A fraction becomes 2 when 1 is added to both the numerator and denominator and it becomes 3 when 1 is subracted from both numerator and denomiator. Find the difference of numerator and denominator.

Options

A. 3
B. 4
C. 6
D. 7

Answer

To solve the problem, we will denote the fraction as \( \frac{x}{y} \), where \( x \) is the numerator and \( y \) is the denominator.

### Step 1: Set up the equations based on the problem statement.
According to the problem:
1. When 1 is added to both the numerator and the denominator, the fraction becomes 2:
\[ \frac{x + 1}{y + 1} = 2 \]
2. When 1 is subtracted from both the numerator and the denominator, the fraction becomes 3:
\[ \frac{x - 1}{y - 1} = 3 \]

### Step 2: Clear the fractions by cross-multiplying.
From the first equation:
\[ x + 1 = 2(y + 1) \implies x + 1 = 2y + 2 \implies x - 2y = 1 \quad \text{(Equation 1)} \]

From the second equation:
\[ x - 1 = 3(y - 1) \implies x - 1 = 3y - 3 \implies x - 3y = -2 \quad \text{(Equation 2)} \]

### Step 3: Solve the system of equations.
We now have the following system of equations:
1. \( x - 2y = 1 \) (Equation 1)
2. \( x - 3y = -2 \) (Equation 2)

To eliminate \( x \), we can subtract Equation 1 from Equation 2:
\[ (x - 3y) - (x - 2y) = -2 - 1 \]
This simplifies to:
\[ -3y + 2y = -3 \implies -y = -3 \implies y = 3 \]

### Step 4: Substitute \( y \) back to find \( x \).
Now that we have \( y = 3 \), we can substitute it back into Equation 1:
\[ x - 2(3) = 1 \implies x - 6 = 1 \implies x = 7 \]

### Step 5: Find the difference between the numerator and the denominator.
Now we have:
- \( x = 7 \)
- \( y = 3 \)

The difference between the numerator and the denominator is:
\[ x - y = 7 - 3 = 4 \]

### Final Answer
The difference between the numerator and the denominator is \( 4 \).

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