Two types of ornaments are made by mixing gold and copper in different ratios. In the first type, 6 grams of gold is combined with 5 grams of copper. In the second type, 5 grams of gold is combined with 3 grams of copper. If the total amount of gold used is 122 grams and the total amount of copper used is 90 grams, how many ornaments of the second type are made? (Assume all gold and copper are used.)

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**Solution:**

1. Let the number of first type ornaments be *a* and the number of second type ornaments be *b*.

2. Each first type ornament contains 6 grams of gold and 5 grams of copper, so total gold and copper in first type are:
   - Gold: $6a$
   - Copper: $5a$

3. Each second type ornament contains 5 grams of gold and 3 grams of copper, so total gold and copper in second type are:
   - Gold: $5b$
   - Copper: $3b$

4. According to the problem, total gold used is 122 grams and total copper used is 90 grams. Therefore, we have the system of equations:
   $$6a + 5b = 122 \quad 	ext{(gold)}$$
   $$5a + 3b = 90 \quad 	ext{(copper)}$$

5. Multiply the second equation by 6 and the first by 5 to eliminate *a*:
   - $6 	imes (5a + 3b) = 6 	imes 90 \Rightarrow 30a + 18b = 540$
   - $5 	imes (6a + 5b) = 5 	imes 122 \Rightarrow 30a + 25b = 610$

6. Subtract the first new equation from the second:
   $$ (30a + 25b) - (30a + 18b) = 610 - 540 $$
   $$ 7b = 70 $$
   $$ b = 10 $$

7. Substitute $b=10$ into the first equation:
   $$ 6a + 5 	imes 10 = 122 $$
   $$ 6a + 50 = 122 $$
   $$ 6a = 72 $$
   $$ a = 12 $$

8. The question asks for the number of second type ornaments, which is $b = 10$.

**Answer:** 10

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