The strategy is to work from a system with three variables and three equations down to something with two variables and two equations. In a two variable-two equation system, you can use either substitution or elimination.
First, let's organize the equations.
(1) x + y = 8
(2) y + z = 7
(3) x + z = 5
Now, we'll solve one of (1), (2), and (3) for a variable. The choice is yours, but we'll use equation (3) and solve it for x.
x + z = 5
x = 5 - z.
Now we put this into equation (1). Why (1)? It's the only one of (1) and (2) with x.
x + y = 8
5 - z + y = 8
-z + y = 3
y - z - 3 to rearrange it.
Now look at this new equation - call it (4) and the original (2).
(4) y - z = 3
(2) y + z = 7
This is a system of two equations and two variables. Either substitution or elimination works here - let's use elimination because of the -z and +z.
y - z = 3
y + z = 7
We add them together and we have that 2y = 10. Divide on both sides and y = 5. One variable down, two to go.
Now we go back to original equation (2). Substitute y = 5 to find z.
y + z = 7
5 + z = 7
z = 2.
Two down, one to go. Since we know z = 2, let's put it into (3) and find x. (Equation (1) with y = 5 works fine as well.)
x + z = 5
x + 2 = 5
x = 3
Thus x = 3, y = 5 and z = 2.
