Question

two years ago father was three times old as his son and two years hence twice his age will be equal to five times that of his son find their present ages

Answer 1

Hi there!
Here's the answer:


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Let Present age of Son = x

2 years Ago,
Son's age = x-2


Given, At this time
Father's age = 3× His son's age
=> Father's age = 3(x-2)


Father's Present age = 3(x-2) +2 = 3x-4

Two years hence,
Father's age = (3x-4)+2 = 3x-2
Son's age = x+2


Given that, At this time
2×(Father's age) = 5×(His son's age)
=> 2(3x-2) = 5(x+2)
=> 6x-4= 5x+10
=> x = 14

•°• Son's Present Age = 14 years
& His Father's Age = 3x-4 = 3(14)-4= 42-4 = 38 years


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Answer 2

HEY Buddy......!! here is ur answer
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Let, the present age of father = x years

and the present age of son = y years

2 years ago the age of father and son (x-2) and (y-2) respectively.

And two years hence their ages will be (x+2) and (y+2) respectively.

Then, According to the question....

(x-2) = 3(y-2).....(1)

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

=> x-3y = -4....(3)

=> 2x-5y = 6....(4)

[On Solving equation (3) and (4) by elimination method]

On multiplying by 2 in equation (3) and by 1 in equation (4)

2x-6y = -8....(5)

2x-5y = 6.....(6)

On subtracting....

-y = -14

=> y = 14

On putting the value of 'y' in equation (3)

x-3×14 = -4

=> x-42 = -4

=> x = 38

Thus, the present age of father = x = 38 years

and the present age of son = y = 14 years.
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I hope it will be helpful for you......!!

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