Determining Solutions for 3x + 4 < 9 and x + 5 > 3

To solve the given inequalities, we will handle each one individually and then find the valid range of $x$ where both are satisfied:

• First Inequality: $3x + 4 < 9$
  Solve for $x$ by first subtracting 4 from both sides:
  $3x + 4 - 4 < 9 - 4 \Rightarrow 3x < 5$.
  Next, divide both sides by 3 to solve for $x$:
  $x < \frac{5}{3}$.

• Second Inequality: $3 < x + 5$
  Solve for $x$ by subtracting 5 from both sides:
  $3 - 5 < x + 5 - 5 \Rightarrow -2 < x$.

Now we combine the solutions from both inequalities:

• From the first inequality, we have: $x < \frac{5}{3}$.
• From the second inequality, we have: $x > -2$.

Therefore, combining these results, the solution is the intersection of the two ranges:

$-2 < x < \frac{5}{3}$, which can be expressed as $-2 < x < 1\frac{2}{3}$.

The final solution is $-2 < x < 1\frac{2}{3}$.