有限数学示例

$3 x - 4 y = 9$ , $- 2 x + 3 = - 6$

解题步骤 1

将所有不包含变量的项移到等式右边。

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解题步骤 1.1

从等式两边同时减去 $3$ 。

$3 x - 4 y = 9$

$- 2 x = - 6 - 3$

解题步骤 1.2

从 $- 6$ 中减去 $3$ 。

$3 x - 4 y = 9$

$- 2 x = - 9$

$3 x - 4 y = 9$

$- 2 x = - 9$

解题步骤 2

Write the system as a matrix.

$[\begin{array}{c} 3 & - 4 & 9 \\ - 2 & 0 & - 9 \end{array}]$

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 $\frac{1}{3}$ 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 $\frac{1}{3}$ to make the entry at $1, 1$ a $1$ .

$[\begin{array}{c} \frac{3}{3} & \frac{- 4}{3} & \frac{9}{3} \\ - 2 & 0 & - 9 \end{array}]$

解题步骤 3.1.2

化简 $R_{1}$ 。

$[\begin{array}{c} 1 & - \frac{4}{3} & 3 \\ - 2 & 0 & - 9 \end{array}]$

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 $0$ .

$[\begin{array}{c} 1 & - \frac{4}{3} & 3 \\ - 2 + 2 \cdot 1 & 0 + 2 (- \frac{4}{3}) & - 9 + 2 \cdot 3 \end{array}]$

解题步骤 3.2.2

化简 $R_{2}$ 。

$[\begin{array}{c} 1 & - \frac{4}{3} & 3 \\ 0 & - \frac{8}{3} & - 3 \end{array}]$

解题步骤 3.3

Multiply each element of $R_{2}$ by $- \frac{3}{8}$ 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 $- \frac{3}{8}$ to make the entry at $2, 2$ a $1$ .

$[\begin{array}{c} 1 & - \frac{4}{3} & 3 \\ - \frac{3}{8} \cdot 0 & - \frac{3}{8} (- \frac{8}{3}) & - \frac{3}{8} \cdot - 3 \end{array}]$

解题步骤 3.3.2

化简 $R_{2}$ 。

$[\begin{array}{c} 1 & - \frac{4}{3} & 3 \\ 0 & 1 & \frac{9}{8} \end{array}]$

解题步骤 3.4

Perform the row operation $R_{1} = R_{1} + \frac{4}{3} R_{2}$ to make the entry at $1, 2$ a $0$ .

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解题步骤 3.4.1

Perform the row operation $R_{1} = R_{1} + \frac{4}{3} R_{2}$ to make the entry at $1, 2$ a $0$ .

$[\begin{array}{c} 1 + \frac{4}{3} \cdot 0 & - \frac{4}{3} + \frac{4}{3} \cdot 1 & 3 + \frac{4}{3} \cdot \frac{9}{8} \\ 0 & 1 & \frac{9}{8} \end{array}]$

解题步骤 3.4.2

化简 $R_{1}$ 。

$[\begin{array}{c} 1 & 0 & \frac{9}{2} \\ 0 & 1 & \frac{9}{8} \end{array}]$

解题步骤 4

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

$x = \frac{9}{2}$

$y = \frac{9}{8}$

解题步骤 5

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

$(\frac{9}{2}, \frac{9}{8})$

有限数学 示例

有限数学示例