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.

Learn how to solve the linear inequalities 17 < x + 10 and 19 ≥ x + 10. This problem covers key algebraic concepts such as solving and combining inequalities. Suitable for students in Grades 6-8.

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