有限数学示例

$x \cdot 3 + x \cdot 6 = 24$ , $x \cdot 2 + x \cdot 3 = 11$

解题步骤 1

Move variables to the left and constant terms to the right.

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解题步骤 1.1

化简每一项。

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解题步骤 1.1.1

将 $3$ 移到 $x$ 的左侧。

$3 \cdot x + x \cdot 6 = 24$

$x \cdot 2 + x \cdot 3 = 11$

解题步骤 1.1.2

将 $6$ 移到 $x$ 的左侧。

$3 x + 6 x = 24$

$x \cdot 2 + x \cdot 3 = 11$

$3 x + 6 x = 24$

$x \cdot 2 + x \cdot 3 = 11$

解题步骤 1.2

将 $3 x$ 和 $6 x$ 相加。

$9 x = 24$

$x \cdot 2 + x \cdot 3 = 11$

解题步骤 1.3

化简每一项。

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解题步骤 1.3.1

将 $2$ 移到 $x$ 的左侧。

$9 x = 24$

$2 \cdot x + x \cdot 3 = 11$

解题步骤 1.3.2

将 $3$ 移到 $x$ 的左侧。

$9 x = 24$

$2 x + 3 x = 11$

$9 x = 24$

$2 x + 3 x = 11$

解题步骤 1.4

将 $2 x$ 和 $3 x$ 相加。

$9 x = 24$

$5 x = 11$

$9 x = 24$

$5 x = 11$

解题步骤 2

Write the system as a matrix.

$[\begin{array}{c} 9 & 24 \\ 5 & 11 \end{array}]$

解题步骤 3

求行简化阶梯形矩阵。

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解题步骤 3.1

Multiply each element of $R_{1}$ by $\frac{1}{9}$ to make the entry at $1, 1$ a $1$ .

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解题步骤 3.1.1

Multiply each element of $R_{1}$ by $\frac{1}{9}$ to make the entry at $1, 1$ a $1$ .

$[\begin{array}{c} \frac{9}{9} & \frac{24}{9} \\ 5 & 11 \end{array}]$

解题步骤 3.1.2

化简 $R_{1}$ 。

$[\begin{array}{c} 1 & \frac{8}{3} \\ 5 & 11 \end{array}]$

解题步骤 3.2

Perform the row operation $R_{2} = R_{2} - 5 R_{1}$ to make the entry at $2, 1$ a $0$ .

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解题步骤 3.2.1

Perform the row operation $R_{2} = R_{2} - 5 R_{1}$ to make the entry at $2, 1$ a $0$ .

$[\begin{array}{c} 1 & \frac{8}{3} \\ 5 - 5 \cdot 1 & 11 - 5 (\frac{8}{3}) \end{array}]$

解题步骤 3.2.2

化简 $R_{2}$ 。

$[\begin{array}{c} 1 & \frac{8}{3} \\ 0 & - \frac{7}{3} \end{array}]$

解题步骤 3.3

Multiply each element of $R_{2}$ by $- \frac{3}{7}$ to make the entry at $2, 2$ a $1$ .

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解题步骤 3.3.1

Multiply each element of $R_{2}$ by $- \frac{3}{7}$ to make the entry at $2, 2$ a $1$ .

$[\begin{array}{c} 1 & \frac{8}{3} \\ - \frac{3}{7} \cdot 0 & - \frac{3}{7} (- \frac{7}{3}) \end{array}]$

解题步骤 3.3.2

化简 $R_{2}$ 。

$[\begin{array}{c} 1 & \frac{8}{3} \\ 0 & 1 \end{array}]$

解题步骤 3.4

Perform the row operation $R_{1} = R_{1} - \frac{8}{3} R_{2}$ to make the entry at $1, 2$ a $0$ .

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解题步骤 3.4.1

Perform the row operation $R_{1} = R_{1} - \frac{8}{3} R_{2}$ to make the entry at $1, 2$ a $0$ .

$[\begin{array}{c} 1 - \frac{8}{3} \cdot 0 & \frac{8}{3} - \frac{8}{3} \cdot 1 \\ 0 & 1 \end{array}]$

解题步骤 3.4.2

化简 $R_{1}$ 。

$[\begin{array}{c} 1 & 0 \\ 0 & 1 \end{array}]$

解题步骤 4

Use the result matrix to declare the final solution to the system of equations.

$x = 0$

$0 = 1$

解题步骤 5

The system is inconsistent so there is no solution.

$无解$

有限数学 示例

有限数学示例