I'll start out being a smarty pants!  

The problem doesn't say Jack doesn't have other coins.  
So he could have had

 3 quarters, 2 dimes and 6 nickels.

Or he could have had

 6 dimes, 10 nickels and 15 pennies.

Either way he'd have $1.25 and 4 more nickels than dimes.

The problem just said he had 4 more nickels than dimes, and the
total money was $1.25.

So he could have even had a quarter, 5 dimes, 9 nickels and 5 pennies.

All those are solutions, and there are a number of others,
since the problems didn't say he didn't have anything but dimes 
and nickels.

But that was just to prove a point!  Now I'll do it assuming there
were no other coins but just dimes and nickels.

Let x = the number of dimes 
Then x+4 = the number of nickels

Then

$.05 times the number of nickels + $.10 times the number of dimes = $1.25

            .05x + .10(x+4) = 1.25

Multiply through by 100:

              5x +  10(x+4) = 125

              5x + 10x + 40 = 125

                   15x + 40 = 125
  
                        15x = 105

                          x = 7 dimes

                        x+4 = 11 nickels.

Now that's probably the answer your teacher wanted.  But
as the problem is stated Jack could have NO dimes at all,
just 4 nickels and 105 pennies!  After all 4 nickels is 4 more 
than 0 dimes. J

Edwin