1. **State the problem:** We are given the system of differential equations: $$\frac{dx}{dt} = 3x + 4y, \quad \frac{dy}{dt} = 2x + y$$ and two solutions: $$x = 2e^{5t}, y = e^{5t}$$ and $$x = e^{-t}, y = -e^{-t}$$ We need to show these two solutions are linearly independent on any interval $a \leq t \leq b$ and write the general solution. 2. **Recall the concept of linear independence for solutions:** Two solutions $\mathbf{X}_1(t)$ and $\mathbf{X}_2(t)$ of a system are linearly independent if their Wronskian determinant is nonzero for all $t$ in the interval. 3. **Write the solutions as vectors:** $$\mathbf{X}_1 = \begin{pmatrix} 2e^{5t} \\ e^{5t} \end{pmatrix}, \quad \mathbf{X}_2 = \begin{pmatrix} e^{-t} \\ -e^{-t} \end{pmatrix}$$ 4. **Compute the Wronskian:** $$W(t) = \det \begin{pmatrix} 2e^{5t} & e^{-t} \\ e^{5t} & -e^{-t} \end{pmatrix} = (2e^{5t})(-e^{-t}) - (e^{5t})(e^{-t}) = -2e^{4t} - e^{4t} = -3e^{4t}$$ 5. **Analyze the Wronskian:** Since $e^{4t} > 0$ for all real $t$, we have $$W(t) = -3e^{4t} \neq 0$$ for all $t$. This means the two solutions are linearly independent on every interval $a \leq t \leq b$. 6. **Write the general solution:** The general solution to the system is a linear combination of the two independent solutions: $$\mathbf{X}(t) = c_1 \begin{pmatrix} 2e^{5t} \\ e^{5t} \end{pmatrix} + c_2 \begin{pmatrix} e^{-t} \\ -e^{-t} \end{pmatrix} = \begin{pmatrix} 2c_1 e^{5t} + c_2 e^{-t} \\ c_1 e^{5t} - c_2 e^{-t} \end{pmatrix}$$ where $c_1$ and $c_2$ are arbitrary constants. **Final answer:** The two given solutions are linearly independent because their Wronskian $W(t) = -3e^{4t} \neq 0$ for all $t$. The general solution is $$\boxed{\mathbf{X}(t) = c_1 \begin{pmatrix} 2e^{5t} \\ e^{5t} \end{pmatrix} + c_2 \begin{pmatrix} e^{-t} \\ -e^{-t} \end{pmatrix}}$$