Explanation:

To solve the problem, let's set a common variable for the given equality: \( k \).

- Determine expressions for each variable:

- \( 2a = k \) ? \( a = \frac{k}{2} \)

- \( 3b = k \) ? \( b = \frac{k}{3} \)

- \( 6c = k \) ? \( c = \frac{k}{6} \)

- \( 9d = k \) ? \( d = \frac{k}{9} \)

- \( 12e = k \) ? \( e = \frac{k}{12} \)

- \( 18f = k \) ? \( f = \frac{k}{18} \)

- Calculate the expressions:

- \( a + b = \frac{k}{2} + \frac{k}{3} = \frac{3k}{6} + \frac{2k}{6} = \frac{5k}{6} \)

- \( c + d + e + f = \frac{k}{6} + \frac{k}{9} + \frac{k}{12} + \frac{k}{18} \)

- Find a common denominator and simplify:

- \( \text{Common denominator: 36} \)

- \( c + d + e + f = \frac{6k}{36} + \frac{4k}{36} + \frac{3k}{36} + \frac{2k}{36} = \frac{15k}{36} = \frac{5k}{12} \)

- Calculate the ratio:

- \( \frac{a + b}{c + d + e + f} = \frac{\frac{5k}{6}}{\frac{5k}{12}} = \frac{5}{6} \times \frac{12}{5} = 2 \)

- The correct option is Option 2: 2