Originally Posted by

**Neverquit** I found the solutions 0.25, -1 after re-arranging pq = 1 to q = 1/p to get 1/p - 4p = 3 to get quadratic 4p^2+3p-1=0.

There is also apparently another solution of 4, -1 which I can’t find.

How do I find it?

I think you are misinterpreting your own answer. When you solve the system of equations, you will find that

$\displaystyle p=1/4 ~~\mbox{ or }~~ p=-1$

These are two separate solutions, not a single solution. You need to find the value of $\displaystyle q$ that pairs with each of these solutions for $\displaystyle p$. So, you need to plug each value of $\displaystyle p$ back into the system of equations and find the corresponding values of $\displaystyle q$. Since $\displaystyle pq=1$, it's a pretty straightforward calculation:

$\displaystyle p=1/4 \implies q=4$

$\displaystyle p=-1 \implies q=-1$

Therefore, the solutions are:

First solution: $\displaystyle p=1/4 \mbox{ and } q=4$

Second solution: $\displaystyle p=-1 \mbox{ and } q=-1$

You might also write this as:

$\displaystyle (p,q) = (1/4,4) \mbox{ or } (p,q) = (-1,-1)$

But you definitely would not say that the solutions are $\displaystyle (1/4,-1)$ and $\displaystyle (4,-1)$.