Question

two years ago a father was five times as old as his son. 10 years later his age will be 8 more than three times the age of his son. Find their present age

Answer 1

Step-by-step explanation:

Let the of the father be X and his age be y

x-2 = 5(y-2)

X= 5y -10 +2

X = 5y + 8 -------- eq1

case 2

X+10 = 8 +3(y+ 10)

X= 8+ 3y +30 -10

X= 3y +28 ------ eq2

From EQ 1 and 2

5y +8 = 3y + 28

5y -3y = 28-8

2y = 20

y = 10

since X= 5y +8

X = 5(10) +8

X = 58

Answer 2

Answer:

Step-by-step explanation:

Let father age be x, his son age be y.

Before 2 years,

⇒(x−2)=5(y−2)−−−−−−−(1)

After 2 years  

⇒(x+2)=8+3(y+2)−−−−−−−−(2)

From (1) x−5y+8=0

From (2)  

⇒x+2−3y−6−8=0

⇒x−3y−12=0

Solving them  

⇒−2y+20=0

⇒y=10

When y=10,x=42

So the present age of the father is 42 and the son age is 10.

Hope it helps!