初级微积分示例

$5 x + 4 = - 2 y$ , $3 x = - y$

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

在等式两边都加上 $2 y$ 。

$5 x + 4 + 2 y = 0$

$3 x = - y$

解题步骤 1.2

从等式两边同时减去 $4$ 。

$5 x + 2 y = - 4$

$3 x = - y$

解题步骤 1.3

在等式两边都加上 $y$ 。

$5 x + 2 y = - 4$

$3 x + y = 0$

$5 x + 2 y = - 4$

$3 x + y = 0$

解题步骤 2

Write the system as a matrix.

$[\begin{array}{c} 5 & 2 & - 4 \\ 3 & 1 & 0 \end{array}]$

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 $1$ .

$[\begin{array}{c} \frac{5}{5} & \frac{2}{5} & \frac{- 4}{5} \\ 3 & 1 & 0 \end{array}]$

解题步骤 3.1.2

化简 $R_{1}$ 。

$[\begin{array}{c} 1 & \frac{2}{5} & - \frac{4}{5} \\ 3 & 1 & 0 \end{array}]$

解题步骤 3.2

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

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解题步骤 3.2.1

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

$[\begin{array}{c} 1 & \frac{2}{5} & - \frac{4}{5} \\ 3 - 3 \cdot 1 & 1 - 3 (\frac{2}{5}) & 0 - 3 (- \frac{4}{5}) \end{array}]$

解题步骤 3.2.2

化简 $R_{2}$ 。

$[\begin{array}{c} 1 & \frac{2}{5} & - \frac{4}{5} \\ 0 & - \frac{1}{5} & \frac{12}{5} \end{array}]$

解题步骤 3.3

Multiply each element of $R_{2}$ by $- 5$ 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 $- 5$ to make the entry at $2, 2$ a $1$ .

$[\begin{array}{c} 1 & \frac{2}{5} & - \frac{4}{5} \\ - 5 \cdot 0 & - 5 (- \frac{1}{5}) & - 5 (\frac{12}{5}) \end{array}]$

解题步骤 3.3.2

化简 $R_{2}$ 。

$[\begin{array}{c} 1 & \frac{2}{5} & - \frac{4}{5} \\ 0 & 1 & - 12 \end{array}]$

解题步骤 3.4

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

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解题步骤 3.4.1

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

$[\begin{array}{c} 1 - \frac{2}{5} \cdot 0 & \frac{2}{5} - \frac{2}{5} \cdot 1 & - \frac{4}{5} - \frac{2}{5} \cdot - 12 \\ 0 & 1 & - 12 \end{array}]$

解题步骤 3.4.2

化简 $R_{1}$ 。

$[\begin{array}{c} 1 & 0 & 4 \\ 0 & 1 & - 12 \end{array}]$

解题步骤 4

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

$x = 4$

$y = - 12$

解题步骤 5

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

$(4, - 12)$

初级微积分 示例

初级微积分示例