Question 697600
If the digits of a two-digit number, 10t+u,
are interchanged, 
<pre>
That's 10u+t
</pre>
the result exceeds the original number
<pre>
That means 10u+t = 10t+u PLUS something
</pre>
by 
<pre>
Now we're about hear what we are supposed to PLUS onto that
</pre>
2 more than the sum of the digits.  
<pre>
Well, the sum of the digits is t+u, and 2 more
than that is t+u+2.
<pre>
So we have that 10u+t = 10t+u PLUS t+u+2

or

10u+t = 10t+u + t+u+2

simplify:

10u+t = 11t+2u+2

   8u = 10t+2

Divide through by 2

   4u = 5t+1

</pre>
The digits differ by 2.  
<pre>
Oh oh! That doesn't tell us which way to subtract them.

It could be t-u = 2 or u-t = 2

So we have two possible systems of equations:

{{{system(4u=5t+1,t-u=2)}}}or{{{system(4u=5t+1,u-t=2)}}}

Solve each one by substitution.

One of them comes out t = -9 and u = -11

But that's impossible.  The other one is t=7, u=9

So the number must be 79

Edwin</pre>