Question

two years ago father was three times as old as his son and two years hence twice his age will be equal to five times that of his son find their present age (solve in one variable)​

Answer 1

Answer:

father's present age = 38

son's present age = 14

Step-by-step explanation:

let father's age be x

and son's age be y

two years ago father age x - 2

two years ago son's age y - 2

x - 2 = 3(y - 2)

x - 2 = 3y - 6

x - 3y = - 4 equ (1)

multiply with 2

2x - 6y = - 8 equ (2)

two years hence father age x + 2

two years hence son's age y + 2

2(x + 2) = 5(y + 2)

2x + 4 = 5y + 10

2x - 5y = 6 equ (3)

equ (2) - (3)

y = 14

substitute y in equ (1)

x - 3y = - 4

x - 3×14 = - 4

x - 42 = - 4

x = - 4 + 42

x = 36

Answer 2

Answer:

Hope it helps you.....