First order system of equations in DE Looking to solve the following system of equations
$$x'_1=4x_1-x_2$$
$$x'_2=-4x_1+4x_2$$
$$x_1(0)=3, x_2(0)=2$$
I need to solve this equation using Laplace and then I can use the coefficients and use Cramer's rule to solve for $x_1$ and then plug back into $x_2$ to solve for that variable  .
The answer is 
$x_1=2e^{2t}+e^{6t}$
$x_2=4e^{2t}-2e^{6t}$
 A: We are given:
$$x'_1=4x_1-x_2, ~x_1(0) = 3 \\ x'_2=-4x_1+4x_2, ~x_2(0)=2$$
Following this link (there are many other approaches), we take the Laplace of each equation and get:
$$s x_1(s) - x_1(0) = 4 x_1(s) - x_2(s) \\ s x_2(s) - x_2(0) = - 4x_1(s) + 4 x_2(s)$$
We substitute the initial conditions and now have:
$$s x_1(s) - 3 = 4 x_1(s) - x_2(s) \\ s x_2(s) - 2 = - 4 x_1(s) + 4 x_2(s)$$
Rearranging like terms, we get:
$$(s - 4)~x_1(s) + x_2(s) = 3 \\ 4~ x_1(s) + (s-4)~ x_2(s) = 2$$
Can you now proceed as you described with Cramer's rule to find $x_1(s)$ and $x_2(s)$ and then find the Inverse Laplace transforms for $x_1(t)$ and $x_2(t)$?
Using Cramer's Rule (and many other ways) and then the Inverse Laplace Transform, you should end up with:
$$x_1(s) = \dfrac{3 s-14}{s^2 - 8 s + 12} \implies x_1(t) = 2 e^{2t} + e^{6t}$$
$$x_2(s) = \dfrac{2 s- 20}{s^2 - 8 s + 12} \implies x_2(t) = 4e^{2t}-2e^{6t}$$
If you need an example, see this link, under the heading "Solution of Systems of Initial Value ODEs".
Update 1 
Our system can be written as:
$$\begin{bmatrix}
(s-4) & 1\\4 & (s-4) \end{bmatrix}\begin{bmatrix}
x_1(s)\\x_2(s) \end{bmatrix} = \begin{bmatrix}
3\\2 \end{bmatrix}$$
Using Cramer's Rule, for $x_1(s)$, we would have:
$$x_1(s) = \dfrac{\begin{vmatrix}
3&1\\
2 & (s-4) \\
\end{vmatrix}}{\begin{vmatrix}
(s-4) &1\\
4 &(s-4) 
\end{vmatrix}} = \dfrac{3(s-4)-2}{(s-4)^2-4} = \dfrac{3s-14}{s^2-8s+12}$$
It is worth noting that you can totally avoid using Cramer's Rule here and just find the inverse of the system's matrix directly to solve for $x_1(s)$ and $x_2(s)$ and then do the Inverse Laplace Transform. You should try this after you get this approach. There are many other approaches to solve these types of  problems (eigenvalues/eigenvectors, matrix exponential, fundamental matrix ...).
Update 2:
Using Cramer's Rule, for $x_2(s)$, we would have:
$$x_2(s) = \dfrac{\begin{vmatrix}
(s-4)&3\\
4 & 2 \\
\end{vmatrix}}{\begin{vmatrix}
(s-4) &1\\
4 &(s-4) 
\end{vmatrix}} = \dfrac{2(s-4)-12}{(s-4)^2-2} = \dfrac{2s-20}{s^2-8s+12}$$
