We are given two inequalities:

1. $17 < x + 10$
2. $19 \geq x + 10$

Let's solve each inequality separately.

Solving the first inequality:

$17 < x + 10$ Subtract 10 from both sides: $17 - 10 < x$ $7 < x$ This simplifies to: $x > 7$

Solving the second inequality:

$19 \geq x + 10$ Subtract 10 from both sides: $19 - 10 \geq x$ $9 \geq x$ This simplifies to: $x \leq 9$

Combined solution:

From the first inequality, we have $x > 7$, and from the second inequality, we have $x \leq 9$.

Thus, the solution is: $7 < x \leq 9$

This means that $x$ is greater than 7 and less than or equal to 9.

Would you like more details or further explanation on any part?

Here are some related questions for further practice:

1. Solve: $5 < 2x + 3 \leq 13$
2. Solve: $3x - 4 > 2$ and $7 - 2x \geq 5$
3. Solve: $4 \leq x - 7 < 10$
4. Solve: $12 \geq 2x + 4$ and $3x - 5 < 7$
5. Solve the system of inequalities: $x + 6 > 8$ and $x + 2 \leq 12$

Tip: When solving inequalities, always reverse the inequality symbol if you multiply or divide both sides by a negative number.