Worksheet for graphing systems of inequalities, including six problems with coordinate grids for visual solutions.

Let’s solve each system of inequalities step by step. We’ll graph each inequality on the coordinate plane and find where they overlap — that overlapping region is the solution.

We’ll go one problem at a time.

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Problem 1:
Inequalities:
- \(3x + y \leq 2\)
- \(6x - 2y > -4\)

Step 1: Graph \(3x + y = 2\) (solid line because ≤)
Find two points:
- If x=0, y=2 → point (0,2)
- If y=0, 3x=2 → x=2/3 ≈ 0.67 → point (0.67, 0)

Shade below the line (because ≤).

Step 2: Graph \(6x - 2y = -4\) (dashed line because >)
Simplify: divide by 2 → \(3x - y = -2\) → or \(y = 3x + 2\)
Points:
- x=0 → y=2 → (0,2)
- x=-1 → y=-1 → (-1,-1)

Since it’s >, shade ABOVE the line.

Step 3: Find overlapping shaded region.
Note: Both lines pass through (0,2). The solution is the region that is below the first line AND above the second line.

Check test point (0,0):
- First inequality: 3(0)+0=0 ≤ 2 → TRUE
- Second: 6(0)-2(0)=0 > -4 → TRUE → so (0,0) is in solution.

So shade the region containing (0,0), bounded by both lines.

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Problem 2:
Inequalities:
- \(6x - 2y \leq 4\)
- \(y < x\)

Step 1: Graph \(6x - 2y = 4\) → simplify to \(3x - y = 2\) → \(y = 3x - 2\) (solid line)
Points:
- x=0 → y=-2 → (0,-2)
- x=1 → y=1 → (1,1)

Shade BELOW the line (since ≤ means y ≤ 3x - 2? Wait — let’s rearrange properly.)

Original: \(6x - 2y \leq 4\)
→ Subtract 6x: \(-2y \leq -6x + 4\)
→ Divide by -2 (flip inequality): \(y \geq 3x - 2\)

Ah! Important: when dividing by negative, flip sign.

So actually, shade ABOVE the line \(y = 3x - 2\).

Step 2: Graph \(y = x\) (dashed line because <)
Shade BELOW the line (since y < x).

Step 3: Overlap region: above \(y = 3x - 2\) and below \(y = x\).

Test point (0,0):
- First: 6(0)-2(0)=0 ≤ 4 → TRUE → and since we have y ≥ 3x-2 → 0 ≥ -2 → TRUE
- Second: 0 < 0? FALSE → so (0,0) not in solution.

Try (1,0):
- First: 6(1)-2(0)=6 ≤ 4? FALSE → no.

Try (0,-1):
- First: 6(0)-2(-1)=2 ≤ 4 → TRUE; y≥3x-2 → -1 ≥ -2 → TRUE
- Second: -1 < 0 → TRUE → so (0,-1) is in solution.

Shade region including (0,-1).

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Problem 3:
Inequalities:
- \(4x + y < 2\)
- \(y > x\)

Step 1: Graph \(4x + y = 2\) → dashed line (because <)
Points:
- x=0 → y=2 → (0,2)
- y=0 → 4x=2 → x=0.5 → (0.5,0)

Shade BELOW the line.

Step 2: Graph \(y = x\) → dashed line (because >)
Shade ABOVE the line.

Step 3: Overlap: below first line, above second line.

Test point (0,1):
- First: 4(0)+1=1 < 2 → TRUE
- Second: 1 > 0 → TRUE → so (0,1) is in solution.

Shade region around (0,1).

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Problem 4:
Inequalities:
- \(y < 2x + 1\)
- \(y > -\frac{1}{3}x + 4\)

Step 1: Graph \(y = 2x + 1\) → dashed line → shade BELOW.

Step 2: Graph \(y = -\frac{1}{3}x + 4\) → dashed line → shade ABOVE.

Step 3: Find where regions overlap.

Find intersection point of the two lines:
Set \(2x + 1 = -\frac{1}{3}x + 4\)

Multiply both sides by 3:
\(6x + 3 = -x + 12\)
→ \(7x = 9\) → \(x = 9/7 ≈ 1.2857\)
Then y = 2*(9/7) + 1 = 18/7 + 7/7 = 25/7 ≈ 3.57

So lines cross at about (1.29, 3.57)

The solution is the region between the two lines — above the downward sloping line and below the upward sloping line.

Test point (0,3):
- First: 3 < 2(0)+1 → 3<1? FALSE → not in solution.

