有限数学示例

$2 y = 14$ , $x = y + 1$

解题步骤 1

从等式两边同时减去 $y$ 。

$2 y = 14$

$x - y = 1$

解题步骤 2

Write the system as a matrix.

$[\begin{array}{c} 0 & 2 & 14 \\ 1 & - 1 & 1 \end{array}]$

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

Swap $R_{2}$ with $R_{1}$ to put a nonzero entry at $1, 1$ .

$[\begin{array}{c} 1 & - 1 & 1 \\ 0 & 2 & 14 \end{array}]$

解题步骤 3.2

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

点击获取更多步骤...

解题步骤 3.2.1

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

$[\begin{array}{c} 1 & - 1 & 1 \\ \frac{0}{2} & \frac{2}{2} & \frac{14}{2} \end{array}]$

解题步骤 3.2.2

化简 $R_{2}$ 。

$[\begin{array}{c} 1 & - 1 & 1 \\ 0 & 1 & 7 \end{array}]$

解题步骤 3.3

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

点击获取更多步骤...

解题步骤 3.3.1

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

$[\begin{array}{c} 1 + 0 & - 1 + 1 \cdot 1 & 1 + 1 \cdot 7 \\ 0 & 1 & 7 \end{array}]$

解题步骤 3.3.2

化简 $R_{1}$ 。

$[\begin{array}{c} 1 & 0 & 8 \\ 0 & 1 & 7 \end{array}]$

解题步骤 4

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

$x = 8$

$y = 7$

解题步骤 5

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

$(8, 7)$

有限数学 示例