What is the largest integer $x$ such that $\frac{x}{3}+\frac{4}{5} < \frac{5}{3}$?
To solve this inequality, I need to isolate $x$ on one side and find its maximum possible value. I can start by subtracting $\frac{4}{5}$ from both sides to get $\frac{x}{3} < \frac{5}{3}-\frac{4}{5}$. Then I can find a common denominator for the fractions on the right and simplify them. I get $\frac{x}{3} < \frac{25}{15}-\frac{12}{15} = \frac{13}{15}$. Next, I can multiply both sides by $3$ to get rid of the denominator on the left. I get $x < \frac{13}{15} \cdot 3 = \frac{39}{15}$. Now, I need to find the largest integer that is less than $\frac{39}{15}$. I can see that $\frac{39}{15}$ is equivalent to $2.6$, so the largest integer less than that is $2$. Therefore, the largest integer $x$ that satisfies the inequality is $2$. 