Question

5. A father is twice as old as his son. 20
years ago, the age of the father was
12 times the age of the son. find the present age of the father?​

Answer 1

Step-by-step explanation:

let son be x years

father = 2x

..

20 years ago

son = x-20

father = 2x-20

..

2x-20=12*(x-20)

2x-20 = 12x -240

add 20 to both sides

2x = 12x-240+20

2x=12x-220

add -12x to both sides

2x-12x= 12x-12x-220

-12x+2x= -240+20

-10x= -22

divide by -10

x= -220/-10

x=22 son's age

Father's age = 2*22 = 44 years

.

hope it helps

have a nice day... dear

Answer 2

Answer:

let's son be X years .

father = 2x - 20

20 years ago

son= x -20

father =2x - 20

12 (x -20)=(y- 20 )

12x - 240 = y - 20

12x - y = 220 2 eq

substitution 1 in 2

12x - 2x = 220

10x = 220

son age = 22

father age =44