Question 996439
Let x be the original number, a be the hundreds' digit, b be the tenths' and c be the ones. 
Form 4 equations from the question. 
Equation 1:100a+10b+c=x
Equation 2:a+b+c=14
Equation 3:c+b=a+12
Equation 4:14+20b=x
1. Find the similarity of these 4 equations. You could realize how equation 2 and 3 is similar
(a+b+c=14)
↓
(c+b=14-a)
c+b=a+12
Both of them are equivalent to c+b, which means 14-a=a+12. Here you got a linear equation. 
2. Solve for a. 
14-a=a+12
-a-a=12-14
-2a=-2
a=1
3. Substitute a=1 to equation 3. 
c+b=12+a
c+b=12+1
c+b=13
4. Choose either to solve b or c first. I personally choose to solve for c. So I can conclude that, 
b=13-c
5. Substitute b=13-c into equation 1 and 4. Now you got another similarity since both equations equal x. 
x=100+10(13-c)+c
x= 14+20(13-c)
6. Now you can find c. 
100+10(13-c)+c=14+20(13-c)
100+130-10c+c=14+260-20c
230-9c=274-20c
20c-9c=274-230
11c=44
c=4
7. Now you can find b. 
a+b+c=14
5+b=14
b=9
8. Now you can see
a=1
b=9
c=4
9. So now you get the final answer. 
x=194