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

Let Five years ago the age of son be x years and age of father be 7x years

Present age of son =x+5

Present age of father =7x+5

5 years later their age will (x+10) and (7x+10)

∴7x+10=3(x+10)

7x−3x=20

4x=20

x=5

So, the present age of son=x+5=5+5=10years

and the present age of father=7x+5=35+5=40years

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Answer 2

Answer:

Father age is 40 years and

son age is 10 years

Step-by-step explanation:

Refer the posted page and solve eqs 1 and equ 2

X-7y = -30

X-3y = 10

-4y = -40

y = -40

-4

therefore, y = 10 and x=40

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