Question

2 yrs ago father was 5 times the age of his son. 2 yrs later the father is 8 more than 3 times the age of his son. Find their present ages ?

Answer 1

Answer:

let x be the age of the father

and

y be the age of son

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

x + 2 = 3(y + 2) +8...eq 2

then solve it by subsitution , elimination or cross multiplication method

Answer 2

Step-by-step explanation:

let the father present age be x and son's be y

then 2 yr ago

father=(x-2)

son=(y_2)

according to question x-2=5(y-2).........i

after 2 yr

father=(x+2)

son=(y+2)

according to question x+2=3(y+2)+8.........ii

x-2=5y-10

x-5y=-8.......iii

for eq(ii)

x+2=3y+6+8

x+2=3y+14

x-3y=12.......iv

sbutract eq iii & iv

we get -2y= -20

2y=20

y=10

putting value of y in iv eq

x-3(10)=12

x-30=12

x=12+30

x=42

father is 42 and his son is 10

...........hope it's clear