Question: The difference of the ages of two brothers is 4 years. 5 years ago, if the product of their ages was 96, find their present ages.

Solution:

Let the age of the elder brother before 5 years be x years and that of the younger brother be y years.

According to the question,

Condition I,
The difference of their ages is 4 years.
or, x -y = 4 - (i)

Condition II,
5 years ago, the product of their ages was 96.
xy = 96
or, y = \frac{96}{x} - (ii)

Put value of y from equation (ii) in equation (i), we get,
or, x - \dfrac{96}{x} = 4

or, \dfrac{x² -96}{x} = 4

or, x² -96 = 4x

or, x² -4x -96 = 0

or, x² -(12-8)x -96 = 0

or, x² - 12x +8x -96 = 0

or, x(x -12) +8(x -12) = 0

or, (x +8)(x -12) = 0

Either,
x +8 = 0
\therefore x = -8

Or,
x -12 = 0
\therefore x = 12

Since, years in age is positive. Put value of x = 12 in equation (ii), we get,
or, y = \frac{96}{12}
\therefore y = 8

We know, x and y represents the ages of two brothers before 5 years, so add 5 to each to get their present ages i.e. (x +5) = (12+5) = 17 and (y +5) = (8+5) = 13

Therefore, the required present ages of the two brothers is 17 years and 23 years.