the sum of numerator and denominator of a fraction is 180. when simplifying the fraction, it becomes 2/3. what is the fraction?

Let the fraction be represented as \( \frac{a}{b} \), where \( a \) is the numerator and \( b \) is the denominator. According to the problem, we know two key things:

1. The sum of the numerator and denominator is 180:
   \[
   a + b = 180
   \]
   
2. When simplified, the fraction becomes \( \frac{2}{3} \):
   \[
   \frac{a}{b} = \frac{2}{3}
   \]

From the second condition, we can express \( a \) in terms of \( b \) as follows:
\[
a = \frac{2}{3}b
\]

### Step 1: Substituting \( a \) in the first equation
Now substitute \( a = \frac{2}{3}b \) into \( a + b = 180 \):
\[
\frac{2}{3}b + b = 180
\]

### Step 2: Solving for \( b \)
To combine the terms, first express \( b \) as \( \frac{3}{3}b \):
\[
\frac{2}{3}b + \frac{3}{3}b = 180
\]
\[
\frac{5}{3}b = 180
\]
Now multiply both sides by 3 to eliminate the fraction:
\[
5b = 540
\]
Finally, divide by 5:
\[
b = 108
\]

### Step 3: Finding \( a \)
Now that we know \( b = 108 \), substitute this value back into \( a = \frac{2}{3}b \):
\[
a = \frac{2}{3} \times 108 = 72
\]

### Final Answer:
The fraction is \( \frac{72}{108} \), which simplifies to \( \frac{2}{3} \).

Would you like further details or have any questions? Here are some related questions for further exploration:

1. How do you simplify other fractions?
2. What is the method to solve equations with two variables?
3. Can you explain the concept of equivalent fractions?
4. How do you find the greatest common divisor (GCD) of two numbers?
5. What are the properties of proportions?

**Tip:** When simplifying fractions, always divide both the numerator and denominator by their greatest common divisor (GCD).

The sum of numerator and denominator of a fraction is 180. When simplifying the fraction, it becomes 2/3. What is the fraction?

This math problem asks to find a fraction where the sum of its numerator and denominator is 180, and when simplified, it becomes 2/3. The solution involves setting up algebraic equations and solving for the unknowns, suitable for students in Grades 6-8.

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