Let x and y be the volumes (in liters, L) to be mixed.


Then one equation is "the volume" equation 

x + y = 60    (liters)     (1)

for the total volume.


The amount of "x" liters of the 2% solution contains 0.02x liters of pure alcohol.

The amount of "y" liters of the 6% solution contains 0.06y liters of pure alcohol.


Thus the mixture contains  0.02x + 0.06y liters of pure alcohol.

From the other side, the mixture contains 0.05*60 liters of pure alcohol.


It gives you the second equation, "pure alcohol contents equation"

0.02x + 0.06y = 0.05*60   (liters of pure alcohol)   (2)


Combining equations (1) and (2), you have this system of two linear equations for two unknowns

    x +     y = 60,     (3)
0.02x + 0.06y =  3.     (4)


Multiply equation (4) by 100 (both sides). You will get an equivalent system in the form

 x +  y =  60,          (5)
2x + 6y = 300           (6)

To solve it, express x = 60 - y from (5) and substitute into (6) replacing x. You will get

2(60-y) + 6y = 300,

120 - 2y + 6y = 300,

4y = 300 - 120   ====>  4y = 180  ====>  y =  = 45.


Answer.  45 liters of the 6% alcohol solution must be mixed with 60-45 = 15 liters of the 2% alcohol solution.