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Solve-p-2-2p-1-q-2p-2q-p-1-2q-2-q-Find-p-q-




Question Number 42528 by rahul 19 last updated on 27/Aug/18
Solve :  ((p+2)/(2p+1)) = ((q+2p)/(2q+p)) = ((1+2q)/(2+q)) = λ.  Find (p,q) ?
$$\mathrm{Solve}\:: \\ $$$$\frac{\mathrm{p}+\mathrm{2}}{\mathrm{2p}+\mathrm{1}}\:=\:\frac{\mathrm{q}+\mathrm{2p}}{\mathrm{2q}+\mathrm{p}}\:=\:\frac{\mathrm{1}+\mathrm{2q}}{\mathrm{2}+\mathrm{q}}\:=\:\lambda. \\ $$$$\mathrm{Find}\:\left(\mathrm{p},\mathrm{q}\right)\:? \\ $$
Answered by math1967 last updated on 27/Aug/18
λ=((p+2+q+2p+1+2q)/(2p+1+2q+p+2+q))=((3(p+q+1))/(3(p+q+1)))=1  ∴((p+2)/(2p+1))=1⇒p=1  ((1+2q)/(2+q))=1⇒q=1
$$\lambda=\frac{{p}+\mathrm{2}+{q}+\mathrm{2}{p}+\mathrm{1}+\mathrm{2}{q}}{\mathrm{2}{p}+\mathrm{1}+\mathrm{2}{q}+{p}+\mathrm{2}+{q}}=\frac{\mathrm{3}\left({p}+{q}+\mathrm{1}\right)}{\mathrm{3}\left({p}+{q}+\mathrm{1}\right)}=\mathrm{1} \\ $$$$\therefore\frac{{p}+\mathrm{2}}{\mathrm{2}{p}+\mathrm{1}}=\mathrm{1}\Rightarrow{p}=\mathrm{1} \\ $$$$\frac{\mathrm{1}+\mathrm{2}{q}}{\mathrm{2}+{q}}=\mathrm{1}\Rightarrow{q}=\mathrm{1} \\ $$
Commented by rahul 19 last updated on 31/Aug/18
thanks sir!

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