有限数学示例

$3 x - 4 y = 13$ , $3 y + 2 x = 3$

解题步骤 1

将 $3 y$ 和 $2 x$ 重新排序。

$3 x - 4 y = 13$

$2 x + 3 y = 3$

解题步骤 2

Write the system as a matrix.

$[\begin{array}{c} 3 & - 4 & 13 \\ 2 & 3 & 3 \end{array}]$

解题步骤 3

求行简化阶梯形矩阵。

点击获取更多步骤...

解题步骤 3.1

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

点击获取更多步骤...

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 $1$ .

$[\begin{array}{c} \frac{3}{3} & \frac{- 4}{3} & \frac{13}{3} \\ 2 & 3 & 3 \end{array}]$

解题步骤 3.1.2

化简 $R_{1}$ 。

$[\begin{array}{c} 1 & - \frac{4}{3} & \frac{13}{3} \\ 2 & 3 & 3 \end{array}]$

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 $0$ .

点击获取更多步骤...

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 $0$ .

$[\begin{array}{c} 1 & - \frac{4}{3} & \frac{13}{3} \\ 2 - 2 \cdot 1 & 3 - 2 (- \frac{4}{3}) & 3 - 2 (\frac{13}{3}) \end{array}]$

解题步骤 3.2.2

化简 $R_{2}$ 。

$[\begin{array}{c} 1 & - \frac{4}{3} & \frac{13}{3} \\ 0 & \frac{17}{3} & - \frac{17}{3} \end{array}]$

解题步骤 3.3

Multiply each element of $R_{2}$ by $\frac{3}{17}$ to make the entry at $2, 2$ a $1$ .

点击获取更多步骤...

解题步骤 3.3.1

Multiply each element of $R_{2}$ by $\frac{3}{17}$ to make the entry at $2, 2$ a $1$ .

$[\begin{array}{c} 1 & - \frac{4}{3} & \frac{13}{3} \\ \frac{3}{17} \cdot 0 & \frac{3}{17} \cdot \frac{17}{3} & \frac{3}{17} (- \frac{17}{3}) \end{array}]$

解题步骤 3.3.2

化简 $R_{2}$ 。

$[\begin{array}{c} 1 & - \frac{4}{3} & \frac{13}{3} \\ 0 & 1 & - 1 \end{array}]$

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 $0$ .

点击获取更多步骤...

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 $0$ .

$[\begin{array}{c} 1 + \frac{4}{3} \cdot 0 & - \frac{4}{3} + \frac{4}{3} \cdot 1 & \frac{13}{3} + \frac{4}{3} \cdot - 1 \\ 0 & 1 & - 1 \end{array}]$

解题步骤 3.4.2

化简 $R_{1}$ 。

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

解题步骤 4

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

$x = 3$

$y = - 1$

解题步骤 5

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

$(3, - 1)$

有限数学 示例

有限数学示例