Question

5 years ago a man's age was 5 times the age of his daughter's. 3 years hence, twice his age will be equal to 6 times his daughter's age. Find their present ages.​

Answer 1

...x= father's age

y = daughter's age

5 years ago, the age of the father was 5 times that of his daughter:

x - 5 = 5(y - 5)

In 3 years, twice the age of the father will be equal to 6 times the age of his daughter:

2(x + 3) = 6(y + 3)

system :

x-5=5(y-5)

2(x+3)=6y+18

we calculate

x-5=5y-25

2x+6=6y+18

we write the system correctly

x-5y=-25+5=-20

2x-6y=18-6=12

x-5y=-20

2x-6y=12

by substitution: x=-20+5y

we replace x in the 2nd equation

2(-20+5y)-6y=12

-40+10y-6y=12

4y=12+40

y = 52/4=13

the girl is 13 years old

we calculate x:

x-5y=-20

x-65=-20

x=20+65=45

the father is 45 years old

Explanation:

Answer 2

Let the present age of father and daughter be x and y respectively,

According to the question,

x-5=(y-5)×5

x-5=5y-25

x=5y-20

2(x+3)=6(y+3)

2x-6y=12

Then,

2(5y-20)-6y=12

10y-40-6y=12

4y-40=12

4y=52

y=13

Now,

x=5×13-20

x=65-20

x=45