有限数学示例

$\frac{1}{2} x - \frac{1}{4} y = 1$ , $\frac{1}{3} x + y = 3$

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

化简每一项。

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解题步骤 1.1.1

组合 $\frac{1}{2}$ 和 $x$ 。

$\frac{x}{2} - \frac{1}{4} y = 1$

$\frac{1}{3} x + y = 3$

解题步骤 1.1.2

组合 $y$ 和 $\frac{1}{4}$ 。

$\frac{x}{2} - \frac{y}{4} = 1$

$\frac{1}{3} x + y = 3$

$\frac{x}{2} - \frac{y}{4} = 1$

$\frac{1}{3} x + y = 3$

解题步骤 1.2

重新排序项。

$\frac{1}{2} x - (\frac{1}{4} y) = 1$

$\frac{1}{3} x + y = 3$

解题步骤 1.3

去掉圆括号。

$\frac{1}{2} x - \frac{1}{4} y = 1$

$\frac{1}{3} x + y = 3$

解题步骤 1.4

组合 $\frac{1}{3}$ 和 $x$ 。

$\frac{1}{2} x - \frac{1}{4} y = 1$

$\frac{x}{3} + y = 3$

解题步骤 1.5

重新排序项。

$\frac{1}{2} x - \frac{1}{4} y = 1$

$\frac{1}{3} x + y = 3$

$\frac{1}{2} x - \frac{1}{4} y = 1$

$\frac{1}{3} x + y = 3$

解题步骤 2

Write the system as a matrix.

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

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 $2$ 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 $1$ .

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

解题步骤 3.1.2

化简 $R_{1}$ 。

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

解题步骤 3.2

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

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解题步骤 3.2.1

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

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

解题步骤 3.2.2

化简 $R_{2}$ 。

$[\begin{array}{c} 1 & - \frac{1}{2} & 2 \\ 0 & \frac{7}{6} & \frac{7}{3} \end{array}]$

解题步骤 3.3

Multiply each element of $R_{2}$ by $\frac{6}{7}$ 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 $\frac{6}{7}$ to make the entry at $2, 2$ a $1$ .

$[\begin{array}{c} 1 & - \frac{1}{2} & 2 \\ \frac{6}{7} \cdot 0 & \frac{6}{7} \cdot \frac{7}{6} & \frac{6}{7} \cdot \frac{7}{3} \end{array}]$

解题步骤 3.3.2

化简 $R_{2}$ 。

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

解题步骤 3.4

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 3.4.1

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

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

解题步骤 3.4.2

化简 $R_{1}$ 。

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

解题步骤 4

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

$x = 3$

$y = 2$

解题步骤 5

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

$(3, 2)$

有限数学 示例

有限数学示例