Question

Two year ago a father has five times as old as his son. two years layer his age will be 8 more than three timesthe age of son . find the present age of both

Answer 1

Let present age of son be x years

and present age of father be y years

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2 years ago ;

son's age = x - 2

father's age = y - 2

but a/q ;

y - 2 = 5 ( x - 2 ) = 5x - 10

or, 5x - y = 8 _______________(i)

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Two years later ;

son's age = x + 2

father's age = y + 2

but a/q ;

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

or, y + 2 = 8 + 3x + 6 = 3x + 14

or, 3x - y = -12 ________________(ii)

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Subtracting eqn (ii) from eqn (i) ; we get

2x = 20

or, x = 20 / 2 = 10 (son's age)

y = 3x + 12 = 3×10 + 12 = 42 (father's age)

Answer 2

Hi friend,

# Age of father is 42 and son is 10

➡Solution :-

Let, the age of father is X and the age of son is Y

Let's Start :-

➡Given in the question ;-

x-2=5(y-2)

Or, x-5y= -10+2

Or, x-5y = -8

Or, x = 5y-8

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

Or, x-3y = 6+8-2

Or, 5y-8-3y = 12

Or, 2y = 12 + 8

Or , y = 20/2
Or, y= 10

➡Therefore, X = 5y-8...............(ii)
Or, x = 50-8

Or, x = 42

So, the age of father is 42 and the age of son is 10

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Thanks !!

Be brainly !!!✌