Finite Mathematik Beispiele

$- 3 x - 4 y = 2$ , $8 y = 6 x - 4$

Schritt 1

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

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Schritt 1.1

Subtrahiere $6 x$ von beiden Seiten der Gleichung.

$- 3 x - 4 y = 2$

$8 y - 6 x = - 4$

Schritt 1.2

Stelle $8 y$ und $- 6 x$ um.

$- 3 x - 4 y = 2$

$- 6 x + 8 y = - 4$

$- 3 x - 4 y = 2$

$- 6 x + 8 y = - 4$

Schritt 2

Write the system as a matrix.

$[\begin{array}{c} - 3 & - 4 & 2 \\ - 6 & 8 & - 4 \end{array}]$

Schritt 3

Ermittele die normierte Zeilenstufenform.

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Schritt 3.1

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

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Schritt 3.1.1

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

$[\begin{array}{c} - \frac{1}{3} \cdot - 3 & - \frac{1}{3} \cdot - 4 & - \frac{1}{3} \cdot 2 \\ - 6 & 8 & - 4 \end{array}]$

Schritt 3.1.2

Vereinfache $R_{1}$ .

$[\begin{array}{c} 1 & \frac{4}{3} & - \frac{2}{3} \\ - 6 & 8 & - 4 \end{array}]$

Schritt 3.2

Perform the row operation $R_{2} = R_{2} + 6 R_{1}$ to make the entry at $2, 1$ a $0$ .

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Schritt 3.2.1

Perform the row operation $R_{2} = R_{2} + 6 R_{1}$ to make the entry at $2, 1$ a $0$ .

$[\begin{array}{c} 1 & \frac{4}{3} & - \frac{2}{3} \\ - 6 + 6 \cdot 1 & 8 + 6 (\frac{4}{3}) & - 4 + 6 (- \frac{2}{3}) \end{array}]$

Schritt 3.2.2

Vereinfache $R_{2}$ .

$[\begin{array}{c} 1 & \frac{4}{3} & - \frac{2}{3} \\ 0 & 16 & - 8 \end{array}]$

Schritt 3.3

Multiply each element of $R_{2}$ by $\frac{1}{16}$ to make the entry at $2, 2$ a $1$ .

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Schritt 3.3.1

Multiply each element of $R_{2}$ by $\frac{1}{16}$ to make the entry at $2, 2$ a $1$ .

$[\begin{array}{c} 1 & \frac{4}{3} & - \frac{2}{3} \\ \frac{0}{16} & \frac{16}{16} & \frac{- 8}{16} \end{array}]$

Schritt 3.3.2

Vereinfache $R_{2}$ .

$[\begin{array}{c} 1 & \frac{4}{3} & - \frac{2}{3} \\ 0 & 1 & - \frac{1}{2} \end{array}]$

Schritt 3.4

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

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Schritt 3.4.1

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

$[\begin{array}{c} 1 - \frac{4}{3} \cdot 0 & \frac{4}{3} - \frac{4}{3} \cdot 1 & - \frac{2}{3} - \frac{4}{3} (- \frac{1}{2}) \\ 0 & 1 & - \frac{1}{2} \end{array}]$

Schritt 3.4.2

Vereinfache $R_{1}$ .

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

Schritt 4

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

$x = 0$

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

Schritt 5

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

$(0, - \frac{1}{2})$