use the Laplace transform to solve the
given system of differential equations.

2x''=-6x+2y
y''=2x-2y+40sin3t

where x(0)=x'(0)=y(0)=y'(0)=0
i applied Laplace transform and got

1- X(s){s^2-s+2}-2Y(s)=0
2- Y(s)(s^2-s+2)-2x(s)= 40{3/(s^2+3^2)}

now how would i solve stuck i know its basic algebra but need help

Question

use the Laplace transform to solve the
given system of differential equations.

2x''=-6x+2y
y''=2x-2y+40sin3t

where x(0)=x'(0)=y(0)=y'(0)=0
i applied Laplace transform and got 

1-> X(s){s^2-s+2}-2Y(s)=0
2-> Y(s)(s^2-s+2)-2x(s)= 40{3/(s^2+3^2)}

now how would i solve stuck i know its basic algebra but need help

primeralph · Answer

Put it in the form AX(s) - 2Y(s) = 0 and BY(s) - 2X(s) = k   and solve the system by elimination or whatever.
Then substitute A and B back into the solution.

abb0t · Answer

factor out the $X(s)$ from equation one, and same with equation 2, except in terms $Y(s)$
then, finally, find the inverse laplace for $X(s)$ and $Y(s)$