Solve the linear system if 3x+4y-z=-6 5x+8y+2z=2. -x+y+z=0

We are given with three equations and three unknowns and we need to solve this problem. The solution is shown below:
Three equations are below:
3x + 4y - z = -6
5x + 8y + 2z = 2
-x + y + z = 0

use the first (multiply by +2) and use the second equation:
2 (3x+4y -z = -6)  => 6x + 8y -2z = -12
                               + ( 5x + 8y +2z = 2)
                                 
                                     11x + 16y = -10  -> this the fourth equation

use the first and third equation:
 3x + 4y -z = -6
+ (-x + y + z =0)

        2x + 5y = -6 -> this is the fifth equaiton

use fourth (multiply by 2) and use fifth (multiply by -11) equations such as:
2 (11x + 16y = -10)   => 22x + 32y = -20  -> this is the sixth equation
-11 (2x + 5y = -6)    => -22x -55y = 46   -> this is the seventh equation

add 6th and 7th equation such as:
   22x + 32y = -20
+(-22x - 55y = 66)

             - 23y = 46
                  y = -2

solving for x, we have:
2x + 5y = -6 
2x = -6 - 5y
2x = -6 - (5*(-2))
2x = -6 +10
2x = 4
x=2

solving for y value, we have:
-x + y + z =0
z = x -y 
z = 2- (-2)
z =4

The answers are the following:
x = 2
y = -2
z = 4

Set square i think so