***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 his son. Find the ages
of father and son.
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Answer :
Fathers age = 42
Sons age = 10
Step-by-step explanation :
Let father age be x, his son age be y.
Before 2 years,
⇒( x − 2 ) = 5 ( y − 2 )−−−−−−−(1)
After 2 years
⇒( x + 2 ) = 8 + 3 ( y + 2 )−−−−−−−−(2)
From (1) .....
x − 5 y + 8 = 0
From (2)
⇒x + 2 − 3 y − 6 − 8 = 0
⇒x − 3 y − 12 = 0
Solving them
⇒ − 2 y + 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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