Worksheet displaying four systems of inequalities with corresponding graphs illustrating their solution sets.

Let’s go step by step to check each system of inequalities and see if the shaded region in the graph matches the solution.

We are solving systems of two linear inequalities. The solution is the region where both inequalities are true — that’s the overlapping shaded area.

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Problem 1:

Inequalities:
① \( y \leq \frac{5}{2}x - 2 \)
② \( y \geq \frac{1}{2}x + 2 \)

Step 1: Graph both lines.

- Line ①: slope = 5/2, y-intercept = -2 → goes through (0, -2), and for every 2 right, up 5.
- Line ②: slope = 1/2, y-intercept = 2 → goes through (0, 2), and for every 2 right, up 1.

Step 2: Shade correctly.

- For ①: “≤” means shade below the line.
- For ②: “≥” means shade above the line.

The solution is where both shadings overlap.

Looking at the graph provided for #1:
→ The red shaded region is above the lower line (which should be y ≥ ½x + 2) and below the steeper line (y ≤ ⁵⁄₂x – 2).
→ That matches! Also, the intersection point can be found by setting them equal:

Set \( \frac{5}{2}x - 2 = \frac{1}{2}x + 2 \)
Subtract ½x from both sides:
\( 2x - 2 = 2 \)
Add 2:
\( 2x = 4 \) → x = 2
Then y = ½(2) + 2 = 1 + 2 = 3 → Point (2, 3)

Graph shows shading starting around there — looks correct.

✔ Problem 1 graph is correct.

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Problem 2:

Inequalities:
① \( y \geq \frac{5}{2}x + 2 \)
② \( y \geq \frac{1}{2}x - 2 \)

Both are “≥”, so we shade above both lines.

Line ①: steep, starts at (0,2)
Line ②: less steep, starts at (0,-2)

Since both are “greater than or equal”, the solution is the region that is above BOTH lines — which will be the area above the higher of the two lines at any given x.

But looking at the graph for #2:
→ It shades a large rectangle on the left side? Wait — actually, it shades everything to the left of some vertical boundary? No — let me look again.

Actually, in the graph for #2, they’ve shaded a big pink region that includes areas BELOW one of the lines? Let’s test a point.

Take origin (0,0):

Check inequality ①: y ≥ ⁄₂x + 2 → 0 ≥ 0 + 2? → 0 ≥ 2? ✘ False
So (0,0) should NOT be in solution.

But in the graph for #2, (0,0) is inside the shaded region? Actually, looking closely — no, wait: the shaded region in #2 seems to be mostly on the LEFT side, including negative x-values.

Try point (-2, 0):

: y ≥ ⁵⁄₂(-2) + 2 = -5 + 2 = -3 → 0 ≥ -3 ✔
②: y ≥ ½(-2) - 2 = -1 - 2 = -3 → 0 ≥ -3 ✔
So (-2,0) should be included — and it appears to be shaded. Good.

Now try point (0,0) again:

As above, fails ① → so should NOT be shaded. In the graph, is (0,0) shaded? Looking at the grid — the shaded region stops before x=0? Actually, in the image, the shaded region for #2 covers from about x=-6 to x=0, and y from bottom to top? But that doesn’t make sense because for large positive y, even if x is negative, it might still satisfy.

Wait — actually, since both inequalities are “≥”, the solution should be unbounded upward. The graph shows a bounded rectangular shade? That can’t be right.

Hold on — I think there's a mistake in the graph for #2.

Let me find the intersection point:

Set \( \frac{5}{2}x + 2 = \frac{1}{2}x - 2 \)
Subtract ½x:
\( 2x + 2 = -2 \)
Subtract 2:
\( 2x = -4 \) → x = -2
Then y = ½(-2) - 2 = -1 - 2 = -3 → Point (-2, -3)

For x > -2, the line y = ⁵⁄₂x + 2 is ABOVE y = ½x - 2, so the stricter condition is y ≥ ⁵₂x + 2.

For x < -2, the line y = ½x - 2 is ABOVE y = ⁵⁄₂x + 2? Let’s check at x = -4:

①: y ≥ ⁵⁄₂(-4)+2 = -10+2 = -8
②: y ≥ ½(-4)-2 = -2-2 = -4
So for x=-4, we need y ≥ -4 (since -4 > -8) — so the binding constraint is the second inequality.

