Solve the system by substitution
-x-y-z=-8
-4x+4y+5z=7
2x+2x=4

Let
-x - y - z  = -8        (1a)
-4x + 4y +5z = 7       (2a)
2x + 2x =  4         (3a)

since (3a) a is an equation solely in x make x the subject of the equation
   2 ( x + x) = 4
  divide both sides by 2
       (x + x)  = 2
             2x  = 2
               x  =  1    (3b)

   substitute x in (3b) for x in (1a)
           -(1) - y - z  = -8  
     make y the subject of the equation
               -y - z  =  -8 + 1
                     -y =  -7 + z
                      y = 7 - z   (1b)

  substitute y in (1b) for y in (2a) as well as the value of x ( x= 1)
         -4(1)+ 4( 7 - z ) +5z = 7
        - 4 + 28 - 4z + 5z  = 7
               -24 + z  = 7
                        z  = 7 + 24
                        z  = 31

By using (1a) solve for y
           -(1) - y - (31)= -8          
           -1  - y - 31 = -8        
                         y = -1 - 31 + 8
                         y = -24
 
 ∴ the solutions of the system are  x = 1 ; y = -24 ; z = 31 

128/3reason being 5 and 1/3 = 16/316/3 • 2/3 = 32/31 and 1/3 = 4/3 32/3 • 4/3 this will equal 128/3