有限数学示例

$0.5 x = - 23 + 4 y$ , $- \frac{4}{3} x + y = - \frac{19}{3}$

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

从等式两边同时减去 $4 y$ 。

$0.5 x - 4 y = - 23$

$- \frac{4}{3} x + y = - \frac{19}{3}$

解题步骤 1.2

化简每一项。

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解题步骤 1.2.1

组合 $x$ 和 $\frac{4}{3}$ 。

$0.5 x - 4 y = - 23$

$- \frac{x \cdot 4}{3} + y = - \frac{19}{3}$

解题步骤 1.2.2

将 $4$ 移到 $x$ 的左侧。

$0.5 x - 4 y = - 23$

$- \frac{4 x}{3} + y = - \frac{19}{3}$

$0.5 x - 4 y = - 23$

$- \frac{4 x}{3} + y = - \frac{19}{3}$

解题步骤 1.3

重新排序项。

$0.5 x - 4 y = - 23$

$- (\frac{4}{3} x) + y = - \frac{19}{3}$

解题步骤 1.4

去掉圆括号。

$0.5 x - 4 y = - 23$

$- \frac{4}{3} x + y = - \frac{19}{3}$

$0.5 x - 4 y = - 23$

$- \frac{4}{3} x + y = - \frac{19}{3}$

解题步骤 2

Write the system as a matrix.

$[\begin{array}{c} 0.5 & - 4 & - 23 \\ - \frac{4}{3} & 1 & - \frac{19}{3} \end{array}]$

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 $\frac{1}{0.5}$ 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 $\frac{1}{0.5}$ to make the entry at $1, 1$ a $1$ .

$[\begin{array}{c} \frac{0.5}{0.5} & \frac{- 4}{0.5} & \frac{- 23}{0.5} \\ - \frac{4}{3} & 1 & - \frac{19}{3} \end{array}]$

解题步骤 3.1.2

化简 $R_{1}$ 。

$[\begin{array}{c} 1 & - 8 & - 46 \\ - \frac{4}{3} & 1 & - \frac{19}{3} \end{array}]$

解题步骤 3.2

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

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解题步骤 3.2.1

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

$[\begin{array}{c} 1 & - 8 & - 46 \\ - \frac{4}{3} + \frac{4}{3} \cdot 1 & 1 + \frac{4}{3} \cdot - 8 & - \frac{19}{3} + \frac{4}{3} \cdot - 46 \end{array}]$

解题步骤 3.2.2

化简 $R_{2}$ 。

$[\begin{array}{c} 1 & - 8 & - 46 \\ 0 & - \frac{29}{3} & - \frac{203}{3} \end{array}]$

解题步骤 3.3

Multiply each element of $R_{2}$ by $- \frac{3}{29}$ 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 $- \frac{3}{29}$ to make the entry at $2, 2$ a $1$ .

$[\begin{array}{c} 1 & - 8 & - 46 \\ - \frac{3}{29} \cdot 0 & - \frac{3}{29} (- \frac{29}{3}) & - \frac{3}{29} (- \frac{203}{3}) \end{array}]$

解题步骤 3.3.2

化简 $R_{2}$ 。

$[\begin{array}{c} 1 & - 8 & - 46 \\ 0 & 1 & 7 \end{array}]$

解题步骤 3.4

Perform the row operation $R_{1} = R_{1} + 8 R_{2}$ to make the entry at $1, 2$ a $0$ .

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解题步骤 3.4.1

Perform the row operation $R_{1} = R_{1} + 8 R_{2}$ to make the entry at $1, 2$ a $0$ .

$[\begin{array}{c} 1 + 8 \cdot 0 & - 8 + 8 \cdot 1 & - 46 + 8 \cdot 7 \\ 0 & 1 & 7 \end{array}]$

解题步骤 3.4.2

化简 $R_{1}$ 。

$[\begin{array}{c} 1 & 0 & 10 \\ 0 & 1 & 7 \end{array}]$

解题步骤 4

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

$x = 10$

$y = 7$

解题步骤 5

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

$(10, 7)$

有限数学 示例

有限数学示例