Question

Five years ago a man was seven times as old as his son. Five years hence, the father will be three times as old as his son. Find their present ages.​

Answer 1

Answer:

let age of man be x

age of son be y

five years ago

x-5=7(y-5)

x-5=7y-35

x-7y= -30 eq 1

five year hence

x+5=3(y+5)

x+5=3y+15

x-3y=10 eq 2

by subtracting eq 1 and 2

we get

x-7y= -30

x-3y= 10

hence we get x=40

y=10

age of man is 40 years

age of son is 10 years

Answer 2

Five years ago a man was seven times as old as his son. Five years hence, the father will be three times as old as his son. Find their present ages.

Solution-.

let ,

Father ' s age => x

His son's age = y

Five years ago

(x - 5) = 7(y - 5 )

x - 5 = 7y - 35

x -7y = - 35+5

x - 7y = 30......(1)

Five years hence ,

(x + 5) = 3( y + 5 )

x + 5 = 3y + 15

x -3y = 15 - 5

x - 3y = 10.........(2)

By elimination method

$x - 7y = - 30 \\ x - 3y = 10 \\ - - - - - - - \\ - \: \: \: \: \: \: \: \: + \: \: \: \: \: \: - \\ - - - - - - - \\ \: \: \: \: \: - 4y = > - 40 \\ \\ y = > \frac{ - 40}{ -4} \\ \\ \boxed{y = > 10} \\ \\ \\ \\ put \: the \: value \: in \: (2) \: equation \\ \\ x - 3y = 10 \\ \\ x - 3(10) = 10 \\ \\ x - 30 = 10 \\ \\ x = > 10 + 30 \\ \\ \boxed{ x = > 40}$

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