To solve the inequality \( 48 \frac{2}{3} \leq -18 + \frac{1}{3}x \), we first convert \( 48 \frac{2}{3} \) to an improper fraction.
\[ 48 \frac{2}{3} = 48 + \frac{2}{3} = \frac{48 \times 3 + 2}{3} = \frac{144 + 2}{3} = \frac{146}{3} \]
Now, we can rewrite the inequality:
\[ \frac{146}{3} \leq -18 + \frac{1}{3}x \]
Next, we will isolate \( \frac{1}{3}x \). First, we add \( 18 \) to both sides:
\[ \frac{146}{3} + 18 \leq \frac{1}{3}x \]
We need to convert \( 18 \) to a fraction with a denominator of 3:
\[ 18 = \frac{18 \times 3}{3} = \frac{54}{3} \]
Now, we can add the two fractions on the left side:
\[ \frac{146}{3} + \frac{54}{3} = \frac{146 + 54}{3} = \frac{200}{3} \]
So the inequality now looks like this:
\[ \frac{200}{3} \leq \frac{1}{3}x \]
Next, we eliminate the fraction \( \frac{1}{3} \) by multiplying both sides by 3 (note that since 3 is positive, the inequality stays the same):
\[ 200 \leq x \]
This can also be written as:
\[ x \geq 200 \]
Thus, the solution to the inequality is:
\[ \boxed{x \geq 200} \]