대수 예제

3 x + 1 = 0 y

$3 x + 1 = 0 y$ ,

y + 2 x > - 9

$y + 2 x > - 9$

단계 1

여유변수

u

$u$ 와

v

$v$ 를 사용하여 부등식을 등식으로 바꿉니다.

y + 2 x - Z = - 9

$y + 2 x - Z = - 9$

3 x + 1 = 0

$3 x + 1 = 0$

단계 2

방정식의 양변에서

1

$1$ 를 뺍니다.

y + 2 x - Z = - 9, 3 x = - 1

$y + 2 x - Z = - 9, 3 x = - 1$

단계 3

연립방정식을 행렬 형태로 씁니다.

[\begin{array}{c} 2 & 1 & 0 & - 9 \\ 3 & 0 & 0 & - 1 \end{array}]

$[\begin{array}{c} 2 & 1 & 0 & - 9 \\ 3 & 0 & 0 & - 1 \end{array}]$

단계 4

기약 행 사다리꼴을 구합니다.

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단계 4.1

Multiply each element of

R_{1}

$R_{1}$ by

\frac{1}{2}

$\frac{1}{2}$ to make the entry at

1, 1

$1, 1$ a

1

$1$ .

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단계 4.1.1

Multiply each element of

R_{1}

$R_{1}$ by

\frac{1}{2}

$\frac{1}{2}$ to make the entry at

1, 1

$1, 1$ a

1

$1$ .

[\begin{array}{c} \frac{2}{2} & \frac{1}{2} & \frac{0}{2} & \frac{- 9}{2} \\ 3 & 0 & 0 & - 1 \end{array}]

$[\begin{array}{c} \frac{2}{2} & \frac{1}{2} & \frac{0}{2} & \frac{- 9}{2} \\ 3 & 0 & 0 & - 1 \end{array}]$

단계 4.1.2

R_{1}

$R_{1}$ 을 간단히 합니다.

[\begin{array}{c} 1 & \frac{1}{2} & 0 & - \frac{9}{2} \\ 3 & 0 & 0 & - 1 \end{array}]

$[\begin{array}{c} 1 & \frac{1}{2} & 0 & - \frac{9}{2} \\ 3 & 0 & 0 & - 1 \end{array}]$

[\begin{array}{c} 1 & \frac{1}{2} & 0 & - \frac{9}{2} \\ 3 & 0 & 0 & - 1 \end{array}]

$[\begin{array}{c} 1 & \frac{1}{2} & 0 & - \frac{9}{2} \\ 3 & 0 & 0 & - 1 \end{array}]$

단계 4.2

Perform the row operation

R_{2} = R_{2} - 3 R_{1}

$R_{2} = R_{2} - 3 R_{1}$ to make the entry at

2, 1

$2, 1$ a

0

$0$ .

자세한 풀이 단계를 보려면 여기를 누르십시오...

단계 4.2.1

Perform the row operation

R_{2} = R_{2} - 3 R_{1}

$R_{2} = R_{2} - 3 R_{1}$ to make the entry at

2, 1

$2, 1$ a

0

$0$ .

[\begin{array}{c} 1 & \frac{1}{2} & 0 & - \frac{9}{2} \\ 3 - 3 \cdot 1 & 0 - 3 (\frac{1}{2}) & 0 - 3 \cdot 0 & - 1 - 3 (- \frac{9}{2}) \end{array}]

$[\begin{array}{c} 1 & \frac{1}{2} & 0 & - \frac{9}{2} \\ 3 - 3 \cdot 1 & 0 - 3 (\frac{1}{2}) & 0 - 3 \cdot 0 & - 1 - 3 (- \frac{9}{2}) \end{array}]$

단계 4.2.2

R_{2}

$R_{2}$ 을 간단히 합니다.

[\begin{array}{c} 1 & \frac{1}{2} & 0 & - \frac{9}{2} \\ 0 & - \frac{3}{2} & 0 & \frac{25}{2} \end{array}]

$[\begin{array}{c} 1 & \frac{1}{2} & 0 & - \frac{9}{2} \\ 0 & - \frac{3}{2} & 0 & \frac{25}{2} \end{array}]$

[\begin{array}{c} 1 & \frac{1}{2} & 0 & - \frac{9}{2} \\ 0 & - \frac{3}{2} & 0 & \frac{25}{2} \end{array}]

$[\begin{array}{c} 1 & \frac{1}{2} & 0 & - \frac{9}{2} \\ 0 & - \frac{3}{2} & 0 & \frac{25}{2} \end{array}]$

단계 4.3

Multiply each element of

R_{2}

$R_{2}$ by

- \frac{2}{3}

$- \frac{2}{3}$ to make the entry at

2, 2

$2, 2$ a

1

$1$ .

