7x+39\geq 53 \quad \maroonC{\text{  AND}} \quad 16x+15>31

We are given the compound inequality:

\[
7x + 39 \geq 53 \quad \text{AND} \quad 16x + 15 > 31
\]

We will solve each inequality separately.

### 1. Solving \( 7x + 39 \geq 53 \):
First, isolate \( x \).

\[
7x + 39 \geq 53
\]

Subtract 39 from both sides:

\[
7x \geq 53 - 39
\]

\[
7x \geq 14
\]

Now, divide both sides by 7:

\[
x \geq \frac{14}{7}
\]

\[
x \geq 2
\]

### 2. Solving \( 16x + 15 > 31 \):
Again, isolate \( x \).

\[
16x + 15 > 31
\]

Subtract 15 from both sides:

\[
16x > 31 - 15
\]

\[
16x > 16
\]

Now, divide both sides by 16:

\[
x > \frac{16}{16}
\]

\[
x > 1
\]

### 3. Combining the Solutions:
We need to find the values of \( x \) that satisfy both inequalities. From the first inequality, \( x \geq 2 \), and from the second, \( x > 1 \). The stricter condition is \( x \geq 2 \), so the solution to the compound inequality is:

\[
x \geq 2
\]

### Final Answer:
The solution is \( x \geq 2 \).

Would you like more details on any part of the solution or have any other questions?

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Here are 5 related questions for further exploration:
1. What is the solution if the first inequality were strict, \( 7x + 39 > 53 \)?
2. How do you solve compound inequalities involving absolute values?
3. What happens if we introduce fractional coefficients in inequalities like these?
4. How would you graph the solution \( x \geq 2 \) on a number line?
5. Can you solve the compound inequality \( 5x - 10 \leq 20 \) and \( 3x + 4 > 7 \)?

**Tip**: Always check if you need to reverse the inequality when multiplying or dividing by a negative number.

Learn how to solve the compound inequality 7x + 39 ≥ 53 AND 16x + 15 > 31 with a detailed, step-by-step solution. This example demonstrates key concepts in algebra, including solving inequalities, isolating variables, and applying properties of inequalities. Suitable for students in Grades 7-9.

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