Let the fraction be represented as $\frac{x}{y}$, where $x$ is the numerator and $y$ is the denominator.

Step 1: Set up the equations

1. When 9 is added to the numerator, the fraction becomes $\frac{2}{3}$: $\frac{x + 9}{y} = \frac{2}{3}$ Cross-multiply to clear the fraction: $3(x + 9) = 2y$ Expanding this, we get: $3x + 27 = 2y$ This simplifies to: 3x - 2y = -27 \tag{Equation 1}

2. Half the sum of the numerator and denominator is 28: $\frac{x + y}{2} = 28$ Multiply both sides by 2: x + y = 56 \tag{Equation 2}

Step 2: Solve the system of equations

Now we have: $3x - 2y = -27$ $x + y = 56$

From Equation 2, we can express $y$ in terms of $x$: $y = 56 - x$

Substitute $y = 56 - x$ into Equation 1: $3x - 2(56 - x) = -27$ Expanding and simplifying: $3x - 112 + 2x = -27$ $5x = 85$ $x = 17$

Now substitute $x = 17$ back into Equation 2 to find $y$: $17 + y = 56$ $y = 39$

Conclusion

The fraction is: $\frac{x}{y} = \frac{17}{39}$

Would you like further details or have any questions?

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Here are some related questions:

1. How would you solve if 9 was subtracted instead of added to the numerator?
2. What if the fraction became $\frac{3}{4}$ when 9 was added?
3. How does the solution change if the half-sum is a different value?
4. What if we know only one equation about the fraction?
5. How can we verify the fraction obtained is correct?

Tip: In word problems involving fractions, translating statements into algebraic equations is key to setting up the problem for solving.