Let the numerator of the fraction be $x$ and the denominator be $y$. We are given the following two conditions:

1. The sum of the numerator and denominator is 3 less than twice the denominator: $x + y = 2y - 3$ Simplifying this equation: x = y - 3 \tag{1}

2. If the numerator and denominator are both decreased by 1, the fraction becomes $\frac{1}{2}$: $\frac{x - 1}{y - 1} = \frac{1}{2}$ Cross-multiplying this equation: $2(x - 1) = y - 1$ Simplifying: $2x - 2 = y - 1$ 2x - y = 1 \tag{2}

Now, substitute $x = y - 3$ from equation (1) into equation (2):

$2(y - 3) - y = 1$ Simplify: $2y - 6 - y = 1$ $y - 6 = 1$ $y = 7$

Substitute $y = 7$ into equation (1) to find $x$: $x = 7 - 3 = 4$

Thus, the fraction is $\frac{4}{7}$.

Answer:

The fraction is $\frac{4}{7}$.

Do you want further details or have any questions about this solution?

Here are 5 related questions:

1. What happens if the numerator and denominator are increased by 1 instead?
2. How do we solve this type of problem using substitution and elimination methods?
3. What are some real-life applications of fractions and equations like this one?
4. How would the solution change if the fraction became $\frac{2}{3}$ instead of $\frac{1}{2}$?
5. What is the general strategy for solving word problems involving fractions?

Tip: Always start by carefully defining variables and translating word problems into equations for easy solving.