Question

Two years ago a father was five times as old his son 2 years later his age will be 8 more than 3 times the age of his son . Find the present ages

Answer 1

Answer:

Present age of father = 42 years

Present age of his son = 10 years

Step-by-step explanation:

given that,

Two years ago

a father was five times as old his son

let the age of father be x

and his son be y

now,

x - 2 = 5(y - 2)

x - 2 = 5y - 10

x - 5y = - 10 + 2

x - 5y = -8. ....(1)

also given that,

2 years later

his age will be 8 more than 3 times

the age of his son

here,

x + 2 = 3(y + 2) + 8

x + 2 = 3y + 6 + 8

x - 3y = 14 - 2

x - 3y = 12. ....(2)

now we have,

x - 5y = -8. ....(1)

x - 3y = 12. ....(2)

substracting (2) from (1)

x - 5y - (x - 3y) = -8 - 12

x - 5y - x + 3y = -20

-2y = -20

y = -20/-2

y = 10

from (1)

x - 5y = -8

x - 5(10) = -8

x - 50 = -8

x = -8 + 50

x = 42

so

Present age of father = 42 years

Present age of his son 10 years

Answer 2

Answer .

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