Question

a number of consists of two digits whose sum is 8 if 18 is added to the number it's digits are reversed .Find the number.

Answer 1

LET UNIT DIGIT BE X AND TENCE DIgit be y

x+y=8                               ..............eq1
 
number=10y+x
reversed number=10x+y 
10x+y=18+10y+x
10x+y-10y-x=18
9x-9y=18 
divide both side by 9
x-y=2                    ...........eq2
 add eq1+eq2
x+y+x-y=8+2
2x=10
x=5 
 
now put x=5 in eq 1
x+y=8
5+y=8
y=3

number
35

Answer 2

Step-by-step explanation:

Let the one number be x and another number be 8 - x.

$\underline{ \large \purple{ \mathscr{\dag\:A \bf{ccourding} \: to \: \mathscr {Q} \bf{uestion} ....}}} </p><p>$

10x + (8 - x) + 28 = 10(8 - x) + x

10x + 8 - x + 18 = 80 - 10x + x

9x + 26 = 80 - 9x

9x + 9x = 80 - 26

18x = 54

x = 3

Therefore,

One number = x = 2

Other number = (8 - x) = 8 - 3 = 5