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Question

Two years ago, a father was five times as old as his son. Two years later, his age will be 8 more than three times the age of the son. Find the present ages of father and son.

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Solution

Let father age be x, his son age be y.
Before 2 years,

(x2)=5(y2)(1)

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

From (1) x5y+8=0

From (2)
x+23y68=0

x3y12=0

Solving them
2y+20=0

y=10

When y=10,x=42

So the present age of the father is 42 and the son age is 10.

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