Add and subtract the first and second equations to obtain:
\begin{align}
(y+z)'' + K(y+z) &= 0 \label{added} \\
(y-z)'' + (K+2L)(y-z) &= 0 \label{subtracted}
\end{align}
Since $K, L > 0$, both equations are of the form $u'' + \omega^2 u = 0$, with general solution $u = A \cos (\omega t) + B \sin (\omega t)$. So the general solution to $(\ref{added})$ is
\begin{equation}
y+z = A \cos (\sqrt{K} t) + B \sin (\sqrt{K} t)
\end{equation}
and the general solution to $(\ref{subtracted})$ is
\begin{equation}
y - z = C \cos (\sqrt{K+2L} t) + D \sin(\sqrt{K+2L}t)
\end{equation}
Using the identities $y = \frac{1}{2}((y+z)+(y-z))$ and $z = \frac{1}{2}((y+z)-(y-z))$ we obtain the general solution to $(\ref{eq-1})$:
\begin{align}
y &= A' \cos(\omega_1 t) + B' \sin(\omega_1 t) + C' \cos(\omega_2 t) + D' \sin(\omega_2 t) \\
z &= A' \cos(\omega_1 t) + B' \sin(\omega_1 t) - C' \cos(\omega_2 t) - D' \sin(\omega_2 t)
\end{align}
where $A' = A/2$ and so on, and the frequencies are $\omega_1 = \sqrt{K}$, and $\omega_2 = \sqrt{K+2L}$.