A wallet contains ₱50 and ₱100 bills amounting ₱450. How many of each kind will produce the given amount? 

Let the number of ₱50 bills be denoted by \( x \) and the number of ₱100 bills be denoted by \( y \).

We are given the following conditions:

1. The total amount is ₱450.
   \[
   50x + 100y = 450
   \]
   
2. We know that both \( x \) and \( y \) must be non-negative integers.

To simplify the equation, divide the entire equation by 50:
\[
x + 2y = 9
\]

Now, solve for \( x \) in terms of \( y \):
\[
x = 9 - 2y
\]

Since \( x \) must be non-negative, we have the condition:
\[
9 - 2y \geq 0
\]
\[
y \leq 4.5
\]
Thus, \( y \) can only be 0, 1, 2, or 3 (since \( y \) must be an integer).

Let's check these values of \( y \) to find the corresponding \( x \):

- If \( y = 0 \), \( x = 9 - 2(0) = 9 \).
- If \( y = 1 \), \( x = 9 - 2(1) = 7 \).
- If \( y = 2 \), \( x = 9 - 2(2) = 5 \).
- If \( y = 3 \), \( x = 9 - 2(3) = 3 \).

So, the possible solutions are:

1. 9 ₱50 bills and 0 ₱100 bills.
2. 7 ₱50 bills and 1 ₱100 bill.
3. 5 ₱50 bills and 2 ₱100 bills.
4. 3 ₱50 bills and 3 ₱100 bills.

Would you like more details or have any questions?

Here are some related questions for further exploration:

1. What if the total amount were ₱600 instead of ₱450?
2. How would the solution change if there were also ₱200 bills in the wallet?
3. What if you are only allowed to have an even number of ₱50 bills?
4. How can this problem be solved graphically?
5. How would the solution be affected if we used different currencies?

**Tip:** Always check if the given problem involves integer solutions, as it can limit or expand the number of valid solutions.

A wallet contains ₱50 and ₱100 bills amounting ₱450. How many of each kind will produce the given amount?

Learn how to solve the problem of determining how many ₱50 and ₱100 bills are needed to total ₱450. This solution involves linear equations and introduces Diophantine equations, making it suitable for students in Grades 7-9.

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