Thus, the solution region is:

- For x ≤ -2: above y = ½x - 2
- For x ≥ -2: above y = ⁵⁄₂x + 2

This forms a V-shaped region opening upwards, with vertex at (-2, -3).

But in the graph for #2, they have shaded a rectangle-like area covering from x≈-6 to x=0, and y from ≈-4 to y=6? That includes points like (0,0) which we saw does NOT satisfy the first inequality.

Also, for example, take point (-3, 0):

①: y ≥ ⁵⁄₂(-3)+2 = -7.5 + 2 = -5.5 → 0 ≥ -5.5 ✔
②: y ≥ ½(-3)-2 = -1.5 -2 = -3.5 → 0 ≥ -3.5 ✔ → should be included.

Point (-1, 0):

①: y ≥ ⁵⁄₂(-1)+2 = -2.5 + 2 = -0.5 → 0 ≥ -0.5 ✔
②: y ≥ ½(-1)-2 = -0.5 -2 = -2.5 → 0 ≥ -2.5 ✔ → should be included.

Wait — earlier I thought (0,0) fails ①:
At x=0, ① requires y ≥ 2 → so (0,0) has y=0 < 2 → fails. So (0,0) should NOT be shaded.

In the graph for #2, is (0,0) shaded? Looking at the image — yes, it appears to be inside the pink region. That’s wrong.

Moreover, the graph shows a hard vertical cutoff at x=0? There’s no reason for that — the inequalities don’t involve x bounds.

✘ So Problem 2 graph is INCORRECT. It incorrectly includes points like (0,0) and possibly cuts off too early.

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Problem 3:

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

Note: Second inequality is strict (>), so dashed line; first is ≤, solid line.

Find intersection:

Set \( \frac{1}{2}x + 2 = 3x - 3 \)
Multiply both sides by 2:
x + 4 = 6x - 6
4 + 6 = 6x - x
10 = 5x → x = 2
Then y = 3(2) - 3 = 6 - 3 = 3 → Point (2,3)

Shading:

- ①: below or on the line y = ½x + 2
- ②: strictly above y = 3x - 3 (dashed line)

Solution is between the two lines, above the steep dashed line and below the flatter solid line, for x < 2? Let’s see.

For x < 2, which line is on top? At x=0:

①: y ≤ 2
: y > -3
So between y=-3 and y=2 — but specifically above y=3x-3 and below y=½x+2.

At x=2, they meet at (2,3). For x>2, say x=4:

①: y ≤ ½(4)+2 = 4
②: y > 3(4)-3 = 9 → so y > 9 and y ≤ 4 → impossible. So no solution for x>2.

Thus, solution is for x < 2, between the two lines.

Graph for #3:
→ Shows shaded region between the two lines, left of x=2, with dashed line for the steeper one (correct, since >), solid for the other.
→ Shading is below the flat line and above the steep dashed line — perfect.

✔ Problem 3 graph is correct.

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Problem 4:

Inequalities:
① \( y < -3 \) → horizontal dashed line at y=-3, shade below
② \( y \geq 4x + 1 \) → solid line, slope 4, y-int 1, shade above

Find intersection: set 4x + 1 = -3 → 4x = -4 → x = -1
Point (-1, -3)

But note: ① is strict (<), so y=-3 is not included. ② is ≥, so includes the line.

At x=-1, y must be ≥ 4(-1)+1 = -3, AND y < -3 → so only possible if y=-3, but y<-3 excludes it. So actually, at x=-1, no solution? Wait — let’s think.

The solution region is where y < -3 AND y ≥ 4x + 1.

So combine: 4x + 1 ≤ y < -3

Which implies 4x + 1 < -3 → 4x < -4 → x < -1

So for x < -1, we can have y between 4x+1 and -3 (not including -3).

Graph for #4:
→ Shows dashed horizontal line at y=-3, shaded below — good for ①.
→ Solid line for y=4x+1, which passes through (-1,-3) and (0,1) — correct.
→ Shaded region is below y=-3 and above y=4x+1 — which for x < -1, that’s a triangular wedge going down-left.

In the graph, they show shading in the bottom-left, bounded above by y=-3 (dashed) and below/right by the line y=4x+1 — and it stops at x=-1? Actually, since for x >= -1, 4x+1 >= -3, so y >= 4x+1 and y < -3 cannot both be true. So shading should only be for x < -1.

In the graph, it looks like they shaded only for x < -1, below y=-3 and above the line — which is correct.

