Maths

unsure on this question and cannot find a solution to help
using g=9.8ms^-2
A box of mass 9kg rests on a smooth horizontal table.
A string attached to the box passes over a smooth pulley at the edge of the table. The other end of the string is attached to a box of mass 3kg, which hangs vertically. The system is released from rest.
Calculate the acceleration of the system, the tension in the string and the time it takes the boxes to move 3m.

Draw a diagram, and label the forces that are felt by each box.

Make a separate F = ma equation for each particle, using only the forces felt by that particle.

You should now have two equations, each of which will include the two unknowns T and a. Solve them simultaneously.

This won't answer the last part about how ling it takes the boxes to move 3 m, but you should now have all of the information you need to do this using suvat equations.

Would the two equations be:
T-9g= 9a
T-3g=3a
-unsure on when you have to add the weight with tension or ignore it/ add it

You need to pay careful attention to the directions of the forces and the acceleration.

For the box on the table, the only force acting in the direction of the acceleration is the tension. It's weight acts straight down, and in any case will be cancelled by the normal reaction from the table, so you don't have to include it.

For the box hanging over the side of the table, the tension acts upwards and its weight acts downwards. It will obviously accelerate downwards. How does this change your equation?

box on table would be T=9a
box hanging would be T-3g=3a?