Ok, so I've got the question:

Use laplace transforms to find x(t) when x and y satisfy the differential equations:

dx/dt -dy/dt -2x = 0

and

dx/dt +dy/dt -2y = t

With x = 1, y = 2 at t = 0

So far i've taken laplace transforms of each side and substituted one into the other to get a single equation for Laplace(x)

and got to the point where

Laplace(x) = {3*(1/s)*[(s-1)^2/(s-1)^2 + 1]} + [3* 1/(s-1)^2 + 1] -{15/6 *(1/s)*(1/(s-1)^2 + 1)}

So now i presume that i just inverse the laplace transforms that i have gotten, but how do you invert the parts where there are two transforms multiplied together such as the 1st and 3rd terms in the above?

is it just a case of : Laplace^-1 {3*(1/s)*[(s-1)^2/(s-1)^2 + 1]} is

simply the product of the individual transforms once they have been inverted ie

3 * 1 * exp(t) * cos(t)

OR is there some other way that i am not aware of?