Finite Mathematik Beispiele

$\frac{5 x}{2} + \frac{5 y}{3} = \frac{145}{6}$ , $\frac{x}{4} + \frac{y}{3} = \frac{31}{12}$

Schritt 1

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

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Schritt 1.1

Stelle die Terme um.

$\frac{5}{2} x + \frac{5}{3} y = \frac{145}{6}$

$\frac{x}{4} + \frac{y}{3} = \frac{31}{12}$

Schritt 1.2

Stelle die Terme um.

$\frac{5}{2} x + \frac{5}{3} y = \frac{145}{6}$

$\frac{1}{4} x + \frac{1}{3} y = \frac{31}{12}$

$\frac{5}{2} x + \frac{5}{3} y = \frac{145}{6}$

$\frac{1}{4} x + \frac{1}{3} y = \frac{31}{12}$

Schritt 2

Write the system as a matrix.

$[\begin{array}{c} \frac{5}{2} & \frac{5}{3} & \frac{145}{6} \\ \frac{1}{4} & \frac{1}{3} & \frac{31}{12} \end{array}]$

Schritt 3

Ermittele die normierte Zeilenstufenform.

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Schritt 3.1

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

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Schritt 3.1.1

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

$[\begin{array}{c} \frac{2}{5} \cdot \frac{5}{2} & \frac{2}{5} \cdot \frac{5}{3} & \frac{2}{5} \cdot \frac{145}{6} \\ \frac{1}{4} & \frac{1}{3} & \frac{31}{12} \end{array}]$

Schritt 3.1.2

Vereinfache $R_{1}$ .

$[\begin{array}{c} 1 & \frac{2}{3} & \frac{29}{3} \\ \frac{1}{4} & \frac{1}{3} & \frac{31}{12} \end{array}]$

Schritt 3.2

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

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Schritt 3.2.1

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

$[\begin{array}{c} 1 & \frac{2}{3} & \frac{29}{3} \\ \frac{1}{4} - \frac{1}{4} \cdot 1 & \frac{1}{3} - \frac{1}{4} \cdot \frac{2}{3} & \frac{31}{12} - \frac{1}{4} \cdot \frac{29}{3} \end{array}]$

Schritt 3.2.2

Vereinfache $R_{2}$ .

$[\begin{array}{c} 1 & \frac{2}{3} & \frac{29}{3} \\ 0 & \frac{1}{6} & \frac{1}{6} \end{array}]$

Schritt 3.3

Multiply each element of $R_{2}$ by $6$ to make the entry at $2, 2$ a $1$ .

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Schritt 3.3.1

Multiply each element of $R_{2}$ by $6$ to make the entry at $2, 2$ a $1$ .

$[\begin{array}{c} 1 & \frac{2}{3} & \frac{29}{3} \\ 6 \cdot 0 & 6 (\frac{1}{6}) & 6 (\frac{1}{6}) \end{array}]$

Schritt 3.3.2

Vereinfache $R_{2}$ .

$[\begin{array}{c} 1 & \frac{2}{3} & \frac{29}{3} \\ 0 & 1 & 1 \end{array}]$

Schritt 3.4

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

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Schritt 3.4.1

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

$[\begin{array}{c} 1 - \frac{2}{3} \cdot 0 & \frac{2}{3} - \frac{2}{3} \cdot 1 & \frac{29}{3} - \frac{2}{3} \cdot 1 \\ 0 & 1 & 1 \end{array}]$

Schritt 3.4.2

Vereinfache $R_{1}$ .

$[\begin{array}{c} 1 & 0 & 9 \\ 0 & 1 & 1 \end{array}]$

Schritt 4

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

$x = 9$

$y = 1$

Schritt 5

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

$(9, 1)$