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Question
Two years ago, a father was five times as old as his son. Two year later, his age will be 8 more than three times the age of the son. Find the present ages of father and son.
Solution
Let the present age of father be x years and the present age of his son be y years.
After 2 years, father’s age will be (x+2) years and the age of son will be (y+2) years. Thus using the given information, we have
`x+2 =3(y+2)+8`
`⇒ x+2 =3y +6+8`
`⇒ x-3y-12=0`
Before 2 years, the age of father was (x-2) years and the age of son was (y-2) years. Thus using the given information, we have
`x-2=5(y-2)`
`⇒ x-2 =5y-10`
`⇒ x-5y+8=0`
So, we have two equations
`x-3y-12=0`
`x-5y+8=0`
Here x and y are unknowns. We have to solve the above equations for x and y.
By using cross-multiplication, we have
`x/((-3)xx8-(-5)xx-12)=(-y)/(1xx8-1xx(-12))=1/(1xx(-5)-1xx(-3))`
`⇒ x/(-24-60)=(-y)/(8+12)=1/(-5+3)`
`⇒ x/(-84)=(-y)/20=1/(-2)`
`⇒ x/84=y/20=1/2`
`⇒ x= 84/2,y=20/2`
`⇒ x=42,y=10`
Hence, the present age of father is 42 years and the present age of son is= 10 years.
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