Bronze which costs $7.05/kg is made by combining copper which costs $6.20/kg with ton which costs $8.70/kg.  Find the number of kg of copper and tin required to make 5kg of bronze.

To solve this problem, we'll use the concept of a weighted average to determine the required amounts of copper and tin.

Let's denote:
- $ x $ as the mass of copper in kg.
- $ y $ as the mass of tin in kg.

We know the following:
1. The total mass of the bronze is 5 kg, so:
   \[
   x + y = 5
   \]

2. The cost per kg of bronze is a weighted average of the costs of copper and tin. The equation for the total cost can be set up as:
   \[
   6.20x + 8.70y = 7.05 \times 5
   \]
   Simplifying the right-hand side:
   \[
   6.20x + 8.70y = 35.25
   \]

We now have a system of linear equations:
\[
x + y = 5
\]
\[
6.20x + 8.70y = 35.25
\]

### Solving the System of Equations

1. From the first equation, solve for $ y $:
   \[
   y = 5 - x
   \]

2. Substitute this expression for $ y $ into the second equation:
   \[
   6.20x + 8.70(5 - x) = 35.25
   \]

3. Distribute the 8.70:
   \[
   6.20x + 43.50 - 8.70x = 35.25
   \]

4. Combine like terms:
   \[
   -2.50x + 43.50 = 35.25
   \]

5. Isolate $ x $:
   \[
   -2.50x = 35.25 - 43.50
   \]
   \[
   -2.50x = -8.25
   \]
   \[
   x = \frac{-8.25}{-2.50}
   \]
   \[
   x = 3.3 \text{ kg}
   \]

6. Now, find $ y $ using $ y = 5 - x $:
   \[
   y = 5 - 3.3 = 1.7 \text{ kg}
   \]

### Conclusion

To make 5 kg of bronze that costs $7.05/kg:
- You need 3.3 kg of copper.
- You need 1.7 kg of tin.

Would you like any further details or have any questions?

Here are 5 related questions you might find interesting:
1. How would the solution change if the cost of copper was $7/kg?
2. What if the total mass of bronze required was 10 kg instead of 5 kg?
3. How would the solution differ if the desired cost of bronze was $7.50/kg?
4. What if the mixture included a third metal, say zinc, costing $9/kg?
5. How do changes in the prices of copper or tin affect the proportions used in the alloy?

**Tip:** When solving systems of equations, substitution is effective if one equation can easily be solved for one variable, while elimination works well when equations are in a similar form.

Bronze which costs $7.05/kg is made by combining copper which costs $6.20/kg with tin which costs $8.70/kg. Find the number of kg of copper and tin required to make 5kg of bronze.

Learn how to solve a system of linear equations to find the required amounts of copper and tin needed to create 5kg of bronze costing $7.05/kg. This problem involves key algebraic concepts such as weighted averages and is suitable for students in Grades 9-10.

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