Understanding the Fraction Problem

The question asks us to find the difference between the denominator and the numerator of an original fraction based on two given conditions. Let's break down the problem and set up equations to solve it.

Setting Up Variables for the Fraction

Let the numerator of the original fraction be \( n \) and the denominator be \( d \). The original fraction is therefore \( \frac{n}{d} \).

Translating Conditions into Equations

The first condition states that "The denominator of a fraction is 4 more than the double of its numerator." We can write this as an equation:

\[ d = 2n + 4 \]

This is our first equation.

The second condition states that "When 3 is added to the numerator and 3 is subtracted from the denominator the fraction becomes 2/3." Let's apply these changes to the original fraction and set up the second equation:

New numerator = \( n + 3 \)

New denominator = \( d - 3 \)

The new fraction is \( \frac{n+3}{d-3} \). According to the condition, this new fraction is equal to \( \frac{2}{3} \).

\[ \frac{n+3}{d-3} = \frac{2}{3} \]

This is our second equation.

Solving the System of Equations

Now we have a system of two linear equations:

1. \( d = 2n + 4 \)
2. \( \frac{n+3}{d-3} = \frac{2}{3} \)

We can substitute the expression for \( d \) from the first equation into the second equation to eliminate \( d \).

Substitute \( d = 2n + 4 \) into the second equation:

\[ \frac{n+3}{(2n + 4) - 3} = \frac{2}{3} \]

Simplify the denominator:

\[ \frac{n+3}{2n + 1} = \frac{2}{3} \]

Now, we can cross-multiply to solve for \( n \):

\[ 3 \times (n+3) = 2 \times (2n+1) \]

Distribute the numbers on both sides:

\[ 3n + 9 = 4n + 2 \]

To isolate \( n \), subtract \( 3n \) from both sides:

\[ 9 = 4n - 3n + 2 \]

\[ 9 = n + 2 \]

Subtract 2 from both sides:

\[ 9 - 2 = n \]

\[ n = 7 \]

Finding the Original Denominator

Now that we have the value of the numerator \( n = 7 \), we can use the first equation \( d = 2n + 4 \) to find the original denominator \( d \).

Substitute \( n = 7 \) into the first equation:

\[ d = 2 \times 7 + 4 \]

\[ d = 14 + 4 \]

\[ d = 18 \]

So, the original numerator is 7 and the original denominator is 18. The original fraction is \( \frac{7}{18} \).

Calculating the Required Difference

The question asks for the difference between the denominator and the numerator of the original fraction. This difference is \( d - n \).

Difference = \( 18 - 7 \)

Difference = \( 11 \)

Thus, the difference between the denominator and numerator of the original fraction is 11.

Verifying the Solution

Let's check if the original fraction \( \frac{7}{18} \) satisfies the second condition. Add 3 to the numerator: \( 7 + 3 = 10 \). Subtract 3 from the denominator: \( 18 - 3 = 15 \). The new fraction is \( \frac{10}{15} \). Simplifying this fraction, we get \( \frac{10 \div 5}{15 \div 5} = \frac{2}{3} \). This matches the condition, so our original fraction is correct.

Revision Table: Solving Fraction Word Problems

Additional Information: Types of Equations

In this problem, we dealt with linear equations. Let's look at what that means:

• A linear equation is an equation where the highest power of the variable is 1. For example, \( 2x + 3 = 7 \) is a linear equation.
• A system of linear equations involves two or more linear equations with the same set of variables. The solution to the system is the set of values for the variables that satisfies all equations simultaneously.
• The first equation \( d = 2n + 4 \) is linear because both \( n \) and \( d \) have a power of 1.
• The second equation \( \frac{n+3}{d-3} = \frac{2}{3} \) becomes linear when we cross-multiply, resulting in \( 3n + 9 = 2d - 6 \), which can be rearranged into the form \( 3n - 2d = -15 \), another linear equation.

Understanding how to form and solve systems of linear equations is fundamental to solving many word problems involving relationships between unknown quantities.