Sustituyendo valores de la ec. 3 en al ec. 2: 1/2 b = (1/3*(3/4 b)) + 5 1/2 b = (((1*3)/(3*4))*b) + 5 1/2 b = (3/12 b) + 5 1/2 b = (1/4 b) + 5 b = 2*((1/4 b) + 5) b = (1/2 b) + 10 b - 1/2 b = 10 1/2 b = 10 b = 10*2 b = 20
De la ec. 3: a = 3/4 b a = (3/4) * 20 a = 60/4 a = 15
2k - k = 5
k = 5
Remplazas k
x=20
y=15
1/2 b = (1/3 a) + 5 ec. 2
De la ec. 1:
a = 3/4 b ec. 3
Sustituyendo valores de la ec. 3 en al ec. 2:
1/2 b = (1/3*(3/4 b)) + 5
1/2 b = (((1*3)/(3*4))*b) + 5
1/2 b = (3/12 b) + 5
1/2 b = (1/4 b) + 5
b = 2*((1/4 b) + 5)
b = (1/2 b) + 10
b - 1/2 b = 10
1/2 b = 10
b = 10*2
b = 20
De la ec. 3:
a = 3/4 b
a = (3/4) * 20
a = 60/4
a = 15
El mayor es:
20
Saludos.............