有限数学示例

$3 x \cdot 1 + 7 x \cdot 2 - x \cdot 3 = 11$ , $x \cdot 1 + 2 x \cdot 2 - x \cdot 3 = 3$ , $2 x \cdot 1 + 4 x \cdot 2 - 2 x \cdot 3 = 10$

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

合并 $3 x \cdot 1 + 7 x \cdot 2 - x \cdot 3$ 中相反的项。

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解题步骤 1.1.1

按照 $3 x \cdot 1$ 和 $- x \cdot 3$ 重新排列因数。

$1 \cdot 3 x + 7 x \cdot 2 - 1 \cdot 3 x = 11$

$x \cdot 1 + 2 x \cdot 2 - x \cdot 3 = 3$

$2 x \cdot 1 + 4 x \cdot 2 - 2 x \cdot 3 = 10$

解题步骤 1.1.2

从 $1 \cdot 3 x$ 中减去 $3 x$ 。

$7 x \cdot 2 + 0 = 11$

$x \cdot 1 + 2 x \cdot 2 - x \cdot 3 = 3$

$2 x \cdot 1 + 4 x \cdot 2 - 2 x \cdot 3 = 10$

解题步骤 1.1.3

将 $7 x \cdot 2$ 和 $0$ 相加。

$7 x \cdot 2 = 11$

$x \cdot 1 + 2 x \cdot 2 - x \cdot 3 = 3$

$2 x \cdot 1 + 4 x \cdot 2 - 2 x \cdot 3 = 10$

$7 x \cdot 2 = 11$

$x \cdot 1 + 2 x \cdot 2 - x \cdot 3 = 3$

$2 x \cdot 1 + 4 x \cdot 2 - 2 x \cdot 3 = 10$

解题步骤 1.2

将 $2$ 乘以 $7$ 。

$14 x = 11$

$x \cdot 1 + 2 x \cdot 2 - x \cdot 3 = 3$

$2 x \cdot 1 + 4 x \cdot 2 - 2 x \cdot 3 = 10$

解题步骤 1.3

化简每一项。

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解题步骤 1.3.1

将 $x$ 乘以 $1$ 。

$14 x = 11$

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

$2 x \cdot 1 + 4 x \cdot 2 - 2 x \cdot 3 = 10$

解题步骤 1.3.2

将 $2$ 乘以 $2$ 。

$14 x = 11$

$x + 4 x - x \cdot 3 = 3$

$2 x \cdot 1 + 4 x \cdot 2 - 2 x \cdot 3 = 10$

解题步骤 1.3.3

将 $3$ 乘以 $- 1$ 。

$14 x = 11$

$x + 4 x - 3 x = 3$

$2 x \cdot 1 + 4 x \cdot 2 - 2 x \cdot 3 = 10$

$14 x = 11$

$x + 4 x - 3 x = 3$

$2 x \cdot 1 + 4 x \cdot 2 - 2 x \cdot 3 = 10$

解题步骤 1.4

将 $x$ 和 $4 x$ 相加。

$14 x = 11$

$5 x - 3 x = 3$

$2 x \cdot 1 + 4 x \cdot 2 - 2 x \cdot 3 = 10$

解题步骤 1.5

从 $5 x$ 中减去 $3 x$ 。

$14 x = 11$

$2 x = 3$

$2 x \cdot 1 + 4 x \cdot 2 - 2 x \cdot 3 = 10$

解题步骤 1.6

化简每一项。

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解题步骤 1.6.1

将 $2$ 乘以 $1$ 。

$14 x = 11$

$2 x = 3$

$2 x + 4 x \cdot 2 - 2 x \cdot 3 = 10$

解题步骤 1.6.2

将 $2$ 乘以 $4$ 。

$14 x = 11$

$2 x = 3$

$2 x + 8 x - 2 x \cdot 3 = 10$

解题步骤 1.6.3

将 $3$ 乘以 $- 2$ 。

$14 x = 11$

$2 x = 3$

$2 x + 8 x - 6 x = 10$

$14 x = 11$

$2 x = 3$

$2 x + 8 x - 6 x = 10$

解题步骤 1.7

将 $2 x$ 和 $8 x$ 相加。

$14 x = 11$

$2 x = 3$

$10 x - 6 x = 10$

解题步骤 1.8

从 $10 x$ 中减去 $6 x$ 。

$14 x = 11$

$2 x = 3$

$4 x = 10$

$14 x = 11$

$2 x = 3$

$4 x = 10$

解题步骤 2

Write the system as a matrix.

