Step 1: Interpret the Inequalities

We are given the inequalities:

1. $4x - y \leq 1$
2. $x + 2y \leq 7$

The solution set consists of all points $(x, y)$ that satisfy both inequalities.

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Step 2: Convert Inequalities to Equations

To sketch the region, we first find the boundary lines by replacing the inequalities with equalities:

1. $4x - y = 1$ → (Equation of Line 1)
2. $x + 2y = 7$ → (Equation of Line 2)

We will graph these lines and determine the region that satisfies both inequalities.

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Step 3: Find Intersection of the Two Lines

To find the corner points, solve the system:

Solve for $y$

1. Solve for $y$ in terms of $x$ from the first equation: $y = 4x - 1$
2. Substitute into the second equation: $x + 2(4x - 1) = 7$ $x + 8x - 2 = 7$ $9x = 9$ $x = 1$

Find $y$

$y = 4(1) - 1 = 3$ So, the intersection point is $(1,3)$.

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Step 4: Find x- and y-Intercepts

For $4x - y = 1$

1. x-intercept ($y = 0$):
   $4x - 0 = 1 \Rightarrow x = \frac{1}{4}$ Point: $\left(\frac{1}{4}, 0\right)$

2. y-intercept ($x = 0$):
   $4(0) - y = 1 \Rightarrow y = -1$ Point: $(0, -1)$

For $x + 2y = 7$

1. x-intercept ($y = 0$):
   $x + 2(0) = 7 \Rightarrow x = 7$ Point: $(7, 0)$

2. y-intercept ($x = 0$):
   $0 + 2y = 7 \Rightarrow y = \frac{7}{2}$ Point: $\left(0, \frac{7}{2}\right)$

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Step 5: Determine the Region

The solution region is the intersection of the two half-planes:

• $4x - y \leq 1$ means the region is below the line $y = 4x - 1$.
• $x + 2y \leq 7$ means the region is below the line $y = \frac{7 - x}{2}$.

By shading the correct region in a graph, we see that the feasible region is bounded by the intersection points.

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Step 6: Find Corner Points

The corner points of the bounded region are the intersections of the lines and axes:

1. $(1,3)$ (Intersection of both lines)
2. $(7,0)$ (x-intercept of $x + 2y = 7$)
3. $\left(\frac{1}{4},0\right)$ (x-intercept of $4x - y = 1$)

Thus, the corner points are: $\left( \frac{1}{4}, 0 \right), (7, 0), (1,3)$

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Final Answers

1. Sketch: Graph the lines $4x - y = 1$ and $x + 2y = 7$. Shade the intersection region.
2. Bounded or Unbounded: Bounded
3. Corner Points:
   $\left( \frac{1}{4}, 0 \right), (7, 0), (1,3)$

Would you like me to generate a graph for visualization?

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Further Exploration Questions

1. How do we determine whether a region is bounded or unbounded?
2. What happens if the inequalities were strict (i.e., $<$ instead of $\leq$)?
3. How does changing the coefficients of $x$ or $y$ in the inequalities affect the graph?
4. Can you find the area of the bounded region?
5. What are some real-world applications of solving systems of linear inequalities?

Tip:

To check whether a point is in the solution region, plug it into both inequalities. If it satisfies both, it's in the feasible region!