Matemática discreta Ejemplos

$8 x \cdot 1 + 2 x \cdot 2 + s \cdot 1 = 400$ , $2 x \cdot 1 + x \cdot 2 + s \cdot 2 = 120$

Paso 1

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

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Paso 1.1

Simplifica cada término.

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Paso 1.1.1

Multiplica $8$ por $1$ .

$8 x + 2 x \cdot 2 + s \cdot 1 = 400$

$2 x \cdot 1 + x \cdot 2 + s \cdot 2 = 120$

Paso 1.1.2

Multiplica $2$ por $2$ .

$8 x + 4 x + s \cdot 1 = 400$

$2 x \cdot 1 + x \cdot 2 + s \cdot 2 = 120$

Paso 1.1.3

Multiplica $s$ por $1$ .

$8 x + 4 x + s = 400$

$2 x \cdot 1 + x \cdot 2 + s \cdot 2 = 120$

$8 x + 4 x + s = 400$

$2 x \cdot 1 + x \cdot 2 + s \cdot 2 = 120$

Paso 1.2

Suma $8 x$ y $4 x$ .

$12 x + s = 400$

$2 x \cdot 1 + x \cdot 2 + s \cdot 2 = 120$

Paso 1.3

Reordena $12 x$ y $s$ .

$s + 12 x = 400$

$2 x \cdot 1 + x \cdot 2 + s \cdot 2 = 120$

Paso 1.4

Simplifica cada término.

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Paso 1.4.1

Multiplica $2$ por $1$ .

$s + 12 x = 400$

$2 x + x \cdot 2 + s \cdot 2 = 120$

Paso 1.4.2

Mueve $2$ a la izquierda de $x$ .

$s + 12 x = 400$

$2 x + 2 \cdot x + s \cdot 2 = 120$

Paso 1.4.3

Mueve $2$ a la izquierda de $s$ .

$s + 12 x = 400$

$2 x + 2 x + 2 s = 120$

$s + 12 x = 400$

$2 x + 2 x + 2 s = 120$

Paso 1.5

Suma $2 x$ y $2 x$ .

$s + 12 x = 400$

$4 x + 2 s = 120$

Paso 1.6

Reordena $4 x$ y $2 s$ .

$s + 12 x = 400$

$2 s + 4 x = 120$

$s + 12 x = 400$

$2 s + 4 x = 120$

Paso 2

Write the system as a matrix.

$[\begin{array}{c} 1 & 12 & 400 \\ 2 & 4 & 120 \end{array}]$

Paso 3

Obtén la forma escalonada reducida por filas.

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Paso 3.1

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

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Paso 3.1.1

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

$[\begin{array}{c} 1 & 12 & 400 \\ 2 - 2 \cdot 1 & 4 - 2 \cdot 12 & 120 - 2 \cdot 400 \end{array}]$

Paso 3.1.2

Simplifica $R_{2}$ .

$[\begin{array}{c} 1 & 12 & 400 \\ 0 & - 20 & - 680 \end{array}]$

$[\begin{array}{c} 1 & 12 & 400 \\ 0 & - 20 & - 680 \end{array}]$

Paso 3.2

Multiply each element of $R_{2}$ by $- \frac{1}{20}$ to make the entry at $2, 2$ a $1$ .

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Paso 3.2.1

Multiply each element of $R_{2}$ by $- \frac{1}{20}$ to make the entry at $2, 2$ a $1$ .

$[\begin{array}{c} 1 & 12 & 400 \\ - \frac{1}{20} \cdot 0 & - \frac{1}{20} \cdot - 20 & - \frac{1}{20} \cdot - 680 \end{array}]$

Paso 3.2.2

Simplifica $R_{2}$ .

$[\begin{array}{c} 1 & 12 & 400 \\ 0 & 1 & 34 \end{array}]$

$[\begin{array}{c} 1 & 12 & 400 \\ 0 & 1 & 34 \end{array}]$

Paso 3.3

Perform the row operation $R_{1} = R_{1} - 12 R_{2}$ to make the entry at $1, 2$ a $0$ .

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Paso 3.3.1

Perform the row operation $R_{1} = R_{1} - 12 R_{2}$ to make the entry at $1, 2$ a $0$ .

$[\begin{array}{c} 1 - 12 \cdot 0 & 12 - 12 \cdot 1 & 400 - 12 \cdot 34 \\ 0 & 1 & 34 \end{array}]$

Paso 3.3.2

Simplifica $R_{1}$ .

$[\begin{array}{c} 1 & 0 & - 8 \\ 0 & 1 & 34 \end{array}]$

$[\begin{array}{c} 1 & 0 & - 8 \\ 0 & 1 & 34 \end{array}]$

$[\begin{array}{c} 1 & 0 & - 8 \\ 0 & 1 & 34 \end{array}]$

Paso 4

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

$s = - 8$

$x = 34$

Paso 5

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

$(- 8, 34)$

$[\begin{array}{c} x^{2} & \frac{1}{2} \\ \sqrt{π} & \int x d x \end{array}]$