Finite Mathematik Beispiele

$y = 3 x - 2$ , $y = - x + 2$

Schritt 1

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

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Schritt 1.1

Subtrahiere $3 x$ von beiden Seiten der Gleichung.

$y - 3 x = - 2$

$y = - x + 2$

Schritt 1.2

Stelle $y$ und $- 3 x$ um.

$- 3 x + y = - 2$

$y = - x + 2$

Schritt 1.3

Addiere $x$ zu beiden Seiten der Gleichung.

$- 3 x + y = - 2$

$y + x = 2$

Schritt 1.4

Stelle $y$ und $x$ um.

$- 3 x + y = - 2$

$x + y = 2$

$- 3 x + y = - 2$

$x + y = 2$

Schritt 2

Write the system as a matrix.

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

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 $1$ .

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

Schritt 3.1.2

Vereinfache $R_{1}$ .

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

Schritt 3.2

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

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Schritt 3.2.1

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

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

Schritt 3.2.2

Vereinfache $R_{2}$ .

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

Schritt 3.3

Multiply each element of $R_{2}$ by $\frac{3}{4}$ to make the entry at $2, 2$ a $1$ .

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Schritt 3.3.1

Multiply each element of $R_{2}$ by $\frac{3}{4}$ to make the entry at $2, 2$ a $1$ .

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

Schritt 3.3.2

Vereinfache $R_{2}$ .

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

Schritt 3.4

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

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Schritt 3.4.1

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

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

Schritt 3.4.2

Vereinfache $R_{1}$ .

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

Schritt 4

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

$x = 1$

$y = 1$

Schritt 5

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

$(1, 1)$