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1) \( 5 x+7 y=45 \) 2) \( 4 x-13 y=10 \)

Ask by Bernard Christensen. in Mexico
Feb 21,2025

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Los valores exactos son: - \( x = \frac{655}{93} \) - \( y = \frac{130}{93} \)

Solución

Vamos a resolver el siguiente sistema de ecuaciones:   (1) 5x + 7y = 45   (2) 4x – 13y = 10 Paso 1. Multiplicamos las ecuaciones para eliminar una variable. Por ejemplo, podemos eliminar x. Multiplicamos la ecuación (1) por 4 y la ecuación (2) por 5:   Ecuación (1) × 4: 20x + 28y = 180   Ecuación (2) × 5: 20x – 65y = 50 Paso 2. Restamos la segunda ecuación de la primera para eliminar x:   (20x + 28y) – (20x – 65y) = 180 – 50   20x – 20x + (28y + 65y) = 130   93y = 130 Despejamos y:   y = 130/93 Esta fracción no se reduce, pues 130 tiene como factores 2, 5 y 13, mientras que 93 es 3 × 31. Paso 3. Sustituimos y en una de las ecuaciones originales para hallar x. Usaremos la ecuación (1):   5x + 7y = 45   5x = 45 – 7y = 45 – 7(130/93) Convierte 45 a número con denominador 93: 45 = (45×93)/93 = 4185/93. Entonces,   5x = (4185 – 910)/93 = 3275/93   x = (3275/93) ÷ 5 = 3275/(93×5) = 3275/465 Simplificamos dividiendo numerador y denominador entre 5:   3275 ÷ 5 = 655   465 ÷ 5 = 93 Por lo tanto,   x = 655/93 Resultado Final:   x = 655/93  y = 130/93 Estos son los valores exactos de x e y.

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To get a feel for solving these equations, think about the historical context of algebra. The study of systems of equations can be traced back to ancient civilizations, such as the Babylonians, who used geometrical methods to solve similar problems. This means the techniques you're using have been polished and refined over thousands of years! When it comes to applying these equations in the real world, consider how they could represent a practical situation, like budgeting for a party. The variables could be the number of two different types of party items (say, pizzas and drinks). By setting up a system of equations, you can determine how many of each type to buy while staying within your budget!

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