Example Question - solving system
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Solving System of Linear Equations by Graphing
The image shows a screenshot of a math homework problem asking to solve the system of linear equations by graphing: 1) y = x - 7 2) y = 2x - 2 To graph these equations, you need to plot at least two points for each line and then draw a straight line through these points. For the first equation, y = x - 7: - When x = 0, y = -7. The point is (0, -7). - When x = 7, y = 0. The point is (7, 0). Plot these two points and draw a line through them. For the second equation, y = 2x - 2: - When x = 0, y = -2. The point is (0, -2). - When x = 1, y = 0. The point is (1, 0). Plot these two points as well and draw a line through them. The solution to the system of equations is the point where the two lines intersect. By examining the graph that would be created, you can visually determine the point of intersection and that will be the solution to the system of equations. In the provided image, there is a small section of the graph visible, but to be precise, you would need to plot these points on a full graph, either on paper or using graphing software. The system's solution will be the coordinates where both lines cross each other on the graph. The exact solution can be found by setting the two equations equal to each other and solving for x, then using that x-value to find the corresponding y-value. However, since the task specifically asks to use the graphing tool, the solution should be obtained graphically.
Solving a System of Linear Equations using Elimination Method
El conjunto de ecuaciones proporcionado es un sistema de ecuaciones lineales. Para resolverlo, utilizaremos el método de eliminación para reducir el sistema a ecuaciones con menos variables y luego resolver esas ecuaciones. Comenzamos con el sistema original: \[ \begin{cases} x + y = 0 \\ x + 2y - 3z = -3 \\ 2x + 3y - 4z = -3 \end{cases} \] Primero, vamos a restar la primera ecuación de la segunda y la tercera para eliminar la variable \(x\): Restando la primera ecuación de la segunda: \[ (x + 2y - 3z) - (x + y) = -3 - 0 \] \[ 2y - y - 3z = -3 \] \[ y - 3z = -3 \] ... (ecuación 2') Restando el doble de la primera ecuación de la tercera: \[ (2x + 3y - 4z) - 2(x + y) = -3 - 2(0) \] \[ 2x + 3y - 4z - 2x - 2y = -3 \] \[ y - 4z = -3 \] ... (ecuación 3') Ahora tenemos un sistema simplificado con las siguientes dos ecuaciones: \[ \begin{cases} y - 3z = -3 \quad \text{(ecuación 2')} \\ y - 4z = -3 \quad \text{(ecuación 3')} \end{cases} \] Restamos la ecuación 2' de la ecuación 3': \[ (y - 4z) - (y - 3z) = -3 - (-3) \] \[ -4z + 3z = 0 \] \[ -z = 0 \] \[ z = 0 \] Una vez que tenemos \(z = 0\), podemos sustituir este valor en la ecuación 2' o 3' para encontrar el valor de \(y\): \[ y - 3(0) = -3 \] \[ y = -3 \] Ahora que conocemos los valores de \(y\) y \(z\), podemos sustituir estos valores en la primera ecuación original para encontrar \(x\): \[ x + (-3) = 0 \] \[ x = 3 \] Por ende, la solución al sistema de ecuaciones es: \[ x = 3, y = -3, z = 0 \]
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