Worksheet titled "Graphing Systems of Inequalities Worksheet" from Math Monks, featuring six systems of inequalities to be graphed on coordinate planes.

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:
3x + y ≤ 2
6x - 2y > -4

Step 1: Graph 3x + y = 2 (solid line because of “≤”)
- When x=0, y=2 → point (0,2)
- When y=0, 3x=2 → x=2/3 ≈ 0.67 → point (0.67, 0)
- Shade BELOW the line (because y ≤ -3x + 2)

Step 2: Graph 6x - 2y = -4 (dashed line because of “>”)
Simplify: divide by 2 → 3x - y = -2 → y = 3x + 2
- When x=0, y=2 → (0,2)
- When x=-1, y=-1 → (-1,-1)
- Since it’s “>”, we shade ABOVE the line (because y < 3x + 2? Wait — let’s rearrange properly.)

Wait! Let’s rearrange 6x - 2y > -4 to slope-intercept form:

6x - 2y > -4
→ -2y > -6x -4
→ Divide by -2 (FLIP inequality sign!)
→ y < 3x + 2

So actually, for the second inequality, we graph y = 3x + 2 as a dashed line, and shade BELOW it.

But wait — first inequality was y ≤ -3x + 2 (from 3x + y ≤ 2 → y ≤ -3x + 2)

So now we have:
- Line 1: y = -3x + 2 (solid), shade below
- Line 2: y = 3x + 2 (dashed), shade below

Find where both shaded regions overlap.

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

So (0,0) is in the solution set.

The two lines intersect when:
-3x + 2 = 3x + 2 → -6x = 0 → x=0 → y=2

So they meet at (0,2). The solution region is below both lines, which forms a V-shape opening downward, with vertex at (0,2), including the solid line but not the dashed one.

Actually, since both are "below", and they cross at (0,2), the solution is the region below both lines — which is the area between them going down from (0,2).

Wait — let me double-check with another point. Try (0,1):

First: 3(0)+1=1 ≤ 2 → yes
Second: 6(0)-2(1)=-2 > -4 → yes → good.

Try (1,0):
First: 3+0=3 ≤ 2? No → not in first region.

Try (-1,0):
First: -3+0=-3 ≤ 2 → yes
Second: -6 -0 = -6 > -4? No → not in second.

So only points near origin work. Actually, the solution is the region below both lines — which is a wedge starting at (0,2) and going down-left and down-right? Wait no — let's think again.

Line 1: y = -3x + 2 → steep negative slope
Line 2: y = 3x + 2 → steep positive slope
They cross at (0,2)

Shading below both means: for any x, y must be less than both lines. So for x>0, line 2 (y=3x+2) is higher than line 1 (y=-3x+2)? No — for x>0, 3x+2 > -3x+2, so the lower boundary is line 1. For x<0, -3x+2 > 3x+2, so lower boundary is line 2.

Actually, the region below both lines is the area that is under the lower of the two lines at each x. Since they cross at x=0, for x<0, line 2 is lower; for x>0, line 1 is lower.

But since we’re shading below both, the solution is the region that is below the minimum of the two lines.

That would be a V-shaped region pointing downward, with vertex at (0,2), bounded by the two lines.

And since (0,0) satisfies both, and it’s below both lines, yes.

Also, note: the first inequality includes the line (solid), second does not (dashed).

So final answer for #1: The solution is the region below both lines, forming a downward-pointing angle with vertex at (0,2), including the line y=-3x+2 but not y=3x+2.

But since this is a sketch, we just need to describe or draw it. Since we can’t draw here, I’ll describe the key features for each.

But perhaps for the purpose of this exercise, since the user expects a final answer, and there are 6 problems, maybe I should provide the description for each.

Wait — the instruction says “Sketch the solution”, but since we’re text-based, I’ll describe the solution region for each.

Alternatively, maybe the expected output is to list the steps or the final graph description.

Looking back at the rules: “Provide only the final correct answer” — but for graphing problems, what is the “final answer”? Probably a description of the solution region.

But to keep it simple, I’ll summarize each solution clearly.

Let me continue solving all six carefully.

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Problem 2:
6x - 2y ≤ 4
y < x

Step 1: 6x - 2y ≤ 4 → simplify: divide by 2 → 3x - y ≤ 2 → y ≥ 3x - 2 (remember to flip when multiplying/dividing by negative? Wait, let's do it right.

6x - 2y ≤ 4
→ -2y ≤ -6x + 4
→ Divide by -2 → FLIP inequality → y ≥ 3x - 2

So graph y = 3x - 2 as solid line (because ≥), shade above.

Step 2: y < x → graph y = x as dashed line, shade below.

Find overlap.

Test point (0,0):
First: 6(0)-2(0)=0 ≤ 4 → true → and y=0 ≥ 3(0)-2 = -2 → true
Second: 0 < 0? False → so (0,0) not in solution.

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

Lines intersect when 3x - 2 = x → 2x=2 → x=1, y=1

So intersection at (1,1)

Solution region: above y=3x-2 and below y=x. This is a wedge between the two lines, starting from their intersection at (1,1) and going down-left? Let's see.

For x<1, say x=0: y≥-2 and y<0 → so between y=-2 and y=0

For x>1, say x=2: y≥4 and y<2 → impossible. So only for x≤1.

Actually, the region is where y < x and y ≥ 3x - 2. Since 3x-2 < x when x<1, so for x<1, the band between the lines exists. At x=1, they meet. For x>1, 3x-2 > x, so no overlap.

So solution is the region between the two lines for x ≤ 1, including the solid line y=3x-2, excluding dashed line y=x.

Vertex at (1,1), extending down to the left.

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This is getting very long. Perhaps for efficiency, since the user might expect concise answers, and given the constraints, I'll provide a summary for each problem.

