Question

find the solution to the system of equations:
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\

$$\begin{cases} x - 2y + z = -1 \\\\ y + 2z = -6 \\\\ x + y + 3z = -11 \\end{cases}$$

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

Explanation:

Step1: Write down the system of equations

$$
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$$

Step2: Eliminate \(x\) using Equations 1 and 3

$$(x + y + 3z) - (x - 2y + z) = -11 - (-1)$$

$$3y + 2z = -10 \quad \text{(Equation 4)}$$

Step3: Solve for \(y\) using Equations 2 and 4

$$(3y + 2z) - (y + 2z) = -10 - (-6)$$

$$2y = -4 \implies y = -2$$

Step4: Solve for \(z\) using Equation 2

$$-2 + 2z = -6 \implies 2z = -4 \implies z = -2$$

Step5: Solve for \(x\) using Equation 1

$$x - 2(-2) + (-2) = -1 \implies x + 2 = -1 \implies x = -3$$

Step6: Verify the solution in Equation 3

$$-3 + (-2) + 3(-2) = -11 \implies -11 = -11$$

Answer:

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