Find a fraction which becomes `(1)/(2)` when the denominatot is increased by 4 , and `(1)/(8)` when the numerator is decreased by 5.

Answer

To find the fraction that meets the conditions given in the problem, let's 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 conditions given.

1. According to the first condition, when the denominator is increased by 4, the fraction becomes \( \frac{1}{2} \):
\[ \frac{x}{y + 4} = \frac{1}{2} \]
Cross-multiplying gives:
\[ 2x = y + 4 \]
Rearranging this, we get our first equation:
\[ 2x - y - 4 = 0 \quad \text{(Equation 1)} \]

2. According to the second condition, when the numerator is decreased by 5, the fraction becomes \( \frac{1}{8} \):
\[ \frac{x - 5}{y} = \frac{1}{8} \]
Cross-multiplying gives:
\[ 8(x - 5) = y \]
Simplifying this, we have:
\[ 8x - y - 40 = 0 \quad \text{(Equation 2)} \]

### Step 2: Solve the system of equations.

Now we have two equations:
1. \( 2x - y - 4 = 0 \) (Equation 1)
2. \( 8x - y - 40 = 0 \) (Equation 2)

We can subtract Equation 1 from Equation 2 to eliminate \( y \):
\[ (8x - y - 40) - (2x - y - 4) = 0 \]
This simplifies to:
\[ 8x - y - 40 - 2x + y + 4 = 0 \]
Combining like terms gives:
\[ 6x - 36 = 0 \]
So, we can solve for \( x \):
\[ 6x = 36 \implies x = 6 \]

### Step 3: Substitute \( x \) back to find \( y \).

Now that we have \( x = 6 \), we can substitute this value back into Equation 1 to find \( y \):
\[ 2(6) - y - 4 = 0 \]
This simplifies to:
\[ 12 - y - 4 = 0 \implies 8 - y = 0 \implies y = 8 \]

### Step 4: Write the final fraction.

Now we have both \( x \) and \( y \):
\[ x = 6, \quad y = 8 \]
Thus, the fraction is:
\[ \frac{x}{y} = \frac{6}{8} = \frac{3}{4} \]

### Final Answer:
The required fraction is \( \frac{3}{4} \).

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