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단계 4.3.1

Multiply each element of

R_{2}

$R_{2}$ by

- \frac{2}{3}

$- \frac{2}{3}$ to make the entry at

2, 2

$2, 2$ a

1

$1$ .

[\begin{array}{c} 1 & \frac{1}{2} & 0 & - \frac{9}{2} \\ - \frac{2}{3} \cdot 0 & - \frac{2}{3} (- \frac{3}{2}) & - \frac{2}{3} \cdot 0 & - \frac{2}{3} \cdot \frac{25}{2} \end{array}]

$[\begin{array}{c} 1 & \frac{1}{2} & 0 & - \frac{9}{2} \\ - \frac{2}{3} \cdot 0 & - \frac{2}{3} (- \frac{3}{2}) & - \frac{2}{3} \cdot 0 & - \frac{2}{3} \cdot \frac{25}{2} \end{array}]$

단계 4.3.2

R_{2}

$R_{2}$ 을 간단히 합니다.

[\begin{array}{c} 1 & \frac{1}{2} & 0 & - \frac{9}{2} \\ 0 & 1 & 0 & - \frac{25}{3} \end{array}]

$[\begin{array}{c} 1 & \frac{1}{2} & 0 & - \frac{9}{2} \\ 0 & 1 & 0 & - \frac{25}{3} \end{array}]$

[\begin{array}{c} 1 & \frac{1}{2} & 0 & - \frac{9}{2} \\ 0 & 1 & 0 & - \frac{25}{3} \end{array}]

$[\begin{array}{c} 1 & \frac{1}{2} & 0 & - \frac{9}{2} \\ 0 & 1 & 0 & - \frac{25}{3} \end{array}]$

단계 4.4

Perform the row operation

R_{1} = R_{1} - \frac{1}{2} R_{2}

$R_{1} = R_{1} - \frac{1}{2} R_{2}$ to make the entry at

1, 2

$1, 2$ a

0

$0$ .

자세한 풀이 단계를 보려면 여기를 누르십시오...

단계 4.4.1

Perform the row operation

R_{1} = R_{1} - \frac{1}{2} R_{2}

$R_{1} = R_{1} - \frac{1}{2} R_{2}$ to make the entry at

1, 2

$1, 2$ a

0

$0$ .

[\begin{array}{c} 1 - \frac{1}{2} \cdot 0 & \frac{1}{2} - \frac{1}{2} \cdot 1 & 0 - \frac{1}{2} \cdot 0 & - \frac{9}{2} - \frac{1}{2} (- \frac{25}{3}) \\ 0 & 1 & 0 & - \frac{25}{3} \end{array}]

$[\begin{array}{c} 1 - \frac{1}{2} \cdot 0 & \frac{1}{2} - \frac{1}{2} \cdot 1 & 0 - \frac{1}{2} \cdot 0 & - \frac{9}{2} - \frac{1}{2} (- \frac{25}{3}) \\ 0 & 1 & 0 & - \frac{25}{3} \end{array}]$

단계 4.4.2

R_{1}

$R_{1}$ 을 간단히 합니다.

[\begin{array}{c} 1 & 0 & 0 & - \frac{1}{3} \\ 0 & 1 & 0 & - \frac{25}{3} \end{array}]

$[\begin{array}{c} 1 & 0 & 0 & - \frac{1}{3} \\ 0 & 1 & 0 & - \frac{25}{3} \end{array}]$

[\begin{array}{c} 1 & 0 & 0 & - \frac{1}{3} \\ 0 & 1 & 0 & - \frac{25}{3} \end{array}]

$[\begin{array}{c} 1 & 0 & 0 & - \frac{1}{3} \\ 0 & 1 & 0 & - \frac{25}{3} \end{array}]$

[\begin{array}{c} 1 & 0 & 0 & - \frac{1}{3} \\ 0 & 1 & 0 & - \frac{25}{3} \end{array}]

$[\begin{array}{c} 1 & 0 & 0 & - \frac{1}{3} \\ 0 & 1 & 0 & - \frac{25}{3} \end{array}]$

단계 5

결과 행렬을 이용해 연립방정식의 최종 해를 구합니다.

x = 0

$x = 0$

y = 0

$y = 0$

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