While arranging certain students of a school in rows containing equal number of students; if three rows are reduced, then three more students have to be arranged in each of the remaining rows. If three more rows are formed, then two students have to be taken off from each previously arranged rows. Find the number of students arranged.

Let the number of rows be x and number of students in each row = y.

⇒ Number of students arranged = x × y

According to the question,

If three rows are reduced, number of rows = (x – 3)

Then three more students have to be arranged in each row, number of students in each row = (y + 3)

Number of students arranged = (x – 3)(y + 3)

If three more rows are formed, number of rows = (x + 3)

Then two students have to be taken off from each row, number of students in each row = (y – 2)

Number of students arranged = (x + 3)(y – 2)

In above both the cases the number of students arranged will remain same.

⇒ (x – 3)(y + 3) = xy and (x + 3)(y – 2) = xy

⇒ xy + 3x – 3y – 9 = xy

And xy – 2x + 3y – 6 = xy

⇒ 3x – 3y – 9 = 0 and – 2x + 3y – 6 = 0

⇒ x – y – 3 = 0 and – 2x + 3y – 6 = 0

On comparing with a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0we get,

a_{1 =} 1, b₁ = – 1, c₁ = – 3; a₂ = – 2, b₂ = 3, c₂ = – 6

Applying cross multiplication method which says,

Putting the given values in the above equation we get,

Similarly,

∴The solution of the pair of equations is (15, 12).

So, Number of students arranged = x×y = 15× 12 = 180.