Finite Mathematik Beispiele

$4 x + 5 y = 34$ , $3 x - 10 = 53$

Schritt 1

Bringe alle Terme, die keine Variable enthalten, auf die rechte Seite der Gleichung.

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Schritt 1.1

Addiere $10$ zu beiden Seiten der Gleichung.

$4 x + 5 y = 34$

$3 x = 53 + 10$

Schritt 1.2

Addiere $53$ und $10$ .

$4 x + 5 y = 34$

$3 x = 63$

$4 x + 5 y = 34$

$3 x = 63$

Schritt 2

Write the system as a matrix.

$[\begin{array}{c} 4 & 5 & 34 \\ 3 & 0 & 63 \end{array}]$

Schritt 3

Ermittele die normierte Zeilenstufenform.

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Schritt 3.1

Multiply each element of $R_{1}$ by $\frac{1}{4}$ to make the entry at $1, 1$ a $1$ .

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Schritt 3.1.1

Multiply each element of $R_{1}$ by $\frac{1}{4}$ to make the entry at $1, 1$ a $1$ .

$[\begin{array}{c} \frac{4}{4} & \frac{5}{4} & \frac{34}{4} \\ 3 & 0 & 63 \end{array}]$

Schritt 3.1.2

Vereinfache $R_{1}$ .

$[\begin{array}{c} 1 & \frac{5}{4} & \frac{17}{2} \\ 3 & 0 & 63 \end{array}]$

Schritt 3.2

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

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Schritt 3.2.1

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

$[\begin{array}{c} 1 & \frac{5}{4} & \frac{17}{2} \\ 3 - 3 \cdot 1 & 0 - 3 (\frac{5}{4}) & 63 - 3 (\frac{17}{2}) \end{array}]$

Schritt 3.2.2

Vereinfache $R_{2}$ .

$[\begin{array}{c} 1 & \frac{5}{4} & \frac{17}{2} \\ 0 & - \frac{15}{4} & \frac{75}{2} \end{array}]$

Schritt 3.3

Multiply each element of $R_{2}$ by $- \frac{4}{15}$ to make the entry at $2, 2$ a $1$ .

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Schritt 3.3.1

Multiply each element of $R_{2}$ by $- \frac{4}{15}$ to make the entry at $2, 2$ a $1$ .

$[\begin{array}{c} 1 & \frac{5}{4} & \frac{17}{2} \\ - \frac{4}{15} \cdot 0 & - \frac{4}{15} (- \frac{15}{4}) & - \frac{4}{15} \cdot \frac{75}{2} \end{array}]$

Schritt 3.3.2

Vereinfache $R_{2}$ .

$[\begin{array}{c} 1 & \frac{5}{4} & \frac{17}{2} \\ 0 & 1 & - 10 \end{array}]$

Schritt 3.4

Perform the row operation $R_{1} = R_{1} - \frac{5}{4} R_{2}$ to make the entry at $1, 2$ a $0$ .

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Schritt 3.4.1

Perform the row operation $R_{1} = R_{1} - \frac{5}{4} R_{2}$ to make the entry at $1, 2$ a $0$ .

$[\begin{array}{c} 1 - \frac{5}{4} \cdot 0 & \frac{5}{4} - \frac{5}{4} \cdot 1 & \frac{17}{2} - \frac{5}{4} \cdot - 10 \\ 0 & 1 & - 10 \end{array}]$

Schritt 3.4.2

Vereinfache $R_{1}$ .

$[\begin{array}{c} 1 & 0 & 21 \\ 0 & 1 & - 10 \end{array}]$

Schritt 4

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

$x = 21$

$y = - 10$

Schritt 5

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

$(21, - 10)$