Question

the product of there present age of the father and his son is 520 years. after 14 years the age of the father will become double of the age of the son.find their present ages.

Answer 1

Let the present ages of
Father = x
Son =y
Then,
xy=520
x=520÷y. ( Equation 1st. )
After 14 years,
Father's age =x+14
Son's age =y+14
According to question,
(x+14)=2(y+14)
(520÷y+14)=2(y+14). ( From equation 1)
(520+14y÷y) =2y +28
520+14y =y(2y+28)

$520 + 14y = 2y {?}^{2} { + 28y}$
28y - 14y +2y square =520
2y square +14y - 520=0
Dividing by 2,
y square +7y - 260=0
y square +20y - 13y - 260=0
y(y+20)-13(y+20)=0
(y+20)(y-13)=0
y=-20,
y=13.
Hence, son's age =13years
Father's age =520÷y
=520÷13=40 yeras

Answer 2

Answer:

Let the present ages of

Father = x

Son =y

Then,

xy=520

x=520÷y. ( Equation 1st. )

After 14 years,

Father's age =x+14

Son's age =y+14

According to question,

(x+14)=2(y+14)

(520÷y+14)=2(y+14). ( From equation 1)

(520+14y÷y) =2y +28

520+14y =y(2y+28)

28y - 14y +2y square =520

2y square +14y - 520=0

Dividing by 2,

y square +7y - 260=0

y square +20y - 13y - 260=0

y(y+20)-13(y+20)=0

(y+20)(y-13)=0

y=-20,

y=13.

Hence, son's age =13years

Father's age =520÷y

=520÷13=40 yeras

Step-by-step explanation: