Question

\\\

$$\begin{cases} x + y + z = 6 \\\\ 4x + 2y - z = 5 \\\\ x + 3y + 2z = 13 \\end{cases}$$

\\

Explanation:

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<reasoning>

列出方程组

设给定的三元一次方程组为：
\[

$$\begin{cases} x + y + z = 6 & \text{--- (1)} \\ 4x + 2y - z = 5 & \text{--- (2)} \\ x + 3y + 2z = 13 & \text{--- (3)} \end{cases}$$

\]

消去未知数 z

将方程 (1) 与方程 (2) 相加：
\[
(x + y + z) + (4x + 2y - z) = 6 + 5
\]
\[
5x + 3y = 11 \quad \text{--- (4)}
\]

将方程 (1) 乘以 2，再与方程 (3) 相减：
\[
2(x + y + z) = 12 \quad \text{--- (5)}
\]
用方程 (3) 减去方程 (5)：
\[
(x + 3y + 2z) - (2x + 2y + 2z) = 13 - 12
\]
\[
-x + y = 1 \quad \text{--- (6)}
\]

求解二元一次方程组

由方程 (6) 可得：
\[
y = x + 1 \quad \text{--- (7)}
\]
将 (7) 代入 (4)：
\[
5x + 3(x + 1) = 11
\]
\[
8x + 3 = 11
\]
\[
8x = 8 \implies x = 1
\]
将 \(x = 1\) 代入 (7) 求得 \(y\)：
\[
y = 1 + 1 = 2
\]

求解未知数 z

将 \(x = 1\) 和 \(y = 2\) 代入方程 (1)：
\[
1 + 2 + z = 6
\]
\[
3 + z = 6 \implies z = 3
\]

检验方程组的解

将 \(x = 1, y = 2, z = 3\) 代入方程 (2) 和 (3) 进行验证：
对于 (2)：\(4(1) + 2(2) - 3 = 4 + 4 - 3 = 5\) （成立）
对于 (3)：\(1 + 3(2) + 2(3) = 1 + 6 + 6 = 13\) （成立）
</reasoning>

<answer>
\[

$$\begin{cases} x = 1 \\ y = 2 \\ z = 3 \end{cases}$$

\]
</answer>

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Answer:

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<reasoning>

列出方程组

设给定的三元一次方程组为：
\[

$$\begin{cases} x + y + z = 6 & \text{--- (1)} \\ 4x + 2y - z = 5 & \text{--- (2)} \\ x + 3y + 2z = 13 & \text{--- (3)} \end{cases}$$

\]

消去未知数 z

将方程 (1) 与方程 (2) 相加：
\[
(x + y + z) + (4x + 2y - z) = 6 + 5
\]
\[
5x + 3y = 11 \quad \text{--- (4)}
\]

将方程 (1) 乘以 2，再与方程 (3) 相减：
\[
2(x + y + z) = 12 \quad \text{--- (5)}
\]
用方程 (3) 减去方程 (5)：
\[
(x + 3y + 2z) - (2x + 2y + 2z) = 13 - 12
\]
\[
-x + y = 1 \quad \text{--- (6)}
\]

求解二元一次方程组

由方程 (6) 可得：
\[
y = x + 1 \quad \text{--- (7)}
\]
将 (7) 代入 (4)：
\[
5x + 3(x + 1) = 11
\]
\[
8x + 3 = 11
\]
\[
8x = 8 \implies x = 1
\]
将 \(x = 1\) 代入 (7) 求得 \(y\)：
\[
y = 1 + 1 = 2
\]

求解未知数 z

将 \(x = 1\) 和 \(y = 2\) 代入方程 (1)：
\[
1 + 2 + z = 6
\]
\[
3 + z = 6 \implies z = 3
\]

检验方程组的解

将 \(x = 1, y = 2, z = 3\) 代入方程 (2) 和 (3) 进行验证：
对于 (2)：\(4(1) + 2(2) - 3 = 4 + 4 - 3 = 5\) （成立）
对于 (3)：\(1 + 3(2) + 2(3) = 1 + 6 + 6 = 13\) （成立）
</reasoning>

<answer>
\[

$$\begin{cases} x = 1 \\ y = 2 \\ z = 3 \end{cases}$$

\]
</answer>

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