Question

Two years ago, a man 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 their
present ages.


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


Please solve. ​

Answer 1

Step-by-step explanation:

Let the present age of father = x

and son = y

Then 2 years ago

ie, (x - 2) = 5(y - 2)

ie, x - 2= 5y-10

x-5y = -8 (eq1)

Then 2years after

ie, (x+2)=8+3(y+2)

ie, x-3y =12 (eq2)

By elimination method

subtract eq2 from eq1

ie, -2y=-20

y=-20/-2

y=10

substitute y=10 on eq2

x-3×10=12

x=12+30

x=42

ie age of father is 42

and age of son is 10

HOPE YOU WILL UNDERSTAND THIS

Answer 2

let mans present age be x.

... sons ... y.

2years ago:

x-2=5(y-2) . (1)

2years hence:

x+2=x+3(y+2) (2)

from eq. (1) x=5y-8

substitute the value x in eq.(2)

5y-x+2 = x+3y+6

2y=20

y = 10

so x= 15-8=42

present age of man 42 years.

...... .... son 10 years.

HOPE IT HELPS U...✌❤