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The problem from the image states:

**b.** There are two numbers. The first number minus the second number is 15. One-third of the sum of the numbers is one-quarter of the first number. What are the two numbers?

Let's define:
- The first number as \( x \)
- The second number as \( y \)

We are given two conditions:
1. The first number minus the second number is 15:  
   \[
   x - y = 15 \quad \text{(Equation 1)}
   \]
2. One-third of the sum of the numbers is one-quarter of the first number:  
   \[
   \frac{1}{3}(x + y) = \frac{1}{4}x \quad \text{(Equation 2)}
   \]

### Solving the system:

From **Equation 1**, we can solve for \( y \):
\[
y = x - 15
\]

Substitute this into **Equation 2**:
\[
\frac{1}{3}(x + (x - 15)) = \frac{1}{4}x
\]
Simplify the left side:
\[
\frac{1}{3}(2x - 15) = \frac{1}{4}x
\]
Multiply both sides by 12 to eliminate the denominators:
\[
12 \times \frac{1}{3}(2x - 15) = 12 \times \frac{1}{4}x
\]
\[
4(2x - 15) = 3x
\]
Expand and solve:
\[
8x - 60 = 3x
\]
\[
8x - 3x = 60
\]
\[
5x = 60
\]
\[
x = 12
\]

Now, substitute \( x = 12 \) back into **Equation 1**:
\[
12 - y = 15
\]
\[
y = 12 - 15 = -3
\]

Thus, the two numbers are **12** and **-3**.

Would you like more details or have any questions?

### Related Questions:
1. How would the solution change if the sum condition were different?
2. Can you graphically represent the two equations to find the intersection?
3. What would happen if the difference between the numbers was another value, like 20?
4. Can you create another riddle using similar relationships between two numbers?
5. How would you set up the problem if the sum of the numbers was twice the first number?

**Tip:** Always check your solutions by substituting the values back into the original equations.

There are two numbers. The first number minus the second number is 15. One-third of the sum of the numbers is one-quarter of the first number. What are the two numbers?

Solve the problem involving two unknowns where the difference between the numbers is 15, and one-third of their sum is one-quarter of the first number. This algebraic challenge uses systems of equations and is suitable for Grades 7-9.

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