What is the best way to solve this system for $x$ and $y$ with $s \in \mathbb{R}$ using the minimum amount of effort?
$$(s^2 - 4)y + x = \dfrac{1}{s+1} + s + 2 \\ (s^2 - 1)x + y = \dfrac{1}{s-2} + s - 1$$
Is using the determinant the best way to solve this? I am interested more, in the steps rather than the result. Can you show me all the procedure? We need the most simplified solution because in the end we want to find the inversed Laplace of x,y. Thank you.
solving the second equation for $y$ we get: $$y=-{\frac {{s}^{3}x-2\,{s}^{2}x-{s}^{2}-sx+3\,s+2\,x-3}{s-2}}$$ with $$s\ne 2$$ plugging this in the first equation we get $$-{\frac { \left( {s}^{2}-4 \right) \left( {s}^{3}x-2\,{s}^{2}x-{s}^{2 }-sx+3\,s+2\,x-3 \right) }{s-2}}+x= \left( s+1 \right) ^{-1}+s+2 $$ solving this for $x$ we obtain $$x=\frac{1}{s+1}$$ for $$s\ne -1$$ if we set this in one equation we get for $y$ $$y=\frac{1}{s-2}$$ for $$s\ne 2$$