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Question Number 42528 by rahul 19 last updated on 27/Aug/18

$$\mathrm{Solve}:$$$$\frac{\mathrm{p}+2}{2\mathrm{p}+1}=\frac{\mathrm{q}+2\mathrm{p}}{2\mathrm{q}+\mathrm{p}}=\frac{1+2\mathrm{q}}{2+\mathrm{q}}=\lambda .$$$$\mathrm{Find}(\mathrm{p},\mathrm{q})?$$

Answered by math1967 last updated on 27/Aug/18

$$\lambda =\frac{p+2+q+2p+1+2q}{2p+1+2q+p+2+q}=\frac{3(p+q+1)}{3(p+q+1)}=1$$$$\therefore \frac{p+2}{2p+1}=1\Rightarrow p=1$$$$\frac{1+2q}{2+q}=1\Rightarrow q=1$$

Commented by rahul 19 last updated on 31/Aug/18

thanks sir!

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