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- Nov 13th 2013, 05:30 AMvegasgunnerhelp
- Nov 13th 2013, 05:35 AMSlipEternalRe: help
In the second equation, solve for $\displaystyle \dfrac{1}{p}$.

First $\displaystyle q = \dfrac{9}{2p}$. Then, multiply both sides by $\displaystyle \dfrac{2}{9}$. So, $\displaystyle \dfrac{2q}{9} = \dfrac{1}{p}$. Then $\displaystyle \dfrac{2q}{9} + \dfrac{1}{q} = 1$. Now, solve for $\displaystyle q$. - Nov 13th 2013, 05:42 AMSlipEternalRe: help
Note: There are a number of ways to solve the system of equations. You first need an expression for $\displaystyle p$ in terms of $\displaystyle q$, so in the first equation, you could multiply both sides by $\displaystyle pq$. Then $\displaystyle q+p = pq$. Now, solve for $\displaystyle p$:

$\displaystyle p-pq = -q$

$\displaystyle p(1-q) = -q$

$\displaystyle p = \dfrac{-q}{1-q} = \dfrac{q}{q-1}$

Plug that into the second equation and solve for $\displaystyle q$. - Nov 13th 2013, 05:46 AMvegasgunnerRe: help
thank you I solved it