有限数学示例

$x - \frac{1}{2} y = 1$ , $\frac{7}{2} x - \frac{1}{7} y = 1$

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

组合 $y$ 和 $\frac{1}{2}$ 。

$x - \frac{y}{2} = 1$

$\frac{7}{2} x - \frac{1}{7} y = 1$

解题步骤 1.2

重新排序项。

$x - (\frac{1}{2} y) = 1$

$\frac{7}{2} x - \frac{1}{7} y = 1$

解题步骤 1.3

去掉圆括号。

$x - \frac{1}{2} y = 1$

$\frac{7}{2} x - \frac{1}{7} y = 1$

解题步骤 1.4

化简每一项。

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解题步骤 1.4.1

组合 $\frac{7}{2}$ 和 $x$ 。

$x - \frac{1}{2} y = 1$

$\frac{7 x}{2} - \frac{1}{7} y = 1$

解题步骤 1.4.2

组合 $y$ 和 $\frac{1}{7}$ 。

$x - \frac{1}{2} y = 1$

$\frac{7 x}{2} - \frac{y}{7} = 1$

$x - \frac{1}{2} y = 1$

$\frac{7 x}{2} - \frac{y}{7} = 1$

解题步骤 1.5

重新排序项。

$x - \frac{1}{2} y = 1$

$\frac{7}{2} x - (\frac{1}{7} y) = 1$

解题步骤 1.6

去掉圆括号。

$x - \frac{1}{2} y = 1$

$\frac{7}{2} x - \frac{1}{7} y = 1$

$x - \frac{1}{2} y = 1$

$\frac{7}{2} x - \frac{1}{7} y = 1$

解题步骤 2

Write the system as a matrix.

$[\begin{array}{c} 1 & - \frac{1}{2} & 1 \\ \frac{7}{2} & - \frac{1}{7} & 1 \end{array}]$

解题步骤 3

求行简化阶梯形矩阵。

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解题步骤 3.1

Perform the row operation $R_{2} = R_{2} - \frac{7}{2} R_{1}$ to make the entry at $2, 1$ a $0$ .

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解题步骤 3.1.1

Perform the row operation $R_{2} = R_{2} - \frac{7}{2} R_{1}$ to make the entry at $2, 1$ a $0$ .

$[\begin{array}{c} 1 & - \frac{1}{2} & 1 \\ \frac{7}{2} - \frac{7}{2} \cdot 1 & - \frac{1}{7} - \frac{7}{2} (- \frac{1}{2}) & 1 - \frac{7}{2} \cdot 1 \end{array}]$

解题步骤 3.1.2

化简 $R_{2}$ 。

$[\begin{array}{c} 1 & - \frac{1}{2} & 1 \\ 0 & \frac{45}{28} & - \frac{5}{2} \end{array}]$

解题步骤 3.2

Multiply each element of $R_{2}$ by $\frac{28}{45}$ to make the entry at $2, 2$ a $1$ .

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解题步骤 3.2.1

Multiply each element of $R_{2}$ by $\frac{28}{45}$ to make the entry at $2, 2$ a $1$ .

$[\begin{array}{c} 1 & - \frac{1}{2} & 1 \\ \frac{28}{45} \cdot 0 & \frac{28}{45} \cdot \frac{45}{28} & \frac{28}{45} (- \frac{5}{2}) \end{array}]$

解题步骤 3.2.2

化简 $R_{2}$ 。

$[\begin{array}{c} 1 & - \frac{1}{2} & 1 \\ 0 & 1 & - \frac{14}{9} \end{array}]$

解题步骤 3.3

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

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解题步骤 3.3.1

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

$[\begin{array}{c} 1 + \frac{1}{2} \cdot 0 & - \frac{1}{2} + \frac{1}{2} \cdot 1 & 1 + \frac{1}{2} (- \frac{14}{9}) \\ 0 & 1 & - \frac{14}{9} \end{array}]$

解题步骤 3.3.2

化简 $R_{1}$ 。

$[\begin{array}{c} 1 & 0 & \frac{2}{9} \\ 0 & 1 & - \frac{14}{9} \end{array}]$

解题步骤 4

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

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

$y = - \frac{14}{9}$

解题步骤 5

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

$(\frac{2}{9}, - \frac{14}{9})$

有限数学 示例

有限数学示例