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Question Number 211509 by RojaTaniya last updated on 11/Sep/24

Answered by A5T last updated on 11/Sep/24

$$\frac{6a+b}{6c+d}=k=\frac{5a+b}{5c+d}$$$$\Rightarrow 6a+b=6ck+dk...\left(i\right);5a+b=5ck+dk...\left(ii\right)$$$$\left(i\right)-\left(ii\right)\Rightarrow a=ck;6\left(ii\right)-5\left(i\right)\Rightarrow b=dk$$$$7a+b=8(7c+d)\Rightarrow 7ck+dk=8(7c+d)$$$$\Rightarrow k=\frac{8(7c+d)}{7c+d}=8$$$$\frac{9a+b}{9c+d}=\frac{9ck+dk}{9c+d}=\frac{k(9c+d)}{9c+d}=k=8$$

Commented by RojaTaniya last updated on 11/Sep/24

$$Sir,thanks.$$

Answered by mr W last updated on 11/Sep/24

$$\frac{6a+b}{6c+d}=\frac{5a+b}{5c+d}$$$$=\frac{(6a+b)-(5a+b)}{(6c+d)-(5c+d)}$$$$=\frac{a}{c}$$$$=\frac{6a+b+a}{6c+d+c}$$$$=\frac{7a+b}{7c+d}$$$$=\frac{na+b}{nc+d}$$$$\Rightarrow \frac{na+b}{nc+d}=constant=8$$$$n\in \mathbb{W}$$

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