Given the system of equations
[tex]\begin{gathered} x+4y-z=20-----1 \\ 3x+2y+z=8-----2 \\ 2x-3y+2z=-16-----3 \end{gathered}[/tex]We can solve for x, y and z below.
Explanation
Step 1: Find the value of z using the substitution method
[tex]\begin{gathered} \begin{bmatrix}x+4y-z=20\\ 3x+2y+z=8\\ 2x-3y+2z=-16\end{bmatrix} \\ Isolate\text{ for x in equation 1} \\ x=20-4y+z \\ \mathrm{Substitute\:}x=20-4y+z\text{ in equation 2 and 3} \\ \begin{bmatrix}3\left(20-4y+z\right)+2y+z=8\\ 2\left(20-4y+z\right)-3y+2z=-16\end{bmatrix} \\ sinplify \\ \begin{bmatrix}-10y+4z+60=8 \\ -11y+4z+40=-16\end{bmatrix} \\ Isolate\text{ for y in}-10y+4z+60=8 \\ -10y=8-4z-60 \\ y=\frac{8-4z-60}{-10} \\ y=\frac{-4z-52}{-10} \\ y=\frac{2\left(z+13\right)}{5} \\ \mathrm{Substitute\:}y=\frac{2\left(z+13\right)}{5}\text{ in }-11y+4z+40=-16 \\ \begin{bmatrix}-11\cdot \frac{2\left(z+13\right)}{5}+4z+40=-16\end{bmatrix} \\ simplify \\ \begin{bmatrix}\frac{-2z-286}{5}+40=-16\end{bmatrix} \\ multiply\text{ through by 5} \\ -2z-286+200=-80 \\ isolate\text{ for z} \\ -2z=-80-200+286 \\ -2z=6 \\ z=\frac{6}{-2} \\ z=-3 \end{gathered}[/tex]Step 2: Find y
[tex]\begin{gathered} \mathrm{Substitute\:}z=-3\text{ in}\mathrm{\:}y=\frac{2\left(z+13\right)}{5} \\ y=\frac{2(-3+13)}{5} \\ y=\frac{2(10)}{5} \\ y=4 \end{gathered}[/tex]Step 3: Find z
[tex]\begin{gathered} \mathrm{Substitute\:}z=-3,\:y=4\text{ in }x=20-4y+z \\ x=20-4\cdot \:4-3 \\ x=1 \end{gathered}[/tex]Answer: The solutions to the system of equations are
[tex]x=1,\:z=-3,\:y=4[/tex]