But let's finish all quickly.

Problem 3:
4x + y < 2
y > x

First: y < -4x + 2 (dashed line, shade below)
Second: y > x (dashed line, shade above)

Intersection: -4x + 2 = x → 2=5x → x=2/5=0.4, y=0.4

Test (0,0): first: 0<2 true; second: 0>0 false → not in solution.

Test (0,1): first: 1<2 true; second: 1>0 true → in solution.

So region above y=x and below y=-4x+2. They cross at (0.4,0.4). For x<0.4, -4x+2 > x, so region between them. For x>0.4, -4x+2 < x, so no overlap.

Solution: triangular-like region above y=x and below y=-4x+2, for x < 0.4, not including either line.

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Problem 4:
y < 2x + 1
y > -1/3 x + 4

Both dashed lines.

Intersection: 2x + 1 = -1/3 x + 4
Multiply by 3: 6x + 3 = -x + 12
7x = 9 → x=9/7≈1.2857, y=2*(9/7)+1=18/7+7/7=25/7≈3.571

Test point (0,0): first: 0<1 true; second: 0>4 false → not in.

Test (0,5): first: 5<1 false.

Test (2,3): first: 3<5 true; second: 3 > -2/3 +4 = 10/3≈3.333? 3>3.333 false.

Test (1,3): first: 3<3 false.

Test (1,2.5): first: 2.5<3 true; second: 2.5 > -1/3 +4 = 11/3≈3.666? 2.5>3.666 false.

Perhaps try x=0: y<1 and y>4 → impossible.

x=3: y<7 and y> -1 +4=3 → so y between 3 and 7.

At x=3, y>3 and y<7.

Intersection at x=9/7≈1.2857, y=25/7≈3.571

For x > 9/7, 2x+1 > -1/3x+4, so region between them.

For example, at x=2, y<5 and y> -2/3+4=10/3≈3.333, so y in (3.333,5)

So solution is the region between the two lines for x > 9/7, not including the lines.

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Problem 5:
2x - y ≥ -6
x > 2

First: 2x - y ≥ -6 → -y ≥ -2x -6 → y ≤ 2x +6 (flip when dividing by -1)

Solid line y=2x+6, shade below.

Second: x>2, vertical dashed line at x=2, shade right.

Intersection: when x=2, y≤2*2+6=10, but x>2, so for x>2, y≤2x+6.

Test (3,0): first: 6-0=6≥-6 true; second: 3>2 true → in solution.

Solution: all points to the right of x=2 and below or on y=2x+6.

It's an infinite region starting from x=2 (not included) going right, under the line y=2x+6.

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Problem 6:
y < 1/2 x + 4
x - 2y ≥ 6

First: y < 0.5x + 4 (dashed, shade below)

Second: x - 2y ≥ 6 → -2y ≥ -x +6 → y ≤ 0.5x - 3 (divide by -2, flip inequality)

Solid line y=0.5x - 3, shade below.

Now, both are "below" their respective lines.

But let's see the lines: y=0.5x+4 and y=0.5x-3 — parallel lines, same slope.

y=0.5x+4 is above y=0.5x-3.

We need y < 0.5x+4 and y ≤ 0.5x-3.

Since 0.5x-3 < 0.5x+4 always, the stricter condition is y ≤ 0.5x-3.

And since y ≤ 0.5x-3 implies y < 0.5x+4 (because 0.5x-3 < 0.5x+4), so the solution is simply y ≤ 0.5x - 3.

Is that right? Let me check.

If y ≤ 0.5x - 3, then certainly y < 0.5x + 4, since 0.5x - 3 < 0.5x + 4.

And the second inequality is satisfied by equality allowed.

So yes, the solution is just the region below or on the line y = 0.5x - 3.

But let's verify with a point.

Take (0,-3): first: -3 < 0 +4 =4 true; second: 0 -2(-3)=6≥6 true → in solution.

Take (0,0): first: 0<4 true; second: 0 -0=0≥6? false → not in, and indeed 0 > 0.5*0 -3 = -3, so not below.

Take (6,0): first: 0<3+4=7 true; second: 6 -0=6≥6 true → in solution, and y=0 ≤ 0.5*6 -3=3-3=0 → on the line.

So yes, solution is y ≤ 0.5x - 3.

The first inequality is redundant because wherever the second is satisfied, the first is automatically satisfied.

So for problem 6, solution is all points on or below the line y = \frac{1}{2}x - 3.

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Now, to compile the final answers concisely.

Since the task is to sketch, and we can't sketch, I'll describe the solution region for each.

Final Answer:

1. The solution is the region below both lines \( y = -3x + 2 \) (solid) and \( y = 3x + 2 \) (dashed), forming a downward-pointing angle with vertex at (0,2). Includes the solid line, excludes the dashed line.

2. The solution is the region above \( y = 3x - 2 \) (solid) and below \( y = x \) (dashed), for \( x \leq 1 \). Vertex at (1,1), includes the solid line, excludes the dashed line.

3. The solution is the region above \( y = x \) (dashed) and below \( y = -4x + 2 \) (dashed), for \( x < 0.4 \). Excludes both lines.

4. The solution is the region between \( y = -\frac{1}{3}x + 4 \) (dashed) and \( y = 2x + 1 \) (dashed), for \( x > \frac{9}{7} \). Excludes both lines.

5. The solution is the region to the right of \( x = 2 \) (dashed) and below or on \( y = 2x + 6 \) (solid). Excludes the line \( x = 2 \), includes the line \( y = 2x + 6 \).

6. The solution is the region below or on the line \( y = \frac{1}{2}x - 3 \) (solid). The first inequality is automatically satisfied in this region.