The problem in the image asks for the range of values for which the following two inequalities are satisfied:

1. $2x + 3 \geq 5$
2. $x^2 < 9$

Step 1: Solve $2x + 3 \geq 5$

$2x + 3 \geq 5$ Subtract 3 from both sides:

$2x \geq 2$ Divide by 2:

$x \geq 1$

Step 2: Solve $x^2 < 9$

$x^2 < 9$ Take the square root of both sides:

$-3 < x < 3$

Step 3: Find the intersection of the two inequalities

The solution for $2x + 3 \geq 5$ is $x \geq 1$, and the solution for $x^2 < 9$ is $-3 < x < 3$. To satisfy both inequalities, we need the range where both conditions hold true.

From $-3 < x < 3$ and $x \geq 1$, the intersection is:

$1 \leq x < 3$

Final Answer:

The values of $x$ that satisfy both inequalities are in the range: $1 \leq x < 3$

Would you like more details or further explanation?

Here are 5 related questions for deeper understanding:

1. How would the solution change if the first inequality were $2x + 3 > 5$ instead?
2. What would happen if the second inequality were $x^2 \leq 9$ instead of $x^2 < 9$?
3. Can you explain how to find the intersection of two inequalities graphically?
4. How do you solve quadratic inequalities like $x^2 \geq 9$?
5. How would the solution be affected if one of the inequalities was non-linear (e.g., $x^3 < 27$)?

Tip: When solving inequalities involving quadratic expressions, it's helpful to remember that the solutions usually fall between or outside the roots of the related equation.