Precalculus, Chapter 7, 7.3, Section 7.3, Problem 13

EQ1: 2x+y-3z=10
EQ2: y+z=12
EQ3: z=2
In this system of equations, the value of variable z is known. So to get the values of the variables substitute z=2 to one of the equations. It is better if it is plug-in to the second equation since it composed of two variables only.
y + z=12
y+2=12
Then, solve for y.
y=12-2
y=10
Now that the values of y and z are known, solve for x. Plug-in them to the first equation.
2x+y-3z = 10
2x+10-3(2)=10
2x+10-6=10
2x+4=10
2x=10-4
2x=6
x=6/2
x=3
Therefore, the solution is (3,10,2).