Fill in the boxes such that the fractions are equivalent: `7square/(15)=7(1)/(square)`

Answer

To solve the problem of filling in the boxes such that the fractions are equivalent, we have the equation:

\[ \frac{7 \text{ (blank)}}{15} = 7 \frac{1}{\text{ (blank)}} \]

Let's denote the first blank as \( x \) and the second blank as \( y \). This gives us the equation:

\[ \frac{7x}{15} = 7 + \frac{1}{y} \]

### Step 1: Simplify the equation
We can simplify the equation by eliminating the common term \( 7 \) from both sides. This means we can focus on the fractions only:

\[ \frac{x}{15} = \frac{1}{y} \]

### Step 2: Cross-multiply
To eliminate the fractions, we can cross-multiply:

\[ x \cdot y = 15 \cdot 1 \]

This simplifies to:

\[ xy = 15 \]

### Step 3: Find pairs of factors
Now, we need to find pairs of integers \( (x, y) \) that multiply to give 15. The pairs of factors of 15 are:

1. \( (1, 15) \)
2. \( (3, 5) \)
3. \( (5, 3) \)
4. \( (15, 1) \)

### Step 4: Write the equivalent fractions
Using the pairs found, we can fill in the blanks:

1. For \( x = 1 \) and \( y = 15 \):
\[ \frac{7 \cdot 1}{15} = 7 \frac{1}{15} \]

2. For \( x = 3 \) and \( y = 5 \):
\[ \frac{7 \cdot 3}{15} = 7 \frac{1}{5} \]

3. For \( x = 5 \) and \( y = 3 \):
\[ \frac{7 \cdot 5}{15} = 7 \frac{1}{3} \]

4. For \( x = 15 \) and \( y = 1 \):
\[ \frac{7 \cdot 15}{15} = 7 \frac{1}{1} \quad \text{(which is just 8)} \]

### Conclusion
Thus, the possible pairs to fill in the blanks are:

1. \( (1, 15) \)
2. \( (3, 5) \)
3. \( (5, 3) \)
4. \( (15, 1) \)