有限数学示例

$\frac{1}{2} x + 2 y = 14$ , $3 x + 16 y = 0$

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 $\frac{1}{2}$ 和 $x$ 。

$\frac{x}{2} + 2 y = 14$

$3 x + 16 y = 0$

解题步骤 1.2

重新排序项。

$\frac{1}{2} x + 2 y = 14$

$3 x + 16 y = 0$

$\frac{1}{2} x + 2 y = 14$

$3 x + 16 y = 0$

解题步骤 2

Write the system as a matrix.

$[\begin{array}{c} \frac{1}{2} & 2 & 14 \\ 3 & 16 & 0 \end{array}]$

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 $2$ 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 $2$ to make the entry at $1, 1$ a $1$ .

$[\begin{array}{c} 2 (\frac{1}{2}) & 2 \cdot 2 & 2 \cdot 14 \\ 3 & 16 & 0 \end{array}]$

解题步骤 3.1.2

化简 $R_{1}$ 。

$[\begin{array}{c} 1 & 4 & 28 \\ 3 & 16 & 0 \end{array}]$

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 $0$ .

$[\begin{array}{c} 1 & 4 & 28 \\ 3 - 3 \cdot 1 & 16 - 3 \cdot 4 & 0 - 3 \cdot 28 \end{array}]$

解题步骤 3.2.2

化简 $R_{2}$ 。

$[\begin{array}{c} 1 & 4 & 28 \\ 0 & 4 & - 84 \end{array}]$

解题步骤 3.3

Multiply each element of $R_{2}$ by $\frac{1}{4}$ 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 $\frac{1}{4}$ to make the entry at $2, 2$ a $1$ .

$[\begin{array}{c} 1 & 4 & 28 \\ \frac{0}{4} & \frac{4}{4} & \frac{- 84}{4} \end{array}]$

解题步骤 3.3.2

化简 $R_{2}$ 。

$[\begin{array}{c} 1 & 4 & 28 \\ 0 & 1 & - 21 \end{array}]$

解题步骤 3.4

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

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解题步骤 3.4.1

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

$[\begin{array}{c} 1 - 4 \cdot 0 & 4 - 4 \cdot 1 & 28 - 4 \cdot - 21 \\ 0 & 1 & - 21 \end{array}]$

解题步骤 3.4.2

化简 $R_{1}$ 。

$[\begin{array}{c} 1 & 0 & 112 \\ 0 & 1 & - 21 \end{array}]$

解题步骤 4

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

$x = 112$

$y = - 21$

解题步骤 5

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

$(112, - 21)$

有限数学 示例

有限数学示例