Question: A number consists of two digits. If the numbers formed by reversing its digits is added to it, the sum is 143 and if the same number is subtracted from it the remainder is 9. Find the number.

Solution:

Let the digit at tens place of a two digit number be x and that at ones place be y. So, the required number is 10x +y.

According to the question,

Condition I,
If the number formed by reversing its digits is added to it, the sum is 143.
(10x +y) + (10y +x) = 143
or, 10x + y +10y +x = 143
or, 11x + 11y = 143
or, 11x = 143 - 11y
or, x = \frac{143 -11y}{11} - (i)

Condition II,
If the number formed by reversing its digits us subtracted from the original number, the remainder is 9.
or, (10x +y) - (10y +x) = 9
or, 10x +y -10y -x = 9
or, 9x -9y = 9
or, 9(x -y) = 9×1
or, x -y = 1 - (ii)

Put value of x from equation (i) in equation (ii), we get,

or, \dfrac{143-11y}{11} - y =1

or, \dfrac{143-11y}{11} = 1-y

or, 143 - 11y = 11(1-y)

or, 143 - 11y = 11 - 11y

or, 143 - 11 = 11y +11y

or,  132 = 22y

or, y = \dfrac{132}{22}

\therefore y = 6

Put value of y in equation (i), we get,

or, x = \dfrac{143 - 11× 6}{11}

or, x = \dfrac{77}{11}

\therefore x = 7

So, (x,y) = (7,6)
10x = 10×7 = 70
10x + y = 70+6 = 76

Therefore, the required two digit number is 76.