uación lineal general (Ax+By+C = 0): (10 puntos) Encuentre la ec Pasa por (-1, 7
ID: 3138813 • Letter: U
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uación lineal general (Ax+By+C = 0): (10 puntos) Encuentre la ec Pasa por (-1, 7) y tiene pendiente -5 II- a) III- Resuelva sistema por Suma: (10 Puntos) -5X-Y+20 = 0 8X + 2Y-24 = 0 IV- Resuelva sistema por Resta: (10 Puntos) 14X-2Y = 70 6X +Y 16 V- Resuelva sistema por Sustitución: (10 puntos) -8X +Y = 36 X+7Y = 54 VI-Sistemas de ecuaciones no-lineales Encuentre los puntos de intercepción: (40 Puntos) 2X +Y 15 X2 - 2X +Y-60 VII- Problema Interpretación: Una empresa enfrenta una curva de Oferta de: p 7/100q - 250 y el mercado de su producto presenta una curva de Demandad de: p -9/100q + 65. ¿Cuál es el precio y la cantidad de equilibrio en ese mercado? (20 puntos)Explanation / Answer
II. a). The point –slope form of the equation of the given line is (y-7)=-5(x+1) or, y-7=-5x-5 or, 5x+y-2= 0.
III. We have -5X –Y +20 = 0 or, 5X+Y = 20…(1) and 8X+2Y -24 = 0 or, 8X+2Y = 24 or, on dividing both the sides by 2, 4X+Y = 12…(2)
On subtracting the 2nd equation from the 1st equation, we get (5X+Y)-( 4X+Y) = 20-12 or, X = 8. On substituting X = 8, in the 2nd equation, we get 4*8+Y = 12 so that Y = 12-32 = -20. Thus, X = 8 and Y = -20.
IV. We have 14X -2Y = 70, on dividing both the sides by 2, 7X-Y = 35…(1) and 6X+Y= 16…(2). On adding these 2 equations, we get 7X-Y + 6X+Y = 35+16 or, 13X = 51 so that X = 51/13. . On substituting X = 51/3, in the 2nd equation, we get 6*51/3 +Y = 16 or, Y = 16-102 = -86. Thus, X = 51/13 and Y = -86.
V. We have -8X +Y = 36…(1) and X+7Y = 54…(2). On multiplying both the sides of the 1st equation by 7, we get -56X +7Y = 252…(3). Now, nn subtracting the 2nd equation from the 3rd equation, we get -56X +7Y –(X+7Y) = 252-54 or, -57X = 198. Hence X = -198/57. On substituting X = -198/57, in the 1st equation, we get -8*(-198/57) + Y = 36 or, Y = 36 -1584/57 = 468/57. Thus, X = -198/57 and Y = 468/57.
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