There are many ways to solve a system of equations. I would solve it by having my calculator invert the coefficient matrix: type in [[ -1 3 1][ 2 5 0 ][ 3 1 -2]]^(-1) * [[1][3][-2]].
Presumably you are studying a particular method and need to practice with that.
If you're learning the elimination method, the fact that one equation already lacks a z is a good place to start.
Add twice the first equation to the third.
Double the resulting equation and subtract the second original equation.
Answers & Comments
Verified answer
-2x + 6y + 2z = 2
3x + y - 2z = - 2 _____ADD
x + 7y = 0
x + 7y = 0
2x + 5y = 3
-2x - 14y = 0
2x + 5y = 3____ADD
-9y = 3
y = - 1/3
2x - 5/3 = 3
2x = 14/3
x = 7/3
-7/3 - 1 + z = 1
z = 2 + 7/3
z = 13/3
x = 7/3 , y = -1/3 , z = 13/3
There are many ways to solve a system of equations. I would solve it by having my calculator invert the coefficient matrix: type in [[ -1 3 1][ 2 5 0 ][ 3 1 -2]]^(-1) * [[1][3][-2]].
Presumably you are studying a particular method and need to practice with that.
If you're learning the elimination method, the fact that one equation already lacks a z is a good place to start.
Add twice the first equation to the third.
Double the resulting equation and subtract the second original equation.