Worksheet for solving systems of linear inequalities using graphing methods.

Let's solve each system of linear inequalities by graphing. We'll go through each problem step-by-step.

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Problem 1:

$$
\begin{cases}
y < 2x - 1 \\
y \geq x - 4
\end{cases}
$$

#### Step 1: Graph the boundary lines
- For $ y < 2x - 1 $:
- The boundary line is $ y = 2x - 1 $.
- Since it's strictly less than, use a dashed line.
- Find two points:
- When $ x = 0 $, $ y = -1 $ → point (0, -1)
- When $ x = 1 $, $ y = 1 $ → point (1, 1)

- For $ y \geq x - 4 $:
- The boundary line is $ y = x - 4 $.
- Since it's greater than or equal to, use a solid line.
- Points:
- $ x = 0 $ → $ y = -4 $ → (0, -4)
- $ x = 4 $ → $ y = 0 $ → (4, 0)

#### Step 2: Shade the regions
- For $ y < 2x - 1 $: shade below the dashed line.
- For $ y \geq x - 4 $: shade above or on the solid line.

#### Step 3: Find the overlapping region
The solution is where both shaded regions overlap.

✔ Final Answer: The solution is the region below the dashed line $ y = 2x - 1 $ and above or on the solid line $ y = x - 4 $. This is a wedge-shaped area between the two lines.

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Problem 2:

$$
\begin{cases}
5x + 1 > y \\
3x - 6y \leq 6
\end{cases}
$$

Rewrite in slope-intercept form:

1. $ 5x + 1 > y $ → $ y < 5x + 1 $
2. $ 3x - 6y \leq 6 $

Solve second inequality:
$$
-6y \leq -3x + 6 \\
y \geq \frac{1}{2}x - 1 \quad \text{(Divide by -6, reverse inequality)}
$$

So now:
$$
\begin{cases}
y < 5x + 1 \\
y \geq \frac{1}{2}x - 1
\end{cases}
$$

#### Graphing:

- $ y < 5x + 1 $: dashed line with slope 5, y-intercept 1 → shade below.
- $ y \geq \frac{1}{2}x - 1 $: solid line with slope 1/2, y-intercept -1 → shade above or on.

#### Shade both regions and find overlap

✔ Solution: The region between the two lines: above the solid line $ y = \frac{1}{2}x - 1 $ and below the dashed line $ y = 5x + 1 $.

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Problem 3:

$$
\begin{cases}
2y < x + 4 \\
y \geq -2 \\
x \leq 4
\end{cases}
$$

Convert first inequality:
$$
2y < x + 4 \Rightarrow y < \frac{1}{2}x + 2
$$

Now:
$$
\begin{cases}
y < \frac{1}{2}x + 2 \\
y \geq -2 \\
x \leq 4
\end{cases}
$$

#### Graphing:

1. $ y < \frac{1}{2}x + 2 $: dashed line, slope 1/2, y-intercept 2 → shade below.
2. $ y \geq -2 $: horizontal solid line at $ y = -2 $ → shade above or on.
3. $ x \leq 4 $: vertical solid line at $ x = 4 $ → shade left of or on.

#### Overlap:
The solution is the region bounded by:
- Below $ y = \frac{1}{2}x + 2 $
- Above $ y = -2 $
- Left of $ x = 4 $

This forms a bounded polygonal region (a trapezoid-like shape) in the left part of the plane.

✔ Solution: The intersection of all three shaded regions — bounded on the right by $ x = 4 $, below by $ y = -2 $, and above by $ y = \frac{1}{2}x + 2 $.

---

Problem 4:

$$
\begin{cases}
4x - y < 4 \\
2x + y \geq -5
\end{cases}
$$

Rewrite in slope-intercept form:

1. $ 4x - y < 4 \Rightarrow -y < -4x + 4 \Rightarrow y > 4x - 4 $
(Note: inequality flips when dividing by negative)

2. $ 2x + y \geq -5 \Rightarrow y \geq -2x - 5 $

So:
$$
\begin{cases}
y > 4x - 4 \\
y \geq -2x - 5
\end{cases}
$$

#### Graphing:

1. $ y > 4x - 4 $: dashed line with slope 4, y-intercept -4 → shade above.
2. $ y \geq -2x - 5 $: solid line with slope -2, y-intercept -5 → shade above or on.

#### Overlap:
The solution is where both conditions are satisfied — above both lines, but since one is strict and one is inclusive, we take the union of shading.

But actually, the region is above both lines, so the final region is the area that lies above the higher of the two lines in each region.

Find intersection point:
Set $ 4x - 4 = -2x - 5 $
$ 6x = -1 \Rightarrow x = -\frac{1}{6} $
Then $ y = 4(-1/6) - 4 = -2/3 - 4 = -14/3 \approx -4.67 $

So the two lines cross at $ (-1/6, -14/3) $

- To the left of this point, $ -2x - 5 $ is higher
- To the right, $ 4x - 4 $ is higher

So the solution region is:
- Above $ y = -2x - 5 $ for $ x < -1/6 $
- Above $ y = 4x - 4 $ for $ x > -1/6 $
- And includes the line $ y = -2x - 5 $, but not $ y = 4x - 4 $

✔ Final Solution: The region above both lines, with the boundary $ y = -2x - 5 $ included, and $ y = 4x - 4 $ excluded.

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✔ Summary of Solutions:

| Problem | System | Solution Region |
|--------|-------|------------------|
| 1 | $ y < 2x - 1 $, $ y \geq x - 4 $ | Between the two lines, below $ y = 2x - 1 $, above $ y = x - 4 $ |
| 2 | $ y < 5x + 1 $, $ y \geq \frac{1}{2}x - 1 $ | Between the two lines, above $ y = \frac{1}{2}x - 1 $, below $ y = 5x + 1 $ |
| 3 | $ y < \frac{1}{2}x + 2 $, $ y \geq -2 $, $ x \leq 4 $ | Bounded region: left of $ x=4 $, above $ y=-2 $, below $ y=\frac{1}{2}x+2 $ |
| 4 | $ y > 4x - 4 $, $ y \geq -2x - 5 $ | Above both lines; includes $ y = -2x - 5 $, excludes $ y = 4x - 4 $ |

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📌 How to Graph on the Grids:

For each problem:

1. Draw the boundary lines (dashed for strict inequalities, solid for ≤ or ≥).
2. Shade the correct side of each line.
3. The solution is the overlapping shaded region.

You can sketch this on the provided grids as follows:

- Use a pencil to draw lines.
- Use different shading patterns if needed (e.g., diagonal lines for one, dots for another).
- Final solution is the intersection of all shaded areas.

Would you like me to generate a visual sketch or describe how to plot each line step-by-step?