Question: A number exceeds another number by 3 and their product is 40. Find the numbers.

Solution:

According to the question,

Condition I,
A number exceeds another number by 3. Let the greater of the two numbers be x. Then, the other number is (x -3).

Condition II,
The product of those numbers is 40.
$or, x (x -3) = 40$
$or, x^2 -3x = 40$
$or, x^2 -3x -40 = 0$
$or, x^2 -(8-5)x -40 = 0$
$or, x^2 -8x +5x -40 = 0$
$or, x(x -8) +5(x -8) = 0$
$or, (x +5)(x-8) = 0$

Either,
$x +5 = 0$
$\therefore x = -5$

Or,
$x -8 = 0$
$\therefore x = 8$

So, when the greater number x=-5, then the other number is (-5-3) = -8.

And, when the greater number x = 8, then the other number is (8-4) = 5.

Hence, the required two numbers are 5 and 8 or -8 and -5.

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