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SUSTITUCIÓN:
=> 2x - 5y =1 --------- (ec.1)
=>-x + 4y = 4 --------- (ec.2)
Resolviendo:
=> Despejamos a "y" en la (ec.2)
=> 4y = 4 + x
=> y = (4/4) + (1/4)x
=> y = 1 + (1/4)x
Con el valor de "y" se reemplaza en la (ec.1)
=> 2x - 5 ( 1 + (1/4)x = 1
=> 2x - 5 - (5/4)x = 1
=> (3/4)x - 5 = 1
=> (3/4) x = 1 + 5
=> (3/4) x = 6
=> x = 6 / (3/4)
=> x = 8 --------- (1° respuesta)
Con el valor de "x" se reemplaza en:
=> y = 1 + (1/4)x
=> y = 1 + (1/4)(8)
=> y = 1 + 2
=> y = 3 --------- (2° respuesta)
IGUALACIÓN:
=> 2x - 5y = 1 ------- (ec.1)
=>-x + 4y = 4 -------- (ec.2)
Resolviendo:
Igualamos a cada ecuación en "y", tenemos:
=> y = 1 + (1/4)x ...... (ec-1)
=> y = -(1/5) + (2/5)x ...... (ec.2)
Ahora se iguala: y = y, entonces:
=> 1 + (1/4)x = -(1/5) + (2/5)x
=> (1/4)x - (2/5)x = -(1/5) - 1
=> - (3/20) x = - (6/5)
=> x = {-(6/5) / -(3/20)}
=> x = 8 --------------- (1° respuesta)
Con el valor de "x" se reemplaza en cualquiera de las ecuaciones despejadas en "y":
=> y = 1 + (1/4)(8)
=> y = 1 + 2
=> y = 3 -------- (2° respuesta)
Suerte.Superdan3chuli