Question

1. Two years ago, a father was five times as old as his son. Two years later, his age will
be 8 years more than three times the age of the son. Find the present ages of father
and son.​

Answer 1

Step-by-step explanation:

let the age of father be x

let the of son be y

two years ago,

x-2=5(y-2)

x-2=5y-10

×-5y= -8 -1

two years later,

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

x+2=8+3y+6

x+2=14+3y

×-3y=12 -2

subtract equation 2 fom 1

-2y=-20

2y=20

y=20÷2

y=10

x-2=5y-10

×-2=5(10)-10

×-2=50-10

×-2=40

×=42

HOPE IT WILL HELP YOU.