$[\begin{array}{c} 14 & 11 \\ 2 & 3 \\ 4 & 10 \end{array}]$

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 $\frac{1}{14}$ 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 $\frac{1}{14}$ to make the entry at $1, 1$ a $1$ .

$[\begin{array}{c} \frac{14}{14} & \frac{11}{14} \\ 2 & 3 \\ 4 & 10 \end{array}]$

解题步骤 3.1.2

化简 $R_{1}$ 。

$[\begin{array}{c} 1 & \frac{11}{14} \\ 2 & 3 \\ 4 & 10 \end{array}]$

解题步骤 3.2

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

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解题步骤 3.2.1

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

$[\begin{array}{c} 1 & \frac{11}{14} \\ 2 - 2 \cdot 1 & 3 - 2 (\frac{11}{14}) \\ 4 & 10 \end{array}]$

解题步骤 3.2.2

化简 $R_{2}$ 。

$[\begin{array}{c} 1 & \frac{11}{14} \\ 0 & \frac{10}{7} \\ 4 & 10 \end{array}]$

解题步骤 3.3

Perform the row operation $R_{3} = R_{3} - 4 R_{1}$ to make the entry at $3, 1$ a $0$ .

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解题步骤 3.3.1

Perform the row operation $R_{3} = R_{3} - 4 R_{1}$ to make the entry at $3, 1$ a $0$ .

$[\begin{array}{c} 1 & \frac{11}{14} \\ 0 & \frac{10}{7} \\ 4 - 4 \cdot 1 & 10 - 4 (\frac{11}{14}) \end{array}]$

解题步骤 3.3.2

化简 $R_{3}$ 。

$[\begin{array}{c} 1 & \frac{11}{14} \\ 0 & \frac{10}{7} \\ 0 & \frac{48}{7} \end{array}]$

解题步骤 3.4

Multiply each element of $R_{2}$ by $\frac{7}{10}$ to make the entry at $2, 2$ a $1$ .

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解题步骤 3.4.1

Multiply each element of $R_{2}$ by $\frac{7}{10}$ to make the entry at $2, 2$ a $1$ .

$[\begin{array}{c} 1 & \frac{11}{14} \\ \frac{7}{10} \cdot 0 & \frac{7}{10} \cdot \frac{10}{7} \\ 0 & \frac{48}{7} \end{array}]$

解题步骤 3.4.2

化简 $R_{2}$ 。

$[\begin{array}{c} 1 & \frac{11}{14} \\ 0 & 1 \\ 0 & \frac{48}{7} \end{array}]$

解题步骤 3.5

Perform the row operation $R_{3} = R_{3} - \frac{48}{7} R_{2}$ to make the entry at $3, 2$ a $0$ .

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解题步骤 3.5.1

Perform the row operation $R_{3} = R_{3} - \frac{48}{7} R_{2}$ to make the entry at $3, 2$ a $0$ .

$[\begin{array}{c} 1 & \frac{11}{14} \\ 0 & 1 \\ 0 - \frac{48}{7} \cdot 0 & \frac{48}{7} - \frac{48}{7} \cdot 1 \end{array}]$

解题步骤 3.5.2

化简 $R_{3}$ 。

$[\begin{array}{c} 1 & \frac{11}{14} \\ 0 & 1 \\ 0 & 0 \end{array}]$

解题步骤 3.6

Perform the row operation $R_{1} = R_{1} - \frac{11}{14} R_{2}$ to make the entry at $1, 2$ a $0$ .

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解题步骤 3.6.1

Perform the row operation $R_{1} = R_{1} - \frac{11}{14} R_{2}$ to make the entry at $1, 2$ a $0$ .

$[\begin{array}{c} 1 - \frac{11}{14} \cdot 0 & \frac{11}{14} - \frac{11}{14} \cdot 1 \\ 0 & 1 \\ 0 & 0 \end{array}]$

解题步骤 3.6.2

化简 $R_{1}$ 。

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

解题步骤 4

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

$x = 0$

$0 = 1$

$0 = 0$

解题步骤 5

The solution is the set of ordered pairs that make the system true.

$(error = error)$

有限数学 示例

有限数学示例