Matemática discreta Exemplos

$3 x \cdot 1 - 5 x \cdot 2 = - 2$ , $- 6 x \cdot 1 + 10 x \cdot 2 = 4$

Etapa 1

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

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Etapa 1.1

Simplifique cada termo.

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Etapa 1.1.1

Multiplique $3$ por $1$ .

$3 x - 5 x \cdot 2 = - 2$

$- 6 x \cdot 1 + 10 x \cdot 2 = 4$

Etapa 1.1.2

Multiplique $2$ por $- 5$ .

$3 x - 10 x = - 2$

$- 6 x \cdot 1 + 10 x \cdot 2 = 4$

$3 x - 10 x = - 2$

$- 6 x \cdot 1 + 10 x \cdot 2 = 4$

Etapa 1.2

Subtraia $10 x$ de $3 x$ .

$- 7 x = - 2$

$- 6 x \cdot 1 + 10 x \cdot 2 = 4$

Etapa 1.3

Simplifique cada termo.

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Etapa 1.3.1

Multiplique $- 6$ por $1$ .

$- 7 x = - 2$

$- 6 x + 10 x \cdot 2 = 4$

Etapa 1.3.2

Multiplique $2$ por $10$ .

$- 7 x = - 2$

$- 6 x + 20 x = 4$

$- 7 x = - 2$

$- 6 x + 20 x = 4$

Etapa 1.4

Some $- 6 x$ e $20 x$ .

$- 7 x = - 2$

$14 x = 4$

$- 7 x = - 2$

$14 x = 4$

Etapa 2

Write the system as a matrix.

$[\begin{array}{c} - 7 & - 2 \\ 14 & 4 \end{array}]$

Etapa 3

Encontre a forma escalonada reduzida por linhas.

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Etapa 3.1

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

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Etapa 3.1.1

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

$[\begin{array}{c} - \frac{1}{7} \cdot - 7 & - \frac{1}{7} \cdot - 2 \\ 14 & 4 \end{array}]$

Etapa 3.1.2

Simplifique $R_{1}$ .

$[\begin{array}{c} 1 & \frac{2}{7} \\ 14 & 4 \end{array}]$

Etapa 3.2

Perform the row operation $R_{2} = R_{2} - 14 R_{1}$ to make the entry at $2, 1$ a $0$ .

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Etapa 3.2.1

Perform the row operation $R_{2} = R_{2} - 14 R_{1}$ to make the entry at $2, 1$ a $0$ .

$[\begin{array}{c} 1 & \frac{2}{7} \\ 14 - 14 \cdot 1 & 4 - 14 (\frac{2}{7}) \end{array}]$

Etapa 3.2.2

Simplifique $R_{2}$ .

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

Etapa 4

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

$x = \frac{2}{7}$

$0 = 0$

Etapa 5

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

$(error = error)$