To graph the solution set of this system of inequalities:

1. Rewrite the inequalities in slope-intercept form, $y = mx + b$, to make graphing easier.

   Inequality 1: $2x + 4y \leq 4$

   • Solve for $y$: $4y \leq -2x + 4$ $y \leq -\frac{1}{2}x + 1$

   Inequality 2: $4x + y \leq 8$

   • Solve for $y$: $y \leq -4x + 8$

2. Graph each inequality:

   • The first inequality $y \leq -\frac{1}{2}x + 1$ has a slope of $-\frac{1}{2}$ and y-intercept $1$. Draw the line $y = -\frac{1}{2}x + 1$ (solid line because of $\leq$), and shade below it.
   • The second inequality $y \leq -4x + 8$ has a slope of $-4$ and y-intercept $8$. Draw the line $y = -4x + 8$ (also solid), and shade below it.

3. Find the intersection region:

   • The solution set is the region where the shaded areas of both inequalities overlap. This represents all points $(x, y)$ that satisfy both inequalities simultaneously.

Would you like a detailed step-by-step on finding the intersection points or graphing tips?

Related Questions

1. How do we solve systems of inequalities with more than two inequalities?
2. What changes if one or both inequalities were strict inequalities ($<$ instead of $\leq$)?
3. How do we interpret the slope and intercept of an inequality line?
4. How can we determine if a particular point lies in the solution region?
5. How would the graph look if one inequality had a positive slope?

Tip:

To verify your graph, test a point in the overlapping region (like $(0,0)$, if it lies in that area) to see if it satisfies both inequalities.