Question

solve the system of equations:
\\\

$$\begin{cases} 3x - 8y + z = 8 \\\\ -x + y - z = 0 \\\\ x - 3y = 3 \\end{cases}$$

\\

\\(x = \\)
\\(y = \\)
\\(z = \\)

Explanation:

Step1: Write down the system of equations

$$
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$$

Step2: Eliminate \(z\) by adding Equation 1 and Equation 2

$$ (3x - 8y + z) + (-x + y - z) = 8 + 0 \implies 2x - 7y = 8 \quad \text{(Equation 4)} $$

Step3: Express \(x\) in terms of \(y\) using Equation 3

$$ x = 3y + 3 $$

Step4: Substitute \(x\) into Equation 4 to solve for \(y\)

$$ 2(3y + 3) - 7y = 8 \implies 6y + 6 - 7y = 8 \implies -y = 2 \implies y = -2 $$

Step5: Substitute \(y\) back to find \(x\)

$$ x = 3(-2) + 3 = -6 + 3 = -3 $$

Step6: Substitute \(x\) and \(y\) into Equation 2 to find \(z\)

$$ -(-3) + (-2) - z = 0 \implies 3 - 2 - z = 0 \implies z = 1 - 0 = 1 $$

Wait, let's re-evaluate:

$$ -(-3) + (-2) - z = 0 \implies 3 - 2 - z = 0 \implies 1 - z = 0 \implies z = 1 $$

Let's double check with Equation 1:

$$ 3(-3) - 8(-2) + 1 = -9 + 16 + 1 = 8 $$

This is correct. Let's re-verify Equation 2:

$$ -(-3) + (-2) - 1 = 3 - 2 - 1 = 0 $$

This is correct. Thus, \(z = 1\).

Let me correct the final answer values:
\(x = -3\)
\(y = -2\)
\(z = 1\)

Answer:

\(x = -3\)
\(y = -2\)
\(z = -1\)