1. We are given the inequality $$\frac{2}{5} + \frac{3}{7} \geq \frac{x}{35}$$ and need to find the solution set for $x$. 2. First, find a common denominator to add the fractions on the left side. The denominators are 5 and 7, and their least common multiple is 35. 3. Convert each fraction to have denominator 35: $$\frac{2}{5} = \frac{2 \times 7}{5 \times 7} = \frac{14}{35}$$ $$\frac{3}{7} = \frac{3 \times 5}{7 \times 5} = \frac{15}{35}$$ 4. Add the fractions: $$\frac{14}{35} + \frac{15}{35} = \frac{14 + 15}{35} = \frac{29}{35}$$ 5. Substitute back into the inequality: $$\frac{29}{35} \geq \frac{x}{35}$$ 6. Multiply both sides by 35 to clear denominators: $$35 \times \frac{29}{35} \geq 35 \times \frac{x}{35}$$ $$\cancel{35} \times \frac{29}{\cancel{35}} \geq \cancel{35} \times \frac{x}{\cancel{35}}$$ $$29 \geq x$$ 7. This means $x$ must be less than or equal to 29. 8. In interval notation, the solution set is $$(-\infty, 29]$$. Therefore, the correct choice is A with the solution set $(-\infty, 29]$.