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Let's break down the inequalities and solve for \( a \).

The given inequalities are:

1. \( a + 10 + 8a > 10 \)
2. \( a + 16 < -2a + 13 \)

### Solving the first inequality:
\[
a + 10 + 8a > 10
\]
Combine like terms:
\[
9a + 10 > 10
\]
Subtract 10 from both sides:
\[
9a > 0
\]
Divide both sides by 9:
\[
a > 0
\]

### Solving the second inequality:
\[
a + 16 < -2a + 13
\]
Add \( 2a \) to both sides to eliminate the negative term:
\[
a + 2a + 16 < 13
\]
Combine like terms:
\[
3a + 16 < 13
\]
Subtract 16 from both sides:
\[
3a < -3
\]
Divide both sides by 3:
\[
a < -1
\]

### Solution:
- From the first inequality: \( a > 0 \)
- From the second inequality: \( a < -1 \)

### Final result:
The solution is \( a > 0 \) or \( a < -1 \).

### Graphing the solution:
On the number line, this represents two disjoint intervals:
- One interval is \( (-\infty, -1) \), represented by an open circle at \( -1 \) and a ray extending to the left.
- The other interval is \( (0, \infty) \), represented by an open circle at \( 0 \) and a ray extending to the right.

Would you like more details on any part of this explanation? 

Here are 5 related questions you can explore:
1. How do you graph compound inequalities on a number line?
2. What is the difference between open and closed intervals?
3. How do you solve inequalities involving absolute values?
4. What is the role of conjunction (AND) vs. disjunction (OR) in compound inequalities?
5. How do you solve inequalities with fractions?

**Tip**: Always be careful to correctly interpret "AND" (both conditions must be true) and "OR" (at least one condition must be true) when solving inequalities.

Solve for a and graph the solution: a + 10 + 8a > 10 or a + 16 < -2a + 13

Learn how to solve the compound inequalities a + 10 + 8a > 10 or a + 16 < -2a + 13 step by step. This problem explores key algebraic concepts such as inequalities and algebraic manipulation. Suitable for Grades 8-10.

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