ANSWER:
[tex]\begin{gathered} x=4 \\ y=2 \\ z=0 \end{gathered}[/tex]STEP-BY-STEP EXPLANATION:
We have the following system of equations:
[tex]\begin{gathered} 4x+4y+z=24\text{ (1)} \\ 2x-4y+z=0\text{ (2)} \\ 5x-4y-5z=12\text{ (3)} \end{gathered}[/tex]We solve by elimination:
[tex]\begin{gathered} \text{ We add (1) and (2)} \\ 4x+4y+z+2x-4y+z=24+0 \\ 6x+2z=24\text{ }\rightarrow x=\frac{24-2z}{6}\text{ (4)} \\ \text{ We add (1) and (3)} \\ 4x+4y+z+5x-4y-5z=24+12 \\ 9x-4z=36\text{ (5)} \\ \text{ replacing (4) in (5)} \\ 9\cdot(\frac{24-2z}{6})-4z=36 \\ 36-3z-4z=36 \\ -7z=36-36 \\ z=\frac{0}{-7} \\ z=0 \end{gathered}[/tex]Now, replacing z in (4):
[tex]\begin{gathered} x=\frac{24-2\cdot0}{6} \\ x=\frac{24}{6} \\ x=4 \end{gathered}[/tex]Then, replacing z and x in (1):
[tex]\begin{gathered} 4\cdot4+4y+0=24 \\ 16+4y=24 \\ 4y=24-16 \\ y=\frac{8}{4} \\ y=2 \end{gathered}[/tex]