Hardy Rose
07/22/2023 · Senior High School
\( \left\{\begin{array}{l}\text { (1) } 3 x+2 y+z=1 \\ 5 x+3 y+4 z=2 \\ \text { (3) } x+y-z=1\end{array}\right. \)
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Step-by-step Solution
Para resolver el sistema de ecuaciones:
\[
\left\{
\begin{array}{l}
(1) \quad 3x + 2y + z = 1 \\
(2) \quad 5x + 3y + 4z = 2 \\
(3) \quad x + y - z = 1
\end{array}
\right.
\]
Podemos usar el método de sustitución o eliminación. Vamos a usar el método de eliminación.
Primero, despejamos \( z \) de la ecuación (3):
\[
z = x + y - 1
\]
Ahora sustituimos \( z \) en las ecuaciones (1) y (2).
Sustituyendo en (1):
\[
3x + 2y + (x + y - 1) = 1
\]
Simplificando:
\[
3x + 2y + x + y - 1 = 1 \\
4x + 3y - 1 = 1 \\
4x + 3y = 2 \quad \text{(4)}
\]
Ahora sustituimos \( z \) en (2):
\[
5x + 3y + 4(x + y - 1) = 2
\]
Simplificando:
\[
5x + 3y + 4x + 4y - 4 = 2 \\
9x + 7y - 4 = 2 \\
9x + 7y = 6 \quad \text{(5)}
\]
Ahora tenemos un nuevo sistema de dos ecuaciones con dos incógnitas:
\[
\left\{
\begin{array}{l}
(4) \quad 4x + 3y = 2 \\
(5) \quad 9x + 7y = 6
\end{array}
\right.
\]
Podemos resolver este sistema. Multiplicamos la ecuación (4) por 7 y la ecuación (5) por 3 para eliminar \( y \):
\[
\begin{array}{l}
(6) \quad 28x + 21y = 14 \\
(7) \quad 27x + 21y = 18
\end{array}
\]
Restamos (7) de (6):
\[
(28x + 21y) - (27x + 21y) = 14 - 18 \\
x = -4
\]
Ahora sustituimos \( x = -4 \) en la ecuación (4):
\[
4(-4) + 3y = 2 \\
-16 + 3y = 2 \\
3y = 18 \\
y = 6
\]
Finalmente, sustituimos \( x \) y \( y \) en la ecuación para \( z \):
\[
z = -4 + 6 - 1 \\
z = 1
\]
Por lo tanto, la solución del sistema es:
\[
(x, y, z) = (-4, 6, 1)
\]
Quick Answer
La solución del sistema es \( (x, y, z) = (-4, 6, 1) \).
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