c}+{d}}=\mathrm{8} \\ $$$$\frac{\mathrm{9}{a}+{b}}{\mathrm{9}{c}+{d}}=\frac{\mathrm{9}{ck}+{dk}}{\mathrm{9}{c}+{d}}=\frac{{k}\left(\mathrm{9}{c}+{d}\right)}{\mathrm{9}{c}+{d}}={k}=\mathrm{8} \\ $$
Commented by RojaTaniya last updated on 11/Sep/24
  Sir, thanks.
$$\:\:{Sir},\:{thanks}. \\ $$
Answered by mr W last updated on 11/Sep/24
((6a+b)/(6c+d))=((5a+b)/(5c+d))  =(((6a+b)−(5a+b))/((6c+d)−(5c+d)))  =(a/c)  =((6a+b+a)/(6c+d+c))  =((7a+b)/(7c+d))  =((na+b)/(nc+d))  ⇒((na+b)/(nc+d))=constant=8  n∈W
$$\frac{\mathrm{6}{a}+{b}}{\mathrm{6}{c}+{d}}=\frac{\mathrm{5}{a}+{b}}{\mathrm{5}{c}+{d}} \\ $$$$=\frac{\left(\mathrm{6}{a}+{b}\right)−\left(\mathrm{5}{a}+{b}\right)}{\left(\mathrm{6}{c}+{d}\right)−\left(\mathrm{5}{c}+{d}\right)} \\ $$$$=\frac{{a}}{{c}} \\ $$$$=\frac{\mathrm{6}{a}+{b}+{a}}{\mathrm{6}{c}+{d}+{c}} \\ $$$$=\frac{\mathrm{7}{a}+{b}}{\mathrm{7}{c}+{d}} \\ $$$$=\frac{{na}+{b}}{{nc}+{d}} \\ $$$$\Rightarrow\frac{{na}+{b}}{{nc}+{d}}={constant}=\mathrm{8} \\ $$$${n}\in\mathbb{W} \\ $$

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