Brewer Clark
01/05/2024 · Junior High School

Si al sistema: \( x-y=1 \) \( y+z=2 \) le añadimos la ecuación \( \dot{x}+z=3 \), entonces el sistema resultante es Compatible determinado Compatible indeterminado incompatible Compatible

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Para determinar si el sistema resultante es compatible, determinado, indeterminado o incompatible, necesitamos resolver el sistema de ecuaciones. Dado que el sistema original es: 1. \( x - y = 1 \) 2. \( y + z = 2 \) 3. \( \dot{x} + z = 3 \) Podemos resolver el sistema de ecuaciones de la siguiente manera: 1. Sumar la primera ecuación con la tercera ecuación para eliminar la variable \( y \): \( (x - y) + (\dot{x} + z) = 1 + 3 \) \( x + \dot{x} + z - y = 4 \) 2. Restar la segunda ecuación de la ecuación resultante para eliminar la variable \( z \): \( (x + \dot{x} + z - y) - (y + z) = 4 - 2 \) \( x + \dot{x} - 2y = 2 \) 3. Ahora tenemos un sistema de dos ecuaciones con dos variables: 1. \( x - y = 1 \) 2. \( x + \dot{x} - 2y = 2 \) Podemos resolver este sistema de ecuaciones para determinar si es compatible, determinado, indeterminado o incompatible. Solve the system of equations by following steps: - step0: Solve using the substitution method: \(\left\{ \begin{array}{l}x-y=1\\x-2y=2\end{array}\right.\) - step1: Solve the equation: \(\left\{ \begin{array}{l}x=1+y\\x-2y=2\end{array}\right.\) - step2: Substitute the value of \(x:\) \(1+y-2y=2\) - step3: Subtract the terms: \(1-y=2\) - step4: Move the constant to the right side: \(-y=2-1\) - step5: Subtract the numbers: \(-y=1\) - step6: Change the signs: \(y=-1\) - step7: Substitute the value of \(y:\) \(x=1-1\) - step8: Simplify: \(x=0\) - step9: Calculate: \(\left\{ \begin{array}{l}x=0\\y=-1\end{array}\right.\) - step10: Check the solution: \(\left\{ \begin{array}{l}x=0\\y=-1\end{array}\right.\) - step11: Rewrite: \(\left(x,y\right) = \left(0,-1\right)\) El sistema de ecuaciones es compatible y tiene una solución única, por lo tanto, el sistema resultante es determinado.

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El sistema resultante es determinado.
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