Show, by shading on the grid, the region defined by all three of the inequalities x  6y  2y  x + 1

To show the region defined by the inequalities $x \leq 6y$, $2y \leq x + 1$, and $y \geq 0$, we can follow these steps:

1. Break Down the Problem

• Identify the inequalities to graph.
• Find the intersection points and boundaries created by each inequality.

2. Relevant Concepts

The inequalities are:

1. $x \leq 6y$
2. $2y \leq x + 1$ or equivalently $x \geq 2y - 1$
3. $y \geq 0$

3. Analyze Each Inequality

Inequality 1: $x \leq 6y$

• This can be rewritten as $y \geq \frac{x}{6}$. It's a line through the origin with a slope of $\frac{1}{6}$, shading below the line.

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

• Rearranging gives $y \leq \frac{x + 1}{2}$. This line has a slope of $\frac{1}{2}$ and intersects the y-axis at $y = 0.5$, shading below the line.

Inequality 3: $y \geq 0$

• This represents the area above the x-axis.

4. Graph the Inequalities

The intersection points of the lines will help determine boundaries.

1. Finding Intersection Points:
   • Set $6y = 2y - 1$: $6y = 2y - 1 \implies 4y = -1 \implies y = -\frac{1}{4} \quad \text{(not in y ≥ 0 region)}$
   • Set $6y = \frac{x + 1}{2}$: $12y = x + 1 \implies x = 12y - 1$
   • Set $2y - 1 = \frac{x + 1}{2}$: $4y - 2 = x + 1 \implies x = 4y - 3$

5. Graphing and Shading

• Graph the lines $y = \frac{x}{6}$, $y = \frac{x + 1}{2}$, and the line $y = 0$ on the coordinate plane.
• Shade the area below the line $y = \frac{x + 1}{2}$, above the line $y = \frac{x}{6}$, and above the x-axis.

Verify and Summarize

• The intersection area should be checked against all the inequalities. The final shaded region represents the solution where all conditions are satisfied.

Final Answer

The shaded region on the grid represents all the points that satisfy the inequalities $x \leq 6y$, $2y \leq x + 1$, and $y \geq 0$.