Let "x" be an amount of the first  alloy to be mixed (in pounds), 
and "y" be an amount of the second alloy to be mixed (in pounds).

So, the "total mass equation" is 

x + y = 30.    (1)

The first alloy contains 2+3 = 5 parts of equal masses in total.
Of them, 2 parts is iron and 3 parts is silver.
So, the mass "x" of the first alloy contains  = 0.6x of silver.


The second alloy contains 7+3 = 10 parts of equal masses in total.
Of them, 7 parts is iron and 3 parts is silver.
So, the mass "y" of the second alloy contains  = 0.3y of silver.

Now, the "silver contents equation" is

0.6x + 0.3y = 0.5*30.   (2)

0.5 on the right side means "half of mass" of the combined alloy.

Let's simplify equations (1) and (2) and write it together as a system.

   x +    y = 30.    (1)
0.6x + 0.3y = 15.    (2)

To solve it, express x = 30-y from (1) and substitute it into (2). You will get a single equation for y:

0.6*(30-y) + 0.3y = 15.

Simplify and solve it:

18 - 0.6y + 0.3y = 15,

3 = 0.3y,

y = 10.

So, 10 pounds of the 2-nd alloy should be taken, and, hence, 
the amount of the 1-st alloy is 30 - 10 = 20 pounds.

Check:  20 pounds of the first alloy contain 3 parts of silver of the total 5 parts, i.e. 12 pounds of silver.

        10 pounds of the second alloy contain 3 parts of silver of the total 10 parts, i.e. 3 pounds of silver.

        12+3 = 15 pounds of silver in the combined/final alloy, which is precisely half of its mass.

The problem is solved.