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 the original number?

Let the original two-digit number be represented as "10a + b," where 'a' represents the tens digit and 'b' represents the units digit.

According to the problem, the digits differ by 3, so we can write this as:

a - b = 3

Now, when the digits are interchanged, the resulting number is "10b + a." If this number is added to the original number, we get 143. So, we can write the equation:

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

Simplify this equation:

11a + 11b = 143

Now, divide both sides of the equation by 11 to solve for 'a' and 'b':

a + b = 13

Now, we have a system of two equations:

a - b = 3
a + b = 13
We can solve this system of equations simultaneously.

Adding the two equations together eliminates 'b':

(a - b) + (a + b) = 3 + 13
2a = 16

Now, divide both sides by 2 to find 'a':

2a/2 = 16/2
a = 8

Now that we have found 'a' as 8, we can substitute it back into either equation to find 'b.' Let's use the second equation:

a + b = 13
8 + b = 13

Subtract 8 from both sides:

b = 13 - 8
b = 5

So, the tens digit 'a' is 8, and the units digit 'b' is 5. Therefore, the original two-digit number is 85.