Question

solve the following system.\

$$\begin{cases}x - 4y + z = -16\\\\y - 3z = 9\\\\3y - 2z = 13\\end{cases}$$

\
\
x = \quad\quad y = \quad\quad z =

Explanation:

Step1: Solve for \( y \) and \( z \) from the last two equations

We have the system:
\[

$$\begin{cases} y - 3z = 9 \quad (1)\\ 3y - 2z = 13 \quad (2) \end{cases}$$

\]
From equation (1), express \( y \) in terms of \( z \): \( y = 3z + 9 \).
Substitute \( y = 3z + 9 \) into equation (2):
\[
3(3z + 9) - 2z = 13
\]

Step2: Simplify and solve for \( z \)

Expand the left - hand side:
\[
9z+27 - 2z=13
\]
Combine like terms:
\[
7z+27 = 13
\]
Subtract 27 from both sides:
\[
7z=13 - 27=- 14
\]
Divide both sides by 7:
\[
z=\frac{-14}{7}=-2
\]

Step3: Solve for \( y \)

Substitute \( z = - 2 \) into \( y = 3z + 9 \):
\[
y=3\times(-2)+9=-6 + 9 = 3
\]

Step4: Solve for \( x \)

Substitute \( y = 3 \) and \( z=-2 \) into the first equation \( x-4y + z=-16 \):
\[
x-4\times3+(-2)=-16
\]
Simplify the left - hand side:
\[
x-12 - 2=-16
\]
\[
x-14=-16
\]
Add 14 to both sides:
\[
x=-16 + 14=-2
\]

Answer:

\( x=-2 \), \( y = 3 \), \( z=-2 \)