I have the following system of 5 equations for 5 unknowns, $p,c_+,c_-,u,v$:
\begin{align} pc_+ - (1-p) c_- &= \mu\sqrt{\delta t}\\ pc_+^2 + (1-p)c_-^2 &= \sigma^2 + \mu^2\delta t\\ \sqrt{\delta t} c_+ &= u - 1\\ \sqrt{\delta t} c_- &= 1 - v\\ uv &= 1 \end{align}
The parameters $\mu,\sigma,\delta t > 0$ are constants. Additionally, the variables $p,c_+,c_-,u,v > 0$. I am trying to solve this system of equations but am having a hard time. I know for sure that it involves creating a new variable to simplify it, but I cannot see what to create. I tried doing $\tilde{u} = (u-1)/\sqrt{\delta t}$ and $\tilde{v} = (1-v)/\sqrt{\delta t}$, which gives the system
\begin{align} p\tilde{u} - (1-p)\tilde{v} &= \mu\sqrt{\delta t}\\ p\tilde{u}^2 + (1-p)\tilde{v}^2 &= \sigma^2 + \mu^2\delta t\\ \sqrt{\delta t} \tilde{u}\tilde{v} &= \tilde{u}-\tilde{v} \end{align}
Then, squaring the first equation gives
$$ p^2\tilde{u}^2 + (1-p)^2\tilde{v}^2 - 2p(1-p)\tilde{u}\tilde{v} = \mu^2\delta t $$
Comparing to the second equation,
$$ p^2\tilde{u}^2 + (1-p)^2\tilde{v}^2 - 2p(1-p)\tilde{u}\tilde{v} = p\tilde{u}^2 + (1-p)\tilde{v}^2 - \sigma^2 $$
Notice that this simplifies greatly. Putting all the variable terms on the right hand side and plugging in Eq 3 gives
$$ \tilde{u}^2 + 2\tilde{u}\tilde{v} + \tilde{v}^2 = (\tilde{u}+\tilde{v})^2 = \frac{\sigma^2}{p(1-p)} \Rightarrow \tilde{u} + \tilde{v} = \frac{\sigma}{\sqrt{p(1-p)}} $$
However, this gives $\tilde{u} + \tilde{v}$ in terms of another variable, so I don't know how to proceed. Any thoughts would be appreciated.
Using the variable substitution $\tilde{u}$ and $\tilde{v}$ as in the original post, we can solve for $p$ from the first two equations. This gives
\begin{align} p &= \frac{\mu\sqrt{\delta t} + \tilde{v}}{\tilde{u}+\tilde{v}}\\ p &= \frac{\mu^2\delta t^2 + \sigma^2 - \tilde{v}^2}{\tilde{u}^2-\tilde{v}^2} \end{align}
Notice that when you equate the two equations, you get
$$ (\mu\sqrt{\delta t} + \tilde{v})(\tilde{u}-\tilde{v}) = \beta^2 - \tilde{v}^2, $$
where $\beta^2 = \sigma^2 + \mu^2 \delta t$. Expanding this out,
$$ \mu\sqrt{\delta t} \tilde{u} - \mu\sqrt{\delta t}\tilde{v} + \tilde{u}\tilde{v} = \beta^2 $$
Lastly, we can use the last equation for $\tilde{u}\tilde{v}$, so that in total we get
$$ \mu\sqrt{\delta t}\tilde{u} - \mu\sqrt{\delta t}\tilde{v} + \frac{1}{\sqrt{\delta t}}(\tilde{u}-\tilde{v}) = \beta^2 \Rightarrow \boxed{\tilde{u} - \tilde{v} = \frac{\sqrt{\delta t}}{1+\mu\delta t}\beta^2} $$
The boxed equation is important because it gives $\tilde{u} - \tilde{v}$ in terms of constants. Notice that in the above, we have $(\mu\sqrt{\delta t} + \tilde{v})(\tilde{u}-\tilde{v}) = \beta^2 - \tilde{v}^2$, so we can plug in the boxed expression into that equation. This leaves one equation for one unknown, $\tilde{v}$, which can be solved using the quadratic formula. The rest is history.