Show that A(t)=300−250e0.2−0.02t satisfies the differential equation ⅆAⅆt=6−0.02A with initial condition A(10)=50 .

So the first thing we do is to multiply the top equation by 3. the reason i want to do this is i notice there will be a 6y on both equations and can possibly add/subtract them. so we get 6y = -3x + 27. adding that and the second equation we get 6x = 27-15 which is 12. divide by 6 and we get 2. ∴ x = 2. substituting 2 into the first equation we get 2y = -2 + 9 which is 7. divide by 2 and we get 7/2. ∴ y = 7/2, x = 2.