Also, at x=-2:
y ≥ 4(-2)+1 = -7
y < -3
So y between -7 and -3 — and graph shows that.

✔ Problem 4 graph is correct.

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## Summary:

Problem 1: Correct
Problem 2: Incorrect — shaded region includes points that don't satisfy both inequalities (like (0,0)), and shape is wrong (should be V-shaped opening up, not a rectangle)
Problem 3: Correct
Problem 4: Correct

But the task was to "sketch the solution" — and the graphs are already drawn. Since this is likely a worksheet where students are to verify or draw, and based on our analysis, only #2 is wrong.

However, the user didn’t ask to identify errors — they just said “solve the problem”. But the problem says “Sketch the solution to each system”, and sketches are already provided. Perhaps the intent is to confirm correctness? Or maybe to describe the solution regions?

Given the context, and since all except #2 are correct, but the instruction is to solve — perhaps we should provide the correct description or correction for #2.

But re-reading the initial request: “Solve the problem accurately.” And the problem is to sketch the solution — which is already done in the image. Maybe the student is supposed to do it themselves, but here the images are given as answers? Confusing.

Alternatively, perhaps the task is to determine which graphs are correct? But the worksheet doesn’t say that.

Another interpretation: maybe the user wants us to solve each system algebraically or describe the solution set.

Given the ambiguity, but following the rule to provide a final answer after reasoning, and since three are correct and one is not, but the problem is presented as four separate sketching tasks, I think the expected output is to state whether each is correct or not — but the format doesn’t specify.

Wait — looking back at the user’s message: they uploaded an image of a worksheet titled “Solving Systems of Inequalities” with four problems, each with a graph. The instruction is “Sketch the solution...”, implying the student is to draw it. But in the image, the graphs are already drawn — so perhaps this is a key or completed worksheet, and the user wants verification.

Given that, and since my job is to help the student, I should point out that #2 is incorrect and explain why.

But the response format requires a “Final Answer” section. What should go there?

Perhaps for each problem, state if the graph is correct or not.

Since the problem says “solve”, and solving means finding the solution region, I could describe the solution region for each.

But to keep it simple and match the format, I’ll assume the task is to verify the given graphs, and report which are correct.

Final decision: After careful checking, Problems 1, 3, and 4 have correct graphs. Problem 2 has an incorrect graph.

But how to box that?

Perhaps list them.

Since the user might expect the solution regions described, but the graphs are visual, I think the best Final Answer is to state the correctness.

Alternatively, since the worksheet is from Kuta Software, and often these are for practice, maybe the student is to compare their sketch to the given one — but again, not specified.

I think safest is to conclude that three are correct, one is not, and specify which.

But let me double-check Problem 2 once more.

In Problem 2, inequalities are both ≥, so solution is intersection of two half-planes above the lines.

The two lines intersect at (-2, -3).

For any point above both lines, it must be above the upper envelope.

The upper envelope is:
- For x <= -2, the line y = (1/2)x - 2 is above y = (5/2)x + 2? At x=-2, both -3. At x=-4, y=(1/2)(-4)-2=-4, y=(5/2)(-4)+2=-10+2=-8, so yes, (1/2)x-2 is higher when x<-2.
- For x>=-2, (5/2)x+2 is higher.

So the solution region is all points (x,y) such that:
if x <= -2, then y >= (1/2)x - 2
if x >= -2, then y >= (5/2)x + 2

This is a convex region extending infinitely upward, with a "corner" at (-2,-3).

The graph in #2 shows a bounded region, roughly from x=-6 to x=0, y=-4 to y=6, which is a rectangle — that is completely wrong. It should be unbounded above, and shaped like a V opening up.

Moreover, as tested, (0,0) is in the shaded region but does not satisfy y >= (5/2)(0)+2 = 2, since 0<2.

So definitely incorrect.

Therefore, the accurate solution is that graphs for 1,3,4 are correct; graph for 2 is incorrect.

But since the problem is to "sketch the solution", and sketches are provided, perhaps the final answer is to identify the error.

Given the constraints, I'll put in Final Answer that Problem 2's graph is incorrect, others are correct.

To be precise:

Final Answer:
Problems 1, 3, and 4 have correctly sketched solutions. Problem 2 has an incorrectly sketched solution; the shaded region should be the area above both lines, forming an unbounded V-shape with vertex at (-2, -3), not a bounded rectangle. Points like (0,0) are incorrectly included in the shaded region for Problem 2.