有限数学示例

$x \cdot 1 - 3 x \cdot 2 = - 10$ , $3 x \cdot 1 - x \cdot 2 = 2$

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 $x$ 乘以 $1$ 。

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

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

解题步骤 1.1.2

将 $2$ 乘以 $- 3$ 。

$x - 6 x = - 10$

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

$x - 6 x = - 10$

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

解题步骤 1.2

从 $x$ 中减去 $6 x$ 。

$- 5 x = - 10$

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

解题步骤 1.3

化简每一项。

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解题步骤 1.3.1

将 $3$ 乘以 $1$ 。

$- 5 x = - 10$

$3 x - x \cdot 2 = 2$

解题步骤 1.3.2

将 $2$ 乘以 $- 1$ 。

$- 5 x = - 10$

$3 x - 2 x = 2$

$- 5 x = - 10$

$3 x - 2 x = 2$

解题步骤 1.4

从 $3 x$ 中减去 $2 x$ 。

$- 5 x = - 10$

$x = 2$

$- 5 x = - 10$

$x = 2$

解题步骤 2

Write the system as a matrix.

$[\begin{array}{c} - 5 & - 10 \\ 1 & 2 \end{array}]$

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 $- \frac{1}{5}$ 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 $- \frac{1}{5}$ to make the entry at $1, 1$ a $1$ .

$[\begin{array}{c} - \frac{1}{5} \cdot - 5 & - \frac{1}{5} \cdot - 10 \\ 1 & 2 \end{array}]$

解题步骤 3.1.2

化简 $R_{1}$ 。

$[\begin{array}{c} 1 & 2 \\ 1 & 2 \end{array}]$

解题步骤 3.2

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

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解题步骤 3.2.1

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

$[\begin{array}{c} 1 & 2 \\ 1 - 1 & 2 - 2 \end{array}]$

解题步骤 3.2.2

化简 $R_{2}$ 。

$[\begin{array}{c} 1 & 2 \\ 0 & 0 \end{array}]$

解题步骤 4

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

$x = 2$

$0 = 0$

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