The condition gives you these three equations for three unknowns n, d and q

 n +   d +   q = 40,       (1)
5n + 10d + 25q = 405,      (2)
 n             = d + 7     (3)

where n = number of nickels, d = number of dimes, q = number of quarters.

To solve it, first reduce (2) by factor (5). You will get (2') instead of (2)

 n +  2d +  5q = 81        (2')

Next, substitute the expression (3) for n into (1) and (2'). You will get

 (d + 7) +  d +  q = 40,   (3)
 (d + 7) + 2d + 5q = 81.   (4)

Simplify (3) and (4):

 2d +  q = 33,             (5)
 3d + 5q = 74.             (6)

Multiply (5) by (-3) (both sides); multiply (6) by 2 and then add. In this way, you will exclude d and will get the single equation for q:

 10q - 3q = 2*74 - 3*33 = 148 - 99 = 49.

It gives

  7q = 49,
   q =  = 7.

Now from (5)

   d = (33 - 7)/2 = 13,

and from (3)

   n = 20.

Answer. n = 20, d = 13, q = 7.