The system of inequalities you provided is:

1. $x + y > 3$
2. $x + y < -7$

To graph this solution set, follow these steps:

Step 1: Rewrite the inequalities in slope-intercept form

• The first inequality, $x + y > 3$, can be written as: $y > -x + 3$ This represents a line with slope $-1$ and y-intercept $3$. The region above this line (since it's a strict inequality) will be shaded.

• The second inequality, $x + y < -7$, can be written as: $y < -x - 7$ This represents a line with slope $-1$ and y-intercept $-7$. The region below this line will be shaded.

Step 2: Graph each boundary line

• Draw the line $y = -x + 3$, using a dashed line because the inequality is strict (no "equal to").
• Draw the line $y = -x - 7$, also using a dashed line for the same reason.

Step 3: Shade the appropriate regions

• For $y > -x + 3$, shade the region above the dashed line $y = -x + 3$.
• For $y < -x - 7$, shade the region below the dashed line $y = -x - 7$.

Step 4: Intersection of the two shaded regions

Since the two inequalities are contradictory, there is no overlap between the two regions. Therefore, the solution set is empty, meaning there is no solution that satisfies both inequalities at the same time.

Let me know if you'd like to see the graph or need further clarifications.

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Would you like more details, or do you have any questions?

Here are 5 related questions to expand on this topic:

1. How do we determine if a system of inequalities has no solution?
2. What is the difference between strict and non-strict inequalities when graphing?
3. How would the solution change if both inequalities used "greater than or equal to" or "less than or equal to"?
4. Can you explain what regions represent the solutions in a system of inequalities?
5. How can systems of inequalities be solved algebraically instead of graphically?

Tip: Always use dashed lines for strict inequalities and solid lines for inclusive inequalities when graphing systems of inequalities.