Question

The excess of two digit number over the number formed by reversing its digits is 18.if sum of its digits is 4 times the difference of the digits find the number

Answer 1

10a+b -(10b+a)=18
9a-9b=18
a-b=2 . . . be it eqn 1

a+b=4(a-b)=4(2)=8 . . . be it eqn 2

elimination method for eqn 1&2
a+b=8
a-b=2
2a=10
a=5
so b=3
number is ab = 53

Answer 2

Let the digit at tens place be x and the digit at units place be y.

Original number = 10x + y
Reversed number = 10y + x

By the first given condition ,

10x + y = 10y + x + 18
10x - x + y - 10y = 18
9x - 9y = 18
[Dividing both sides by 9]
x - y = 2.....................................(1)

By the second given condition ,

x + y = 4 (x - y)
x + y = 4(2) .....................from (1)
x + y = 8.......................................(2)

Adding (1) & (2)

x - y = 2
x + y = 8
________
2x = 10
x = 10/2
x = 5

Substitute,
x=2 in equation no.(2)

x + y = 8
5 + y = 8
y = 8 -5
y = 3

so,
x = 5 ; y = 3

Number:
10x + y
=10(5) + (3)
=50 + 3
=53

Therefore,
The required number is 53.

$hope \: this \: helps$