Problem:
Two years ago, a father's age was three times that of his son. Two years from now, twice the father's age will be equal to five times that of his son. Find their present ages.

Solution:

Let's assume the present age of the son as x years and the present age of the father as y years.

Statement 1: Two years ago, the father's age was three times that of his son.

Two years ago, the son's age was (x - 2) years and the father's age was (y - 2) years.

According to the problem, the father's age was three times that of his son:

(y - 2) = 3(x - 2)

Statement 2: Two years hence, twice the father's age will be equal to five times that of his son.

Two years from now, the son's age will be (x + 2) years and the father's age will be (y + 2) years.

According to the problem, twice the father's age will be equal to five times that of his son:

2(y + 2) = 5(x + 2)

Simplifying the above equation, we get:

2y + 4 = 5x + 10

Solving the Equations:

We now have a system of two equations with two variables. We can solve these equations simultaneously to find the values of x and y.

Step 1: Solve the first equation for y in terms of x:

y - 2 = 3x - 6
y = 3x - 4

Step 2: Substitute the value of y in the second equation:

2(3x - 4) + 4 = 5x + 10
6x - 8 + 4 = 5x + 10
6x - 4 = 5x + 10
x = 14

Step 3: Substitute the value of x in the first equation to find y:

y = 3(14) - 4
y = 38

Conclusion:
Hence, the present age of the son is 14 years and the present age of the father is 38 years.

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