i need help on this please :(

We have more variables than equations....so.....we will have  infinite solutions

Subtracting the first equation from the second  we get that  3y + 5z  =  16  → 

 3y = [ 16 - 5z]  →     y =  [ 16 - 5z ] / 3 

And using the first equation

x + y + z =  8

x + [ 16 - 5z]/3 + z  = 8       multiply through by 3

3x + 16 -5z + 3z = 24

3x - 2z = 8

3x = 8 - 2z

x = [ 8-2z] /3

So....the solutions for  {x, y , z }  =   {  [8-2z]/3 , [16-5z]/3 , z }

For example...if z = 0,  then x = 8/3  and y = 16/3

So....check that these solutions work :

x + y + z  = 8  ???

8/3 + 16/3 + 0  = 8 ???

24/3 = 8  ???

True

And

x + 4y + 6z = 24  ???

(8/3) + 4 [16/3]  + 0  = 24??

8/3 + 64/3  = 24 ???

72/3  = 24   ???

True

So.....   one possible solution is   { x, y , z}  =  (8/3, 16/3 , 0 }

x=2n, y=12-5n, z=3n-4 for any n