Test point (2,3):
- First: 3 < 2(2)+1=5 → TRUE
- Second: 3 > -1/3*2 +4 = -2/3 + 4 = 10/3 ≈ 3.33? 3 > 3.33? FALSE

Test point (1,3):
- First: 3 < 2(1)+1=3 → 3<3? FALSE

Test point (1,2.5):
- First: 2.5 < 3 → TRUE
- Second: 2.5 > -1/3 +4 = 11/3 ≈ 3.666? 2.5 > 3.666? FALSE

Wait — maybe try x=3:

At x=3:
First line: y < 7
Second line: y > -1 +4 = 3

So for x=3, y between 3 and 7 → say (3,4)

Check:
- 4 < 2(3)+1=7 → TRUE
- 4 > -1/3*3 +4 = -1+4=3 → TRUE → yes!

So (3,4) works. Shade region around there.

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Problem 5:
Inequalities:
- \(2x - y \geq -6\)
- \(x > 2\)

Step 1: Graph \(2x - y = -6\) → solid line (≥)
Rearrange: y ≤ 2x + 6? Let's see:

2x - y ≥ -6
→ -y ≥ -2x -6
→ multiply by -1 (flip): y ≤ 2x + 6

So shade BELOW the line y = 2x + 6.

Points:
- x=0 → y=6 → (0,6)
- x=-3 → y=0 → (-3,0)

Step 2: Graph x=2 → vertical dashed line (because >) → shade to the RIGHT.

Step 3: Overlap: right of x=2 AND below y=2x+6.

Test point (3,0):
- First: 2(3)-0=6 ≥ -6 → TRUE
- Second: 3>2 → TRUE → good.

Shade region to the right of x=2 and under the line.

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Problem 6:
Inequalities:
- \(y < \frac{1}{2}x + 4\)
- \(x - 2y \geq 6\)

Step 1: Graph \(y = \frac{1}{2}x + 4\) → dashed line → shade BELOW.

Step 2: Graph \(x - 2y = 6\) → solid line (≥)
Rearrange: -2y ≥ -x + 6 → divide by -2 (flip): y ≤ (1/2)x - 3

Wait — let’s do it carefully:

x - 2y ≥ 6
→ -2y ≥ -x + 6
→ divide by -2 → y ≤ (1/2)x - 3

So shade BELOW the line y = (1/2)x - 3.

But wait — now we have:
- y < (1/2)x + 4
- y ≤ (1/2)x - 3

These are parallel lines! Same slope 1/2.

The second line is lower: y = (1/2)x - 3 vs y = (1/2)x + 4

So the solution must satisfy BOTH: y less than the upper line AND y less than or equal to the lower line.

That means the stricter condition is y ≤ (1/2)x - 3, because if y is ≤ lower line, it’s automatically < upper line.

But check: is that always true?

Suppose y = (1/2)x - 3 → then compare to (1/2)x + 4 → clearly -3 < +4, so yes.

So the solution is just the region below or on y = (1/2)x - 3.

But wait — let’s verify with a point.

Take point (6,0):
- First: 0 < (1/2)(6)+4 = 3+4=7 → TRUE
- Second: 6 - 2(0)=6 ≥ 6 → TRUE → good.

Now take point (6,1):
- First: 1 < 7 → TRUE
- Second: 6 - 2(1)=4 ≥ 6? FALSE → not in solution.

Point (6,-1):
- First: -1 < 7 → TRUE
- Second: 6 - 2(-1)=8 ≥ 6 → TRUE → good.

Now, what about a point above y=(1/2)x -3 but below y=(1/2)x+4? Say x=0, y=0:

- First: 0 < 0+4=4 → TRUE
- Second: 0 - 0 =0 ≥6? FALSE → not in solution.

So indeed, only points satisfying y ≤ (1/2)x -3 will work.

Thus, graph y = (1/2)x -3 (solid line), shade below it.

And since y < (1/2)x +4 is automatically satisfied when y ≤ (1/2)x -3, we don’t need to worry about it separately.

Final Answer: For each problem, sketch the graphs as described and shade the overlapping region.

Since this is a worksheet asking to "sketch", and we can't draw here, I'll describe the key features for each.

But per instructions, I must provide final answer after reasoning.

Actually, re-reading: the user wants me to solve the problem accurately. Since it's graphing, perhaps list the boundary lines and shading directions.

But the instruction says: “Sketch the solution” — so maybe the expected answer is the description of how to graph.

However, looking back at the format, it says “Final Answer: [Provide only the final correct answer]”

For graphing problems, sometimes the answer is the set of inequalities graphed correctly.

But since I can't output images, I think the best is to state for each problem the region defined.

Alternatively, perhaps the problem expects us to identify the solution region verbally.

Given the constraints, I’ll summarize the solution for each as the intersection of half-planes.

But let me check if any have no solution or something.

All seem to have solutions.

Perhaps for the purpose of this, since it's a worksheet, the "answer" is the sketched graph, but since I can't do that, I'll describe the key points.

Wait — maybe I should just say: "Graph each pair of inequalities and shade the overlapping region as described in the steps."

But that seems vague.

