Two ingots, one containing 35% silver and the other 65%, are melted to produce a 20 gram

Two ingots, one containing 35% silver and the other 65%, are melted to produce a 20 gram bar containing 47% silver. What is the mass of each of these bars?

We introduce the variables x, y and denote so the masses of the first and second ingots.
According to the condition of the problem, the parts of silver in each ingot are known:
x * 35% = 0.35x (g) – silver in the first ingot;
y * 65% = 0.65y (g) – silver in the second ingot.
20 * 47% = 9.4 (g) – silver in a new ingot.
We compose a system of two equations:
x + y = 20
0.35x + 0.65y = 9.4
Multiply the first equation by (-0.35)
-0.35x – 0.35y = -7
0.35x + 0.65y = 9.4
0.3y = 2.4
y = 8 (g) is the mass of the second ingot.
x = 20 – y = 12 (g) is the mass of the first ingot.
Answer: 12 g and 8 g of bar mass.