Matematica discreta Esempi

$x + y = 1$ , $0.7 x + 0.3 y = x$ , $0.3 x + 0.7 y = y$

Passaggio 1

Move variables to the left and constant terms to the right.

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Passaggio 1.1

Sottrai $x$ da entrambi i lati dell'equazione.

$x + y = 1$

$0.7 x + 0.3 y - x = 0$

$0.3 x + 0.7 y = y$

Passaggio 1.2

Sottrai $x$ da $0.7 x$ .

$x + y = 1$

$- 0.3 x + 0.3 y = 0$

$0.3 x + 0.7 y = y$

Passaggio 1.3

Sottrai $y$ da entrambi i lati dell'equazione.

$x + y = 1$

$- 0.3 x + 0.3 y = 0$

$0.3 x + 0.7 y - y = 0$

Passaggio 1.4

Sottrai $y$ da $0.7 y$ .

$x + y = 1$

$- 0.3 x + 0.3 y = 0$

$0.3 x - 0.3 y = 0$

$x + y = 1$

$- 0.3 x + 0.3 y = 0$

$0.3 x - 0.3 y = 0$

Passaggio 2

Write the system as a matrix.

$[\begin{array}{c} 1 & 1 & 1 \\ - 0.3 & 0.3 & 0 \\ 0.3 & - 0.3 & 0 \end{array}]$

Passaggio 3

Trova la forma ridotta a scala per righe di Echelon.

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Passaggio 3.1

Perform the row operation $R_{2} = R_{2} + 0.3 R_{1}$ to make the entry at $2, 1$ a $0$ .

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Passaggio 3.1.1

Perform the row operation $R_{2} = R_{2} + 0.3 R_{1}$ to make the entry at $2, 1$ a $0$ .

$[\begin{array}{c} 1 & 1 & 1 \\ - 0.3 + 0.3 \cdot 1 & 0.3 + 0.3 \cdot 1 & 0 + 0.3 \cdot 1 \\ 0.3 & - 0.3 & 0 \end{array}]$

Passaggio 3.1.2

Semplifica $R_{2}$ .

$[\begin{array}{c} 1 & 1 & 1 \\ 0 & 0.6 & 0.3 \\ 0.3 & - 0.3 & 0 \end{array}]$

Passaggio 3.2

Perform the row operation $R_{3} = R_{3} - 0.3 R_{1}$ to make the entry at $3, 1$ a $0$ .

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Passaggio 3.2.1

Perform the row operation $R_{3} = R_{3} - 0.3 R_{1}$ to make the entry at $3, 1$ a $0$ .

$[\begin{array}{c} 1 & 1 & 1 \\ 0 & 0.6 & 0.3 \\ 0.3 - 0.3 \cdot 1 & - 0.3 - 0.3 \cdot 1 & 0 - 0.3 \cdot 1 \end{array}]$

Passaggio 3.2.2

Semplifica $R_{3}$ .

$[\begin{array}{c} 1 & 1 & 1 \\ 0 & 0.6 & 0.3 \\ 0 & - 0.6 & - 0.3 \end{array}]$

Passaggio 3.3

Multiply each element of $R_{2}$ by $\frac{1}{0.6}$ to make the entry at $2, 2$ a $1$ .

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Passaggio 3.3.1

Multiply each element of $R_{2}$ by $\frac{1}{0.6}$ to make the entry at $2, 2$ a $1$ .

$[\begin{array}{c} 1 & 1 & 1 \\ \frac{0}{0.6} & \frac{0.6}{0.6} & \frac{0.3}{0.6} \\ 0 & - 0.6 & - 0.3 \end{array}]$

Passaggio 3.3.2

Semplifica $R_{2}$ .

$[\begin{array}{c} 1 & 1 & 1 \\ 0 & 1 & 0.5 \\ 0 & - 0.6 & - 0.3 \end{array}]$

Passaggio 3.4

Perform the row operation $R_{3} = R_{3} + 0.6 R_{2}$ to make the entry at $3, 2$ a $0$ .

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Passaggio 3.4.1

Perform the row operation $R_{3} = R_{3} + 0.6 R_{2}$ to make the entry at $3, 2$ a $0$ .

$[\begin{array}{c} 1 & 1 & 1 \\ 0 & 1 & 0.5 \\ 0 + 0.6 \cdot 0 & - 0.6 + 0.6 \cdot 1 & - 0.3 + 0.6 \cdot 0.5 \end{array}]$

Passaggio 3.4.2

Semplifica $R_{3}$ .

$[\begin{array}{c} 1 & 1 & 1 \\ 0 & 1 & 0.5 \\ 0 & 0 & 0 \end{array}]$

Passaggio 3.5

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

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Passaggio 3.5.1

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

$[\begin{array}{c} 1 - 0 & 1 - 1 & 1 - 0.5 \\ 0 & 1 & 0.5 \\ 0 & 0 & 0 \end{array}]$

Passaggio 3.5.2

Semplifica $R_{1}$ .

$[\begin{array}{c} 1 & 0 & 0.5 \\ 0 & 1 & 0.5 \\ 0 & 0 & 0 \end{array}]$

Passaggio 4

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

$x = 0.5$

$y = 0.5$

$0 = 0$

Passaggio 5

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

$(0.5, 0.5)$