one more confusion!pls explain this very nicely becoz i am very bad at solving these kind of questions!

10(a+b) + (a+b) = 143

Rosala, you know that      10x+x=10x+1x=11x       right!

This is because 10 lots of something plus one more lots of that same thing = 11 lots of it

Well in the first line the something is (a+b)

10 lots of (a+b) plus one more lot of (a+b) will equal 11 lots of (a+b)

10(a+b) + (a+b) = 143

10(a+b) + 1(a+b) = 143

(10+1)(a+b) = 143

11(a+b)

Will you be alright now.  I assume that was the main problem.   

The digits of a two-digit number differ by 3 . If the digits are interchanged and the resulting number is added to the original number, we get 143 . What can be he original number ?

digit 1 = a

digit 2 = b

a - b = 3

Number 1   ab    is a*10 + b

Number 2   ba    is b*10 + a

Number 1 + Number 2 =  (a*10 + b) + (b*10 + a) = 143

                                                 10(a+b) + (a+b) = 143

                                                              11(a+b) = 143

                                                                   (a+b) = 143/11 = 13

I. a + b = 13

II.  a - b = 3

----------------

I+II: 2a = 16  or  a = 8

I-II:  2b = 10  or  b = 5

The Number is 85 or 58

heureka , can u please explain this more nicely after this step!

 10(a+b) + (a+b) = 143

 i cant understand it now also!

i think im getting into confusion more n more but lets forget this question as now exam is over!phew!

I've made it easier for you Rosala.

Now it is 10 lots of BLUE  plus 1 lot of BLUE

Try again to work it out or I will be upset.   

Melody i will work it out but can u just tell me how did we reach to that formula or equation becoz thats what im not getting!

I hadn't read the whole thing, I thought you were having problems with the bit that I did.

Yes I can see the set up really is tricky!

I did mine without thinking about Heureka's and it looks a  bit different.   (But it is really the same)

Anyway see if it makes sense.   I have put numbers next to many of the lines so you can say which one you don't understand. 

The digits of a two-digit number differ by 3 . If the digits are interchanged and the resulting number is added to the original number, we get 143 . What can be he original number ?

Heureka let the first digit be a and the second digit be b.

So the original number is     10a+b          (1)

The digits are different by 3      That is  |a-b|=3      I put the absolute signs in because I don't know which digit is the biggest one     (2)

If the digits are interchanged then the new number is  10b+a       (3)

When we add these 2 numbers the answer is 143   SO     10a+b+10b+a=143           (4)

11a+11b=143                             (5)

Lets assume for the moment that a is the biggest digit  so  a-b=3

We have 2 equations that have to be solved simultaneously.          (6)

a-b=3   (i)      this can be rewritten as   a=3+b     (iii)

11a+11b=143    (ii)                                                                    (7)

Sub (iii) into (ii)

11(3+b)+11b=143                                                                    (8)

33+11b+11b=143

33+22b=143

22b=110

b= 5                                                                                      (9)

sub into (i)

a-5=3

a=8                                                                                        (10)

So the original number may have been     85

Now if you can follow all this then  you can see what happens if you let b be the big one.