Another idea: perhaps for each problem, give the vertices of the solution region if bounded, but most are unbounded.

Looking at Problem 4, it has an intersection point, but still unbounded.

I think the intended answer is the graphical representation, but since I'm text-based, I'll provide the description.

But the instruction says: "Solve the problem accurately." and "Final Answer: [only the final correct answer]"

Perhaps for such worksheets, the answer is not a single number, so I need to box the descriptions.

Maybe I made a mistake — let me see the original request.

User said: "Solve the problem accurately." and uploaded the worksheet.

In many contexts, for graphing systems, the solution is the graph itself.

But since I can't produce images, I'll describe the solution regions briefly for each.

To comply with "Final Answer", I'll list them.

After careful thought, I realize that for educational purposes, providing the method is part of solving, but the final answer should be concise.

Perhaps the problem is to recognize that each system defines a region, and the answer is to graph it, so in text, I can say:

For each numbered system, the solution is the set of points (x,y) that satisfy both inequalities simultaneously, which can be found by graphing the boundary lines and shading the appropriate half-planes as detailed in the explanation.

But that's not specific.

Let me try to give for each the inequality that defines the solution region after simplification.

For example, in Problem 6, we saw that the solution is simply y ≤ (1/2)x - 3, since the other inequality is redundant.

Similarly, for others.

Let's do that.

Final Summary of Solutions:

1. Region where y ≤ -3x + 2 and y < 3x + 2 (after rearranging)

From earlier:
- 3x + y ≤ 2 → y ≤ -3x + 2
- 6x - 2y > -4 → y < 3x + 2 (since divided by -2 and flipped)

So solution: y ≤ -3x + 2 and y < 3x + 2

2. From earlier:
- 6x - 2y ≤ 4 → y ≥ 3x - 2
- y < x

So solution: y ≥ 3x - 2 and y < x

3.
- 4x + y < 2 → y < -4x + 2
- y > x

Solution: y < -4x + 2 and y > x

4.
- y < 2x + 1
- y > - (1/3)x + 4

Solution: - (1/3)x + 4 < y < 2x + 1

5.
- 2x - y ≥ -6 → y ≤ 2x + 6
- x > 2

Solution: x > 2 and y ≤ 2x + 6

6.
- y < (1/2)x + 4
- x - 2y ≥ 6 → y ≤ (1/2)x - 3

As established, y ≤ (1/2)x - 3 implies y < (1/2)x + 4, so solution is y ≤ (1/2)x - 3

Now, for the Final Answer, since it's multiple parts, I'll list them.

But the format might expect a single box, but I think for clarity, I'll write:

Final Answer:
1. Shade below y = -3x + 2 and above y = 3x + 2 (but wait, from calculation, for problem 1, we have y ≤ -3x+2 and y < 3x+2, and they intersect at (0,2), and solution includes (0,0), which is below both? No.

Let's double-check problem 1.

Inequality 1: 3x + y ≤ 2 → y ≤ -3x + 2

Inequality 2: 6x - 2y > -4 → let's solve for y:

6x - 2y > -4
-2y > -6x -4
Divide by -2, flip: y < 3x + 2

Yes.

Now, the line y = -3x + 2 and y = 3x + 2 intersect when -3x+2 = 3x+2 → -6x=0 → x=0, y=2.

For x<0, say x=-1:
y ≤ -3(-1)+2=5
y < 3(-1)+2= -1

So y < -1 and y ≤ 5, so y < -1

For x>0, say x=1:
y ≤ -3+2= -1
y < 3+2=5
So y ≤ -1

At x=0, y≤2 and y<2, so y<2

The region is below both lines, but since for x>0, y≤ -1 is stricter, and for x<0, y< -1 is stricter, and at x=0, y<2.

Actually, the binding constraint is the lower of the two upper bounds.

Since y must be less than or equal to -3x+2 and less than 3x+2, the effective upper bound is the minimum of the two lines.

The two lines cross at (0,2), and for x>0, -3x+2 < 3x+2, so y ≤ -3x+2 is tighter.

For x<0, 3x+2 < -3x+2, so y < 3x+2 is tighter.

So the solution region is bounded by the V-shape opening downwards with vertex at (0,2).

And since at (0,0) it satisfies, and (0,0) is below both, yes.

So in summary, for each, the solution is the intersection.

For the Final Answer, I'll state the simplified system for each.

So:

Final Answer:
1. \( y \leq -3x + 2 \) and \( y < 3x + 2 \)
2. \( y \geq 3x - 2 \) and \( y < x \)
3. \( y < -4x + 2 \) and \( y > x \)
4. \( -\frac{1}{3}x + 4 < y < 2x + 1 \)
5. \( x > 2 \) and \( y \leq 2x + 6 \)
6. \( y \leq \frac{1}{2}x - 3 \)

This should be accurate based on our calculations.