Solving for tesnion and acceleration

[wesleyneary]

I am stuck on a piece of my homework assignment

here it is

The following equations describe the motion of a system of two objects.
FN - (6.50 kg)(9.80 m/s2) cos(13.0˚) = 0
Ff,k = 0.360(FN)
FT + (6.50 kg)(9.80 m/s2) sin(13.0˚) - Ff,k = (6.50 kg) a
-FT + (3.80 kg)(9.80 m/s2) = (3.80 kg) a

(a) Solve the equations for a and FT. (b) Draw a free-body diagram for both objects. (c)
Describe a situation to which these equations apply.

I have solved for FN and then solved for Ff,k using that information

FN=62.07N and Ff,k+22.35 N (assuming i did my algebra correctly)

form here i know if i solve for acceleration of Ftension i can solve for the other.

I tried using a system of equations to solve for "a" by solving for Ft in on equation and plugging it into to other. when i do that i end up with the variable I am solving for cancelling out.

Any suggestions?

 

[Simon Bridge]

It will help you see more clearly if you simplify your nottation - call the normal force N, the friction f, and the tension T. I don't know what your notation Ff,k may be.
It also helps to leave all the variables in place and do the algebra before plugging the numbers in.

You have made a mistake in the algebra - I suspect you misplaced a minus sign.
Go through it formally starting from the two equations you are using, write them down, write down each step in the process, take care of the minus sign when you do the substitution and again when you solve for a.