Question

1. \\(\

$$\begin{cases} -x + 3y - z = -9 \\\\ x - 2y + 2z = 8 \\\\ 2x - y + z = 7 \\end{cases}$$

\\)

Explanation:

Write the system of equations

Using the System of Linear Equations knowledge point, we represent the given system:

$$
LATEXBLOCK0
$$

Eliminate x from Equations 2 and 3

Add Equation 1 to Equation 2:

$$ (-x + 3y - z) + (x - 2y + 2z) = -9 + 8 \implies y + z = -1 \quad \text{(Equation 4)} $$

Multiply Equation 1 by 2 and add to Equation 3:

$$ 2(-x + 3y - z) + (2x - y + z) = 2(-9) + 7 \implies 5y - z = -11 \quad \text{(Equation 5)} $$

Solve for y and z

Add Equation 4 and Equation 5 to eliminate \(z\):

$$ (y + z) + (5y - z) = -1 + (-11) \implies 6y = -12 \implies y = -2 $$

Substitute \(y = -2\) back into Equation 4:

$$ -2 + z = -1 \implies z = 1 $$

Solve for x

Substitute \(y = -2\) and \(z = 1\) into Equation 2:

$$ x - 2(-2) + 2(1) = 8 \implies x + 4 + 2 = 8 \implies x = 2 $$

Verify the solution

Check the values in Equation 3:

$$ 2(2) - (-2) + 1 = 4 + 2 + 1 = 7 $$

The solution is correct.

Answer:

\(x = 2\), \(y = -2\), \(z = 1\)