Three years ago, the age of a person was six times the age of his younger brother. Eight years later, the age of the person will be three years less than twice the age of his younger brother. Find the present ages of the person and his younger brother.

Question

Accepted Answer

Let the present age of the person be P years and the present age of his younger brother be B years.

Step 1: Translate the first condition:  
Three years ago, the person's age was six times the younger brother's age.  
(P - 3) = 6 * (B - 3)  
=> P - 3 = 6B - 18  
=> P - 6B = -15  ...(1)

Step 2: Translate the second condition:  
Eight years later, the person's age will be three years less than twice the younger brother's age.  
(P + 8) = 2 * (B + 8) - 3  
=> P + 8 = 2B + 16 - 3  
=> P - 2B = 5  ...(2)

Step 3: Solve the system of equations:  
From (1): P - 6B = -15  
From (2): P - 2B = 5

Subtract (2) from (1):  
(P - 6B) - (P - 2B) = -15 - 5  
P - 6B - P + 2B = -20  
-4B = -20  
B = 5

Step 4: Substitute B = 5 into (2):  
P - 2*5 = 5  
P - 10 = 5  
P = 15

Therefore, the present ages are:  
Person = 15 years  
Younger brother = 5 years

Answer

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