Matematica discreta Esempi

$4 x + 12 y = - 12$ , $- x - 3 y = 3$ , $- \frac{1}{2} x - \frac{3}{2} y = \frac{3}{2}$

Passaggio 1

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

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Passaggio 1.1

Semplifica ciascun termine.

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Passaggio 1.1.1

$x$ e $\frac{1}{2}$ .

$4 x + 12 y = - 12$

$- x - 3 y = 3$

$- \frac{x}{2} - \frac{3}{2} y = \frac{3}{2}$

Passaggio 1.1.2

$y$ e $\frac{3}{2}$ .

$4 x + 12 y = - 12$

$- x - 3 y = 3$

$- \frac{x}{2} - \frac{y \cdot 3}{2} = \frac{3}{2}$

Passaggio 1.1.3

Sposta $3$ alla sinistra di $y$ .

$4 x + 12 y = - 12$

$- x - 3 y = 3$

$- \frac{x}{2} - \frac{3 y}{2} = \frac{3}{2}$

$4 x + 12 y = - 12$

$- x - 3 y = 3$

$- \frac{x}{2} - \frac{3 y}{2} = \frac{3}{2}$

Passaggio 1.2

Riordina i termini.

$4 x + 12 y = - 12$

$- x - 3 y = 3$

$- (\frac{1}{2} x) - (\frac{3}{2} y) = \frac{3}{2}$

Passaggio 1.3

Rimuovi le parentesi.

$4 x + 12 y = - 12$

$- x - 3 y = 3$

$- \frac{1}{2} x - (\frac{3}{2} y) = \frac{3}{2}$

Passaggio 1.4

Rimuovi le parentesi.

$4 x + 12 y = - 12$

$- x - 3 y = 3$

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

$4 x + 12 y = - 12$

$- x - 3 y = 3$

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

Passaggio 2

Write the system as a matrix.

$[\begin{array}{c} 4 & 12 & - 12 \\ - 1 & - 3 & 3 \\ - \frac{1}{2} & - \frac{3}{2} & \frac{3}{2} \end{array}]$

Passaggio 3

Trova la forma ridotta a scala per righe di Echelon.

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Passaggio 3.1

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

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Passaggio 3.1.1

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

$[\begin{array}{c} \frac{4}{4} & \frac{12}{4} & \frac{- 12}{4} \\ - 1 & - 3 & 3 \\ - \frac{1}{2} & - \frac{3}{2} & \frac{3}{2} \end{array}]$

Passaggio 3.1.2

Semplifica $R_{1}$ .

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

Passaggio 3.2

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

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Passaggio 3.2.1

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

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

Passaggio 3.2.2

Semplifica $R_{2}$ .

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

Passaggio 3.3

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

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Passaggio 3.3.1

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

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

Passaggio 3.3.2

Semplifica $R_{3}$ .

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

Passaggio 4

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

$x + 3 y = - 3$

$0 = 0$

Passaggio 5

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

$(- 3 - 3 y, y)$