Let the number of liters of the stronger solution be x
Let the number of liters of the weaker solution be y


                      % pure     Liters
 Type      Liters       of         of
  of         of        each       PURE
solution   liquid    solution     HNO3
-------------------------------------------
stronger      x       0.40       0.40x
weaker        y       0.25       0.25y
-------------------------------------------
mixture      58       0.35       0.35(58) = 20.3 liters

 The first equation comes from the liters of liquid column.

  

                 x + y = 58

 The second equation comes from the last column.
  

         0.40x + 0.25y = 20.3

Get rid of decimals by multiplying every term by 100:

             40x + 25y = 2030

 So we have the system of equations:
           .

We solve by substitution.  Solve the first equation for y:

           x + y = 58
               y = 58 - x

Substitute (58 - x) for y in 40x + 25y = 2030

    40x + 25(58 - x) = 2030
    40x + 1450 - 25x = 2030
          15x + 1450 = 2030
                 15x = 580
                   x = 38.6666667 = the number of liters of the stronger.

Substitute in y = 58 - x
              y = 58 - (38.6666667)
              y = 19.3333333 = the number of liters of the weaker.
Edwin