Matemática discreta Exemplos

$- 25 x - 3 y + 69 = 0$ , $10 x + y - 28 = 0$

Etapa 1

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

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Etapa 1.1

Subtraia $69$ dos dois lados da equação.

$- 25 x - 3 y = - 69$

$10 x + y - 28 = 0$

Etapa 1.2

Some $28$ aos dois lados da equação.

$- 25 x - 3 y = - 69$

$10 x + y = 28$

$- 25 x - 3 y = - 69$

$10 x + y = 28$

Etapa 2

Write the system as a matrix.

$[\begin{array}{c} - 25 & - 3 & - 69 \\ 10 & 1 & 28 \end{array}]$

Etapa 3

Encontre a forma escalonada reduzida por linhas.

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Etapa 3.1

Multiply each element of $R_{1}$ by $- \frac{1}{25}$ to make the entry at $1, 1$ a $1$ .

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Etapa 3.1.1

Multiply each element of $R_{1}$ by $- \frac{1}{25}$ to make the entry at $1, 1$ a $1$ .

$[\begin{array}{c} - \frac{1}{25} \cdot - 25 & - \frac{1}{25} \cdot - 3 & - \frac{1}{25} \cdot - 69 \\ 10 & 1 & 28 \end{array}]$

Etapa 3.1.2

Simplifique $R_{1}$ .

$[\begin{array}{c} 1 & \frac{3}{25} & \frac{69}{25} \\ 10 & 1 & 28 \end{array}]$

Etapa 3.2

Perform the row operation $R_{2} = R_{2} - 10 R_{1}$ to make the entry at $2, 1$ a $0$ .

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Etapa 3.2.1

Perform the row operation $R_{2} = R_{2} - 10 R_{1}$ to make the entry at $2, 1$ a $0$ .

$[\begin{array}{c} 1 & \frac{3}{25} & \frac{69}{25} \\ 10 - 10 \cdot 1 & 1 - 10 (\frac{3}{25}) & 28 - 10 (\frac{69}{25}) \end{array}]$

Etapa 3.2.2

Simplifique $R_{2}$ .

$[\begin{array}{c} 1 & \frac{3}{25} & \frac{69}{25} \\ 0 & - \frac{1}{5} & \frac{2}{5} \end{array}]$

Etapa 3.3

Multiply each element of $R_{2}$ by $- 5$ to make the entry at $2, 2$ a $1$ .

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Etapa 3.3.1

Multiply each element of $R_{2}$ by $- 5$ to make the entry at $2, 2$ a $1$ .

$[\begin{array}{c} 1 & \frac{3}{25} & \frac{69}{25} \\ - 5 \cdot 0 & - 5 (- \frac{1}{5}) & - 5 (\frac{2}{5}) \end{array}]$

Etapa 3.3.2

Simplifique $R_{2}$ .

$[\begin{array}{c} 1 & \frac{3}{25} & \frac{69}{25} \\ 0 & 1 & - 2 \end{array}]$

Etapa 3.4

Perform the row operation $R_{1} = R_{1} - \frac{3}{25} R_{2}$ to make the entry at $1, 2$ a $0$ .

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Etapa 3.4.1

Perform the row operation $R_{1} = R_{1} - \frac{3}{25} R_{2}$ to make the entry at $1, 2$ a $0$ .

$[\begin{array}{c} 1 - \frac{3}{25} \cdot 0 & \frac{3}{25} - \frac{3}{25} \cdot 1 & \frac{69}{25} - \frac{3}{25} \cdot - 2 \\ 0 & 1 & - 2 \end{array}]$

Etapa 3.4.2

Simplifique $R_{1}$ .

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

Etapa 4

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

$x = 3$

$y = - 2$

Etapa 5

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

$(3, - 2)$