Systems of equations with three variables Consider:
$a + b = 5$
$2a + b + c = 4$
$a - b - c = 5$
I like to use substituiton for solving systems of equations, so I firstly look at equation 1 and solve for $a$
$a + b = 5$
$a = 5 - b$
I substitute this into equation 2
$2(5-b) + b + c = 4$
$10 -2b + b + c = 4$
$-b = 4 - c - 10$
$b = -4 + c + 10$
What next? Am I on the right track? Can I use substitution method for any simultaneous equation?

You have the right concept, and we can see from three equations and three unknowns that these variables are solvable (taking into account equations are linearly independent)
equation 1: $\ a+b=5$
equation 2:$\ 2a+b+c= 4$
equation 3:$\ a-b-c =5$
lets solve for b first:
from equation 1 and 3, we say
$a=5-b$ and $\ c = a-b-5$ 
Now we plug these two 'new' equations into equation 2:
$\ 2(5-b)  + b + ((5-b) - b -5) = 4 $
simplifying for b gives that $ b=2$.
Now taking that b=2, we can sub that back into equation 1 and solve for a and once we have a we can use a and b to solve for c. Which I'm sure you'll be able to solve for it.
