A system of 1st-order ODEs I am trying to solve $\begin{aligned} \dfrac{dx}{dt} &amp;= -2 x - 6 y \\ \dfrac{dy}{dt} &amp;= x - 7 y \end{aligned}$.

Let $P = \begin{bmatrix}-2&amp;-6\\1&amp;-7\end{bmatrix}$. My approach is to take $\operatorname{det}(P - \lambda I) =0 \implies \lambda = - 5 \vee -4$.

So then I get $\left[\begin{array}{c}
2\cr
1
\end{array}\right]$ as an eigenvector corresponding to $\lambda = -5$ and $ \left[\begin{array}{c}
3\cr
1
\end{array}\right]$ as an eigenvector corresponding to $\lambda = -4$.

So it seems to me that a solution should be: 
$\begin{aligned} x_1 = A2e^{-5t}+3Be^{-4t} \\ x_2 = Ae^{-5t}+Be^{-4t}\end{aligned}$ however this is apparently incorrect. Any help anyone can give on where I went wrong is appreciated.
From second equation
$$x=y'+7y$$
derive it to get
$$x'=y''+7y'$$
substitute in equ.1
$$y''+7y'=-2(y'+7y)-6y$$
$$y''+9y'+20y=0$$
the general solution is
$$y=Ae^{-5t}+Be^{-4t}$$
now derive this equation and substitute in equ 2 to get
$$x=2Ae^{-5t}+3Be^{-4t}$$
I guess there is an $e$ missing in $x_2$ (that is, in the solution for $y$). If you take $x_2=Ae^{-5t}+Be^{-4t}$, your solution is correct. Just derive both soolutions and check that they satisfy the system.