The sum of numerator and denominator of a fraction is 3 less than twice the denominator. If 
the numerator and denominator both are decreased by 1, then the fraction becomes half. 
Determine the fraction

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.

The sum of numerator and denominator of a fraction is 3 less than twice the denominator. If the numerator and denominator both are decreased by 1, then the fraction becomes half. Determine the fraction.

Solve the algebraic problem where the sum of the numerator and denominator is 3 less than twice the denominator, and decreasing both by 1 results in a fraction of 1/2. This detailed solution uses linear equations and cross-multiplication, perfect for students in Grades 7-9.

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