I'll go out on a limb and guess the system is
[tex]\mathbf x'=\begin{bmatrix}\frac12&0\\1&-\frac12\end{bmatrix}\mathbf x[/tex]
with initial condition [tex]\mathbf x(0)=\begin{bmatrix}2&7\end{bmatrix}^\top[/tex]. The coefficient matrix has eigenvalues [tex]\lambda[/tex] such that
[tex]\begin{vmatrix}\frac12-\lambda&0\\1&-\frac12-\lambda\end{vmatrix}=\lambda^2-\dfrac14=0\implies\lambda=\pm\dfrac12[/tex]
The corresponding eigenvectors [tex]\eta[/tex] are such that
[tex]\lambda=\dfrac12\implies\begin{bmatrix}\frac12-\frac12&0\\1&-\frac12-\frac12\end{bmatrix}\eta=\begin{bmatrix}0&0\\1&-1\end{bmatrix}\eta=\begin{bmatrix}0\\0\end{bmatrix}[/tex]
[tex]\implies\eta=\begin{bmatrix}1\\1\end{bmatrix}[/tex]
[tex]\lambda=-\dfrac12\implies\begin{bmatrix}\frac12+\frac12&0\\1&-\frac12+\frac12\end{bmatrix}\eta=\begin{bmatrix}1&0\\1&0\end{bmatrix}\eta=\begin{bmatrix}0\\0\end{bmatrix}[/tex]
[tex]\implies\eta=\begin{bmatrix}0\\1\end{bmatrix}[/tex]
So the characteristic solution to the ODE system is
[tex]\mathbf x(t)=C_1\begin{bmatrix}1\\1\end{bmatrix}e^{t/2}+C_2\begin{bmatrix}0\\1\end{bmatrix}e^{-t/2}[/tex]
When [tex]t=0[/tex], we have
[tex]\begin{bmatrix}2\\7\end{bmatrix}=C_1\begin{bmatrix}1\\1\end{bmatrix}+C_2\begin{bmatrix}0\\1\end{bmatrix}=\begin{bmatrix}C_1\\C_1+C_2\end{bmatrix}[/tex]
from which it follows that [tex]C_1=2[/tex] and [tex]C_2=5[/tex], making the particular solution to the IVP
[tex]\mathbf x(t)=2\begin{bmatrix}1\\1\end{bmatrix}e^{t/2}+5\begin{bmatrix}0\\1\end{bmatrix}e^{-t/2}[/tex]
[tex]\mathbf x(t)=\begin{bmatrix}2e^{t/2}\\2e^{t/2}+5e^{-t/2}\end{bmatrix}[/tex]
