Question

solve the following system.\

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

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

Explanation:

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

We have the system:

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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:

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