Finite Mathematik Beispiele

$5 p + 9 q + 13 = 0$ , $3 p - 4 q - 11 = 11$

Schritt 1

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

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Schritt 1.1

Subtrahiere $13$ von beiden Seiten der Gleichung.

$5 p + 9 q = - 13$

$3 p - 4 q - 11 = 11$

Schritt 1.2

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

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Schritt 1.2.1

Addiere $11$ zu beiden Seiten der Gleichung.

$5 p + 9 q = - 13$

$3 p - 4 q = 11 + 11$

Schritt 1.2.2

Addiere $11$ und $11$ .

$5 p + 9 q = - 13$

$3 p - 4 q = 22$

$5 p + 9 q = - 13$

$3 p - 4 q = 22$

$5 p + 9 q = - 13$

$3 p - 4 q = 22$

Schritt 2

Write the system as a matrix.

$[\begin{array}{c} 5 & 9 & - 13 \\ 3 & - 4 & 22 \end{array}]$

Schritt 3

Ermittele die normierte Zeilenstufenform.

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Schritt 3.1

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

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Schritt 3.1.1

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

$[\begin{array}{c} \frac{5}{5} & \frac{9}{5} & \frac{- 13}{5} \\ 3 & - 4 & 22 \end{array}]$

Schritt 3.1.2

Vereinfache $R_{1}$ .

$[\begin{array}{c} 1 & \frac{9}{5} & - \frac{13}{5} \\ 3 & - 4 & 22 \end{array}]$

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 $0$ .

$[\begin{array}{c} 1 & \frac{9}{5} & - \frac{13}{5} \\ 3 - 3 \cdot 1 & - 4 - 3 (\frac{9}{5}) & 22 - 3 (- \frac{13}{5}) \end{array}]$

Schritt 3.2.2

Vereinfache $R_{2}$ .

$[\begin{array}{c} 1 & \frac{9}{5} & - \frac{13}{5} \\ 0 & - \frac{47}{5} & \frac{149}{5} \end{array}]$

Schritt 3.3

Multiply each element of $R_{2}$ by $- \frac{5}{47}$ to make the entry at $2, 2$ a $1$ .

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Schritt 3.3.1

Multiply each element of $R_{2}$ by $- \frac{5}{47}$ to make the entry at $2, 2$ a $1$ .

$[\begin{array}{c} 1 & \frac{9}{5} & - \frac{13}{5} \\ - \frac{5}{47} \cdot 0 & - \frac{5}{47} (- \frac{47}{5}) & - \frac{5}{47} \cdot \frac{149}{5} \end{array}]$

Schritt 3.3.2

Vereinfache $R_{2}$ .

$[\begin{array}{c} 1 & \frac{9}{5} & - \frac{13}{5} \\ 0 & 1 & - \frac{149}{47} \end{array}]$

Schritt 3.4

Perform the row operation $R_{1} = R_{1} - \frac{9}{5} R_{2}$ to make the entry at $1, 2$ a $0$ .

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Schritt 3.4.1

Perform the row operation $R_{1} = R_{1} - \frac{9}{5} R_{2}$ to make the entry at $1, 2$ a $0$ .

$[\begin{array}{c} 1 - \frac{9}{5} \cdot 0 & \frac{9}{5} - \frac{9}{5} \cdot 1 & - \frac{13}{5} - \frac{9}{5} (- \frac{149}{47}) \\ 0 & 1 & - \frac{149}{47} \end{array}]$

Schritt 3.4.2

Vereinfache $R_{1}$ .

$[\begin{array}{c} 1 & 0 & \frac{146}{47} \\ 0 & 1 & - \frac{149}{47} \end{array}]$

Schritt 4

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

$p = \frac{146}{47}$

$q = - \frac{149}{47}$

Schritt 5

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

$(\frac{146}{47}, - \frac{149}{47})$