Question

The sum of the 2 digit of a two digit number is 9. if 63 is subtracted from the number, its digits are reversed . find the number

Answer 1

Let us assume, x and y are the two digits of a two-digit number
Therefore, The number = 10x + y and reversed number = 10y + x

Given:
x + y = 9 ---------------1

Also given:
10x + y - 63 = 10y + x
9x - 9y = 63
x - y = 7 -----------------2

Adding equation 1 and equation 2
2x = 16
x = 8

Therefore, y = 9 - x = 9 - 8 = 1
Therefore, The two digit number = 10x + y = 10*8 + 1 = 81

Answer 2

$\huge\bf{Answer:-}$

Let p be the product and n be the number.

Product × Number = 18 Equation - (1)

$10 + n - 63 = 10n + p$

$9p - 9n - 63 = 0$

$9p - 9n = 63$

$p - n = 7 Equation - (2)$

$p = 7 + n$

Adding values of p for Equation - (1)

$7 + n × n = 18$

n² + 7n = 18

n² + 7n - 18 = 0

n² + 9n - 2n - 18 = 0

$n*n + 9 - 2*n + 9 = 0$

$n + 9*n - 2 = 0$

$n = -9$

$n = 2$

This negative values is not correct so,

2 = number

7 + 2 = 9 is the product

Therefore, product = 9

The Number =

$= 10*9 + 2 = 92$

Therefore, 92 is the two digit number