Matemática discreta Exemplos

$0.5 x = - 23 + 4 y$ , $- \frac{4}{3} x + y = - \frac{19}{3}$

Etapa 1

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

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Etapa 1.1

Subtraia $4 y$ dos dois lados da equação.

$0.5 x - 4 y = - 23$

$- \frac{4}{3} x + y = - \frac{19}{3}$

Etapa 1.2

Simplifique cada termo.

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Etapa 1.2.1

Combine $x$ e $\frac{4}{3}$ .

$0.5 x - 4 y = - 23$

$- \frac{x \cdot 4}{3} + y = - \frac{19}{3}$

Etapa 1.2.2

Mova $4$ para a esquerda de $x$ .

$0.5 x - 4 y = - 23$

$- \frac{4 x}{3} + y = - \frac{19}{3}$

$0.5 x - 4 y = - 23$

$- \frac{4 x}{3} + y = - \frac{19}{3}$

Etapa 1.3

Reordene os termos.

$0.5 x - 4 y = - 23$

$- (\frac{4}{3} x) + y = - \frac{19}{3}$

Etapa 1.4

Remova os parênteses.

$0.5 x - 4 y = - 23$

$- \frac{4}{3} x + y = - \frac{19}{3}$

$0.5 x - 4 y = - 23$

$- \frac{4}{3} x + y = - \frac{19}{3}$

Etapa 2

Write the system as a matrix.

$[\begin{array}{c} 0.5 & - 4 & - 23 \\ - \frac{4}{3} & 1 & - \frac{19}{3} \end{array}]$

Etapa 3

Encontre a forma escalonada reduzida por linhas.

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Etapa 3.1

Multiply each element of $R_{1}$ by $\frac{1}{0.5}$ to make the entry at $1, 1$ a $1$ .

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Etapa 3.1.1

Multiply each element of $R_{1}$ by $\frac{1}{0.5}$ to make the entry at $1, 1$ a $1$ .

$[\begin{array}{c} \frac{0.5}{0.5} & \frac{- 4}{0.5} & \frac{- 23}{0.5} \\ - \frac{4}{3} & 1 & - \frac{19}{3} \end{array}]$

Etapa 3.1.2

Simplifique $R_{1}$ .

$[\begin{array}{c} 1 & - 8 & - 46 \\ - \frac{4}{3} & 1 & - \frac{19}{3} \end{array}]$

Etapa 3.2

Perform the row operation $R_{2} = R_{2} + \frac{4}{3} R_{1}$ to make the entry at $2, 1$ a $0$ .

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Etapa 3.2.1

Perform the row operation $R_{2} = R_{2} + \frac{4}{3} R_{1}$ to make the entry at $2, 1$ a $0$ .

$[\begin{array}{c} 1 & - 8 & - 46 \\ - \frac{4}{3} + \frac{4}{3} \cdot 1 & 1 + \frac{4}{3} \cdot - 8 & - \frac{19}{3} + \frac{4}{3} \cdot - 46 \end{array}]$

Etapa 3.2.2

Simplifique $R_{2}$ .

$[\begin{array}{c} 1 & - 8 & - 46 \\ 0 & - \frac{29}{3} & - \frac{203}{3} \end{array}]$

Etapa 3.3

Multiply each element of $R_{2}$ by $- \frac{3}{29}$ to make the entry at $2, 2$ a $1$ .

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Etapa 3.3.1

Multiply each element of $R_{2}$ by $- \frac{3}{29}$ to make the entry at $2, 2$ a $1$ .

$[\begin{array}{c} 1 & - 8 & - 46 \\ - \frac{3}{29} \cdot 0 & - \frac{3}{29} (- \frac{29}{3}) & - \frac{3}{29} (- \frac{203}{3}) \end{array}]$

Etapa 3.3.2

Simplifique $R_{2}$ .

$[\begin{array}{c} 1 & - 8 & - 46 \\ 0 & 1 & 7 \end{array}]$

Etapa 3.4

Perform the row operation $R_{1} = R_{1} + 8 R_{2}$ to make the entry at $1, 2$ a $0$ .

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Etapa 3.4.1

Perform the row operation $R_{1} = R_{1} + 8 R_{2}$ to make the entry at $1, 2$ a $0$ .

$[\begin{array}{c} 1 + 8 \cdot 0 & - 8 + 8 \cdot 1 & - 46 + 8 \cdot 7 \\ 0 & 1 & 7 \end{array}]$

Etapa 3.4.2

Simplifique $R_{1}$ .

$[\begin{array}{c} 1 & 0 & 10 \\ 0 & 1 & 7 \end{array}]$

Etapa 4

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

$x = 10$

$y = 7$

Etapa 5

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

$(10, 7)$