有限数学示例

$6 x - 5 Y = 30$ , $12 x - 10 y = 6$

解题步骤 1

将 $6 x$ 和 $- 5 Y$ 重新排序。

$- 5 Y + 6 x = 30$

$12 x - 10 y = 6$

解题步骤 2

Write the system as a matrix.

$[\begin{array}{c} - 5 & 6 & 0 & 30 \\ 0 & 12 & - 10 & 6 \end{array}]$

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 $1$ .

$[\begin{array}{c} - \frac{1}{5} \cdot - 5 & - \frac{1}{5} \cdot 6 & - \frac{1}{5} \cdot 0 & - \frac{1}{5} \cdot 30 \\ 0 & 12 & - 10 & 6 \end{array}]$

解题步骤 3.1.2

化简 $R_{1}$ 。

$[\begin{array}{c} 1 & - \frac{6}{5} & 0 & - 6 \\ 0 & 12 & - 10 & 6 \end{array}]$

解题步骤 3.2

Multiply each element of $R_{2}$ by $\frac{1}{12}$ to make the entry at $2, 2$ a $1$ .

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解题步骤 3.2.1

Multiply each element of $R_{2}$ by $\frac{1}{12}$ to make the entry at $2, 2$ a $1$ .

$[\begin{array}{c} 1 & - \frac{6}{5} & 0 & - 6 \\ \frac{0}{12} & \frac{12}{12} & \frac{- 10}{12} & \frac{6}{12} \end{array}]$

解题步骤 3.2.2

化简 $R_{2}$ 。

$[\begin{array}{c} 1 & - \frac{6}{5} & 0 & - 6 \\ 0 & 1 & - \frac{5}{6} & \frac{1}{2} \end{array}]$

解题步骤 3.3

Perform the row operation $R_{1} = R_{1} + \frac{6}{5} R_{2}$ to make the entry at $1, 2$ a $0$ .

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解题步骤 3.3.1

Perform the row operation $R_{1} = R_{1} + \frac{6}{5} R_{2}$ to make the entry at $1, 2$ a $0$ .

$[\begin{array}{c} 1 + \frac{6}{5} \cdot 0 & - \frac{6}{5} + \frac{6}{5} \cdot 1 & 0 + \frac{6}{5} (- \frac{5}{6}) & - 6 + \frac{6}{5} \cdot \frac{1}{2} \\ 0 & 1 & - \frac{5}{6} & \frac{1}{2} \end{array}]$

解题步骤 3.3.2

化简 $R_{1}$ 。

$[\begin{array}{c} 1 & 0 & - 1 & - \frac{27}{5} \\ 0 & 1 & - \frac{5}{6} & \frac{1}{2} \end{array}]$

解题步骤 4

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

$Y - y = - \frac{27}{5}$

$x - \frac{5}{6} y = \frac{1}{2}$

解题步骤 5

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

$(- \frac{27}{5} + y, \frac{1}{2} + \frac{5 y}{6}, y)$

有限数学 示例

有限数学示例