Question

7 years ago, a man was 7 times as old as his son was. three years hence, he will be three times as old as his son.find the present ages​

Answer 1

Man: 42 years

Son: 12 years

Step-by-step explanation:

h-7 = 7(n-7). Eq.1

h+3 = 3(n+3). Eq.2

From Eq. 1

h-7 = 7*n - 7*7

h-7 = 7n-49

h = 7n - 49+7

h = 7n - 42. Eq. 3

Replace Eq. 3 on Eq. 2

(7n - 42) + 3 = 3(n+3)

7n - 39 = 3*n + 3*3

7n - 39 = 3n + 9

7n - 3n = 9 + 39

4n = 48

n = 48/4

n = 12

From Eq. 3

h = 7n - 42

h = 7*12 - 42

h = 84 - 42

h = 42

Answer 2

Son's age = 14

Father's age = 56

Step-by-step explanation:

Age of the son = x

Age of the father = 7x

Present ages = x+7 and 7x+7

Future age of the son = x + 14

Father = 7x+14

7x + 14 = 3 (x + 14)

7x + 14 = 3x + 42

4x + 14 = 42

4x = 28

x = 7

7x = 35

x + 7 = 7 + 7 = 14

Son's age = 14

Father's age = 7x + 7 = 7 × 7 + 7 = 56