How to find rational solutions to a set of equations I have a set of equations to satisfy and I would like to find if they have rational solutions, and if they do, what they are. The equations are:
\begin{equation}
1 + \alpha = 3 a \alpha\\
1 + \beta = 3 b \beta\\
1 + \gamma = 3 c \gamma\\
1 + \delta = (a + b + c)\delta\\
\gamma = - \frac{\alpha \sigma^2}{2}\\
2\beta + 3 \delta - \frac{2 \gamma}{\sigma} = 0\\
\sigma > 0
\end{equation}
After some algebra I arrived at this equation:
\begin{equation}
- \alpha^3 \sigma^3 - \alpha^2\beta\sigma^3 - 11 \alpha^2\beta\sigma^2 + 2 \alpha^2\beta\sigma - 2 \alpha\beta^2\sigma^2 + 4\alpha\beta^2 = 0
\end{equation}
I am seeking a solution with all variables $a, b, c, \alpha, \beta, \gamma, \delta, \sigma$ rational, with a preference for $\sigma = 1$.
Update: we can add in another parameter $d$:
\begin{equation}
1 + d\alpha = 3 a \alpha\\
1 + d\beta = 3 b \beta\\
1 + d\gamma = 3 c \gamma\\
1 + d\delta = (a + b + c)\delta
\end{equation}
Again, we require d to be rational.
 A: After noting $\alpha \ne 0$, and writing $t = \alpha/\beta$, your last equation is equivalent to
$$ \sigma^3 t^2 + (\sigma^3 + 11 \sigma^2 - 2 \sigma) t + 2 \sigma^2-4 = 0 \tag{1}$$
which has rational solutions if and only if $\sigma \ne 0$ and the discriminant
$$ \sigma^2 (\sigma^4 + 14 \sigma^3 + 117 \sigma^2 - 28 \sigma + 4)$$ is a square.  Thus we are led to the equation
$$ \sigma^4 + 14 \sigma^3 + 117 \sigma^2 - 28 \sigma + 4 - s^2 = 0$$
This is an elliptic curve, which has Weierstrass form $-x^3 + y^2 + 4971 x - 134246$.  According to Sage, it has rank $1$, and a generator is $x=-29, y=504$.  This corresponds to $\sigma=0, s=2$, but we needed $\sigma \ne 0$.  OK, another rational point is
$x=-29, y=-504$, which corresponds to $\sigma = 7/8$, $s = 569/64$.
With $\sigma = 7/8$, the solutions of equation (1) are 
$t = 16/49$ and $t = -79/7$.  You can then solve the original equations: e.g. with $t = 16/49$ I find $b$ is arbitrary (but not $1/3$) with
$$ \eqalign{\alpha &= \frac{16}{49 (3b-1)}\cr
            \beta &= \frac{1}{3b-1}\cr
            \gamma &= - \frac{1}{8(3b-1)}\cr
            \delta &= - \frac{16}{21(3b-1)}\cr
            a &= \frac{49 b - 11}{16} \cr
            c &= 3 - 8 b\cr}$$
