This worksheet provides practice on graphing linear inequalities, including converting standard form to slope-intercept form.

Let’s work through each problem one by one. I’ll solve them step by step, check my math, and then give the final answers.

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Problem 1: Graph: 4x – 6y ≤ –12

First, let’s rewrite this in slope-intercept form (y = mx + b) so it’s easier to graph.

Start with:
4x – 6y ≤ –12

Subtract 4x from both sides:
–6y ≤ –4x – 12

Now divide by –6. Remember: when you divide or multiply by a negative number, flip the inequality sign!

y ≥ (–4/–6)x + (–12/–6)

Simplify:
y ≥ (2/3)x + 2

So we graph a solid line (because of “≥”) with slope 2/3 and y-intercept 2. Then shade above the line.

But since the question just says “Graph”, and there’s no multiple choice, we’re done with the setup. For now, we’ll note the correct form for checking later.

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Problem 2: Write the slope-intercept form of the graphed inequality.

Looking at the graph:

- The line is dashed → so it’s either < or > (not ≤ or ≥)
- The line goes down from left to right → negative slope
- It crosses the y-axis at (0, 1) → y-intercept = 1
- Let’s find the slope: from (0,1) to (2,0): rise = -1, run = 2 → slope = -1/2
- Shading is below the line → so y < ...

So the inequality is: y < –(1/2)x + 1

Check: pick a point in shaded region, say (0,0). Plug in: 0 < –(1/2)(0) + 1 → 0 < 1 → true. Good.

Final answer for #2: y < –½x + 1

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Problem 3: Which inequality best represents the graph?

Graph shows:

- Solid line → includes equality (≥ or ≤)
- Shaded above the line → so y ≥ ... or equivalent
- Line passes through (0, -2) and (4, 0)

Find slope: (0 - (-2)) / (4 - 0) = 2/4 = 1/2

So equation of line: y = (1/2)x – 2

Multiply both sides by 10 to match answer choices:

10y = 5x – 20 → rearrange: 5x – 10y = 20

Since shading is above, and original line is y ≥ (1/2)x – 2, multiplying by 10 (positive) doesn’t change inequality:

→ 5x – 10y ≤ 20? Wait, let’s test.

Original: y ≥ (1/2)x – 2

Multiply both sides by 10: 10y ≥ 5x – 20

Bring all terms to one side: –5x + 10y ≥ –20 → multiply by –1 (flip inequality): 5x – 10y ≤ 20

Yes! So answer is b. 5x – 10y ≤ 20

Test point: (0,0) is NOT in shaded region. Plug into b: 5(0) – 10(0) = 0 ≤ 20 → true, but (0,0) is not shaded? Wait — that’s a problem.

Wait — if (0,0) gives 0 ≤ 20 (true), but (0,0) is NOT in the shaded region, then our inequality must be wrong.

Actually, looking again: the shaded region is ABOVE the line. At x=0, y=-2 is on the line. Above that would be y > -2.

At (0,0): y=0 > -2 → should be shaded. But in the graph, is (0,0) shaded? Looking back — yes, actually, the shading covers the top half, including (0,0). So (0,0) SHOULD satisfy the inequality.

In option b: 5x – 10y ≤ 20 → at (0,0): 0 ≤ 20 → true → good.

Option d: 5x – 10y < 20 → also true at (0,0), but line is solid, so must include equality → so d is out.

Option a: 10x – 5y ≥ 20 → at (0,0): 0 ≥ 20? False → not shaded → good, but let’s see if it matches the line.

Line: from (0,-2) to (4,0). Plug into a: 10x – 5y = 20?

At (0,-2): 0 – 5(-2) = 10 ≠ 20 → not on line → wrong.

Option c: 10x – 5y > 20 → same issue, and strict inequality.

So only b and d have the right line. Since line is solid, b is correct.

Answer: b. 5x – 10y ≤ 20

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Problem 4: Graph the inequality: y > (5/2)x – 3

This is already in slope-intercept form.

- Dashed line (because >, not ≥)
- Slope = 5/2 → up 5, right 2
- y-intercept = -3 → start at (0, -3)
- Shade above the line

No calculation needed — just graphing. We’re done.

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Problem 5: Graph: –8x + 9y ≥ 72

Rewrite in slope-intercept form:

–8x + 9y ≥ 72

Add 8x to both sides:

9y ≥ 8x + 72

Divide by 9:

y ≥ (8/9)x + 8

Solid line (≥), slope 8/9, y-intercept 8. Shade above.

Again, just graphing — no final answer needed beyond the graph.

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Problem 6: Write the slope-intercept form of the graphed inequality.

Graph shows:

- Dashed line → < or >
- Shaded below → so y < ...
- Line passes through (0,1) and (1,2)? Wait, let’s check.

From graph: when x=0, y=1; when x=2, y=3? Actually, looks like it goes through (0,1) and (2,3).

Slope = (3-1)/(2-0) = 2/2 = 1

So line: y = x + 1

Shaded below → y < x + 1

Test point: (0,0) is in shaded region. 0 < 0 + 1 → 0 < 1 → true. Good.

Answer: y < x + 1

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Problem 7: Graph the inequality: y < 3

Horizontal dashed line at y=3. Shade below.

Simple — no calculation.

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Problem 8: Which inequality best represents the graph?

Graph shows:

- Vertical line at x = -1
- Solid line → includes equality
- Shaded to the right → x ≥ -1

Options:

a. x ≥ –1 → matches

b. x < –1 → wrong direction

c. y ≥ –1 → horizontal, wrong

d. y > –1 → horizontal, wrong

Answer: a. x ≥ –1

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Now, compiling all final answers:

1. Graphed correctly as y ≥ (2/3)x + 2 (but since it’s a graphing task, no written answer needed)

2. y < –½x + 1

3. b. 5x – 10y ≤ 20

4. Graphed correctly (no written answer)

5. Graphed correctly (no written answer)

6. y < x + 1

7. Graphed correctly (no written answer)

8. a. x ≥ –1

Since the user didn’t specify which problems to answer, and the format requires a final answer, I’ll list the answers for the problems that require a written response (2, 3, 6, 8).

Final Answer:
2. y < –½x + 1
3. b
6. y < x + 1
8. a