Answer :
To solve for the fraction, let's use variables to represent its numerator and denominator. Let's denote the numerator by [tex]\( n \)[/tex] and the denominator by [tex]\( d \)[/tex].
We are given two conditions:
1. When 1 is subtracted from the numerator, the fraction becomes [tex]\( \frac{1}{3} \)[/tex].
2. When 8 is added to the denominator, the fraction becomes [tex]\( \frac{1}{4} \)[/tex].
### Step 1: Set Up the Equations from the Conditions
First condition: When 1 is subtracted from the numerator:
[tex]\[ \frac{n - 1}{d} = \frac{1}{3} \][/tex]
Cross-multiplying to eliminate the fraction:
[tex]\[ 3(n - 1) = d \quad \Rightarrow \quad 3n - 3 = d \quad \Rightarrow \quad d = 3n - 3 \quad \text{(Equation 1)} \][/tex]
Second condition: When 8 is added to the denominator:
[tex]\[ \frac{n}{d + 8} = \frac{1}{4} \][/tex]
Cross-multiplying to eliminate the fraction:
[tex]\[ 4n = d + 8 \quad \Rightarrow \quad d = 4n - 8 \quad \text{(Equation 2)} \][/tex]
### Step 2: Solve the System of Equations
Now, we have two equations:
1. [tex]\( d = 3n - 3 \)[/tex]
2. [tex]\( d = 4n - 8 \)[/tex]
Since both expressions equal [tex]\( d \)[/tex], we can set them equal to each other:
[tex]\[ 3n - 3 = 4n - 8 \][/tex]
Isolate [tex]\( n \)[/tex]:
[tex]\[ 3n - 4n = -8 + 3 \quad \Rightarrow \quad -n = -5 \quad \Rightarrow \quad n = 5 \][/tex]
Substitute [tex]\( n = 5 \)[/tex] back into one of the original equations to find [tex]\( d \)[/tex]. We can use either equation, but let’s use Equation 1:
[tex]\[ d = 3n - 3 \quad \Rightarrow \quad d = 3(5) - 3 \quad \Rightarrow \quad d = 15 - 3 \quad \Rightarrow \quad d = 12 \][/tex]
### Step 3: Verify the Solution
The fraction is [tex]\( \frac{n}{d} = \frac{5}{12} \)[/tex].
Check the first condition:
[tex]\[ \frac{n - 1}{d} = \frac{5 - 1}{12} = \frac{4}{12} = \frac{1}{3} \][/tex]
This is correct.
Check the second condition:
[tex]\[ \frac{n}{d + 8} = \frac{5}{12 + 8} = \frac{5}{20} = \frac{1}{4} \][/tex]
This is also correct.
### Final Answer
The fraction is:
[tex]\[ \frac{5}{12} \][/tex]
We are given two conditions:
1. When 1 is subtracted from the numerator, the fraction becomes [tex]\( \frac{1}{3} \)[/tex].
2. When 8 is added to the denominator, the fraction becomes [tex]\( \frac{1}{4} \)[/tex].
### Step 1: Set Up the Equations from the Conditions
First condition: When 1 is subtracted from the numerator:
[tex]\[ \frac{n - 1}{d} = \frac{1}{3} \][/tex]
Cross-multiplying to eliminate the fraction:
[tex]\[ 3(n - 1) = d \quad \Rightarrow \quad 3n - 3 = d \quad \Rightarrow \quad d = 3n - 3 \quad \text{(Equation 1)} \][/tex]
Second condition: When 8 is added to the denominator:
[tex]\[ \frac{n}{d + 8} = \frac{1}{4} \][/tex]
Cross-multiplying to eliminate the fraction:
[tex]\[ 4n = d + 8 \quad \Rightarrow \quad d = 4n - 8 \quad \text{(Equation 2)} \][/tex]
### Step 2: Solve the System of Equations
Now, we have two equations:
1. [tex]\( d = 3n - 3 \)[/tex]
2. [tex]\( d = 4n - 8 \)[/tex]
Since both expressions equal [tex]\( d \)[/tex], we can set them equal to each other:
[tex]\[ 3n - 3 = 4n - 8 \][/tex]
Isolate [tex]\( n \)[/tex]:
[tex]\[ 3n - 4n = -8 + 3 \quad \Rightarrow \quad -n = -5 \quad \Rightarrow \quad n = 5 \][/tex]
Substitute [tex]\( n = 5 \)[/tex] back into one of the original equations to find [tex]\( d \)[/tex]. We can use either equation, but let’s use Equation 1:
[tex]\[ d = 3n - 3 \quad \Rightarrow \quad d = 3(5) - 3 \quad \Rightarrow \quad d = 15 - 3 \quad \Rightarrow \quad d = 12 \][/tex]
### Step 3: Verify the Solution
The fraction is [tex]\( \frac{n}{d} = \frac{5}{12} \)[/tex].
Check the first condition:
[tex]\[ \frac{n - 1}{d} = \frac{5 - 1}{12} = \frac{4}{12} = \frac{1}{3} \][/tex]
This is correct.
Check the second condition:
[tex]\[ \frac{n}{d + 8} = \frac{5}{12 + 8} = \frac{5}{20} = \frac{1}{4} \][/tex]
This is also correct.
### Final Answer
The fraction is:
[tex]\[ \frac{5}{12} \][/tex]
