The inequality $y \leq 2x + 1$ represents the region of the coordinate plane that includes all points $(x, y)$ where $y$ is less than or equal to the line defined by $y = 2x + 1$. Let’s analyze this inequality step-by-step:

1. The Line $y = 2x + 1$:

• The equation $y = 2x + 1$ is a straight line with:
  • Slope: $2$ (rise/run = 2/1, meaning it increases by 2 units vertically for every 1 unit horizontally).
  • y-Intercept: $(0, 1)$, where the line crosses the y-axis.
• This line divides the plane into two regions.

2. The Inequality $y \leq 2x + 1$:

• The $\leq$ symbol means the solution includes the line $y = 2x + 1$ itself (solid boundary) and all points below or on the line.

3. Graphing the Inequality:

• Draw the line $y = 2x + 1$.
• Shade the region below the line, as the inequality includes all $y$-values less than or equal to $2x + 1$.
• Use test points, e.g., $(0, 0)$:
  • Substitute into $y \leq 2x + 1$: $0 \leq 2(0) + 1$, which is true.
  • This confirms that $(0, 0)$ is in the solution region.

4. Boundary Type:

• The boundary $y = 2x + 1$ is solid because the inequality includes $y = 2x + 1$.

Would you like me to sketch this graph, provide further analysis, or solve a related problem? Let me know!

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Expanding Thoughts:

1. What happens if the inequality is $y > 2x + 1$?
2. How would the graph change if the slope or y-intercept were different?
3. How can we check whether a point satisfies the inequality?
4. What are some applications of inequalities in real-world scenarios?
5. How do systems of inequalities affect solution regions?

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Tip: Test points like $(0, 0)$ or others away from the boundary are quick ways to verify which region satisfies the inequality.