Question

in two digit number the sum of the digit is 9. if the number with the order of the digit reserved is 28 greater than twice the unit digit of the original number. find the number.

Answer 1

First, lets figure it out by "looking": 

2 digits add to 11, so they must be one of the following 4 possibilities:
29: 2+9 --> reversed is 92
38: 3+8 --> reversed is 83
47: 4+7 --> reversed is 74
56: 5+6 --> reversed is 65 

and twice 38 is 76. Then add 7 is 83. So the answer is 38. 

Right then...mathematically: 

Let the first number be xy and we are told that (x+y) = 11. 

Also, x is the number of tens and y is the number of units. By this we mean that xy really means (10x+y). 

If we reverse it, to yx, when then have the number (10y+x). 

We are told that "reversed number is twice original + 7", so we need to convert this English to maths: 

(reversed number) = twice(original number) + 7
(10y+x) = 2(10x+y) + 7
10y+x = 20x+2y + 7
8y-19x = 7 


So we have 2 equations now:
x+y=11
8y-19x = 7 

scale up the first one by 8 --> 8x+8y = 88, giving us
8x+8y = 88
8y-19x = 7 

Re-write the order:
8y+8x = 88
8y-19x = 7 

Now lets subtract the equations, to get rid of the y terms.
27x = 81
x = 81/27
x = 3 

so from x+y=11 we have
3+y=11
--> y=8 

So the original number, xy is 38. 

jon.