有限数学示例

$5 x + 6 y = 3$ , $- x - \frac{6}{5} y = - \frac{3}{5}$

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

组合 $y$ 和 $\frac{6}{5}$ 。

$5 x + 6 y = 3$

$- x - \frac{y \cdot 6}{5} = - \frac{3}{5}$

解题步骤 1.1.2

将 $6$ 移到 $y$ 的左侧。

$5 x + 6 y = 3$

$- x - \frac{6 y}{5} = - \frac{3}{5}$

$5 x + 6 y = 3$

$- x - \frac{6 y}{5} = - \frac{3}{5}$

解题步骤 1.2

重新排序项。

$5 x + 6 y = 3$

$- x - (\frac{6}{5} y) = - \frac{3}{5}$

解题步骤 1.3

去掉圆括号。

$5 x + 6 y = 3$

$- x - \frac{6}{5} y = - \frac{3}{5}$

$5 x + 6 y = 3$

$- x - \frac{6}{5} y = - \frac{3}{5}$

解题步骤 2

Write the system as a matrix.

$[\begin{array}{c} 5 & 6 & 3 \\ - 1 & - \frac{6}{5} & - \frac{3}{5} \end{array}]$

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 $1$ .

$[\begin{array}{c} \frac{5}{5} & \frac{6}{5} & \frac{3}{5} \\ - 1 & - \frac{6}{5} & - \frac{3}{5} \end{array}]$

解题步骤 3.1.2

化简 $R_{1}$ 。

$[\begin{array}{c} 1 & \frac{6}{5} & \frac{3}{5} \\ - 1 & - \frac{6}{5} & - \frac{3}{5} \end{array}]$

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 $0$ .

$[\begin{array}{c} 1 & \frac{6}{5} & \frac{3}{5} \\ - 1 + 1 \cdot 1 & - \frac{6}{5} + \frac{6}{5} & - \frac{3}{5} + \frac{3}{5} \end{array}]$

解题步骤 3.2.2

化简 $R_{2}$ 。

$[\begin{array}{c} 1 & \frac{6}{5} & \frac{3}{5} \\ 0 & 0 & 0 \end{array}]$

解题步骤 4

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

$x + \frac{6}{5} y = \frac{3}{5}$

$0 = 0$

解题步骤 5

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

$(\frac{3}{5} - \frac{6 y}{5}, y)$

有限数学 示例

有限数学示例