5. There are two lead-tin alloys. In the first alloy, the mass of lead is to the mass of tin as $1: 2$; in the second - as $2: 3$. How many grams of the first alloy are needed to obtain 22 g of a new alloy with the mass ratio of lead to tin being 4:7?
Solution. Let the first alloy contain x g of lead and 2x g of tin. In the second alloy, there are 2y g of lead and 3y g of tin. Then $k \cdot 3x + n \cdot 5y = 22 ; \frac{kx + n \cdot 2y}{k \cdot 2x + n \cdot 3y} = \frac{4}{7} ; \quad$ we need to find $k \cdot 3x$ and $5ny$. Let $ny = b ; kx = a \cdot \frac{a + 2b}{2a + 3b} = \frac{4}{7} ; 7a + 14b = 8a + 12b ; 2b = a$. Using the first equation $3a + 5b = 22 ; 11b = 22 ; b = 2 ; a = 4$. Therefore, $k \cdot 3x = 12 ; 5ny = 10$.

Answer: 12.