Question

Ten years ago a father was five time as old as son two year later his age will be eight more than three times the age of the son find the present age of both

Answer 1

Let father's age be= x and son's age = y

So, 10 years ago,

Father's age - 10 = 5 × son's age

》x - 10 = 5(y - 10)

》x = 5y - 50 + 10

》x = 5y - 40

2 years later,

x + 2 = 3y + 8

substituting value of x

5y - 40 + 2 = 3y + 8

》5y - 3y = 40 - 2 + 8

》2y = 46

》y = 46÷2

》y = 23 years

so,

age of father = 5 × 23 - 40

= 115 - 40

= 75 years

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