$2d+7<-9$2d+7<−9​   or   $4d-1>-3$4d−1>−3​

Let's solve the given inequalities one by one.

### First inequality: 
\[
2d + 7 < -9
\]
1. Subtract 7 from both sides:
\[
2d < -9 - 7
\]
\[
2d < -16
\]

2. Divide both sides by 2:
\[
d < \frac{-16}{2}
\]
\[
d < -8
\]

### Second inequality:
\[
4d - 1 > -3
\]
1. Add 1 to both sides:
\[
4d > -3 + 1
\]
\[
4d > -2
\]

2. Divide both sides by 4:
\[
d > \frac{-2}{4}
\]
\[
d > -\frac{1}{2}
\]

### Final Answer:
The solution for the compound inequality is:
\[
d < -8 \quad \text{or} \quad d > -\frac{1}{2}
\]

This means that $d$ can be any number less than $-8$ or greater than $-\frac{1}{2}$.

Would you like further details or have any questions?

Here are 5 related questions to expand on this:
1. How do we solve inequalities that involve multiplication or division by negative numbers?
2. How would the solution change if we had a conjunction (and) instead of a disjunction (or)?
3. What would be the solution for a similar inequality involving fractions?
4. How do we represent this solution on a number line?
5. Can we express the final answer in interval notation?

**Tip:** Remember, when multiplying or dividing both sides of an inequality by a negative number, the inequality sign reverses!

Learn how to solve the compound inequalities 2d + 7 < -9 or 4d - 1 > -3 with step-by-step solutions. This problem involves basic algebraic techniques for inequalities and covers important concepts such as solving inequalities and handling disjunctions. Suitable for students in Grades 7-9.

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