Matematica discreta Esempi

$x - \frac{1}{2} y = 1$ , $\frac{7}{2} x - \frac{1}{7} y = 1$

Passaggio 1

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

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Passaggio 1.1

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

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

$\frac{7}{2} x - \frac{1}{7} y = 1$

Passaggio 1.2

Riordina i termini.

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

$\frac{7}{2} x - \frac{1}{7} y = 1$

Passaggio 1.3

Rimuovi le parentesi.

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

$\frac{7}{2} x - \frac{1}{7} y = 1$

Passaggio 1.4

Semplifica ciascun termine.

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Passaggio 1.4.1

$\frac{7}{2}$ e $x$ .

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

$\frac{7 x}{2} - \frac{1}{7} y = 1$

Passaggio 1.4.2

$y$ e $\frac{1}{7}$ .

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

$\frac{7 x}{2} - \frac{y}{7} = 1$

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

$\frac{7 x}{2} - \frac{y}{7} = 1$

Passaggio 1.5

Riordina i termini.

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

$\frac{7}{2} x - (\frac{1}{7} y) = 1$

Passaggio 1.6

Rimuovi le parentesi.

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

$\frac{7}{2} x - \frac{1}{7} y = 1$

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

$\frac{7}{2} x - \frac{1}{7} y = 1$

Passaggio 2

Write the system as a matrix.

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

Passaggio 3

Trova la forma ridotta a scala per righe di Echelon.

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Passaggio 3.1

Perform the row operation $R_{2} = R_{2} - \frac{7}{2} R_{1}$ to make the entry at $2, 1$ a $0$ .

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Passaggio 3.1.1

Perform the row operation $R_{2} = R_{2} - \frac{7}{2} R_{1}$ to make the entry at $2, 1$ a $0$ .

$[\begin{array}{c} 1 & - \frac{1}{2} & 1 \\ \frac{7}{2} - \frac{7}{2} \cdot 1 & - \frac{1}{7} - \frac{7}{2} (- \frac{1}{2}) & 1 - \frac{7}{2} \cdot 1 \end{array}]$

Passaggio 3.1.2

Semplifica $R_{2}$ .

$[\begin{array}{c} 1 & - \frac{1}{2} & 1 \\ 0 & \frac{45}{28} & - \frac{5}{2} \end{array}]$

Passaggio 3.2

Multiply each element of $R_{2}$ by $\frac{28}{45}$ to make the entry at $2, 2$ a $1$ .

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Passaggio 3.2.1

Multiply each element of $R_{2}$ by $\frac{28}{45}$ to make the entry at $2, 2$ a $1$ .

$[\begin{array}{c} 1 & - \frac{1}{2} & 1 \\ \frac{28}{45} \cdot 0 & \frac{28}{45} \cdot \frac{45}{28} & \frac{28}{45} (- \frac{5}{2}) \end{array}]$

Passaggio 3.2.2

Semplifica $R_{2}$ .

$[\begin{array}{c} 1 & - \frac{1}{2} & 1 \\ 0 & 1 & - \frac{14}{9} \end{array}]$

Passaggio 3.3

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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Passaggio 3.3.1

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

$[\begin{array}{c} 1 + \frac{1}{2} \cdot 0 & - \frac{1}{2} + \frac{1}{2} \cdot 1 & 1 + \frac{1}{2} (- \frac{14}{9}) \\ 0 & 1 & - \frac{14}{9} \end{array}]$

Passaggio 3.3.2

Semplifica $R_{1}$ .

$[\begin{array}{c} 1 & 0 & \frac{2}{9} \\ 0 & 1 & - \frac{14}{9} \end{array}]$

Passaggio 4

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

$x = \frac{2}{9}$

$y = - \frac{14}{9}$

Passaggio 5

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

$(\frac{2}{9}, - \frac{14}{9})$