Graphing Linear Inequalities Worksheet Answers - Free Printable

Let’s solve each system of inequalities step by step. We’ll graph each inequality and find where the shaded regions overlap — that’s the solution.

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Problem 1:

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

Step 1: Graph the first line \( y = \frac{5}{2}x - 2 \).
Slope = 5/2, y-intercept = -2.
Since it’s “≤”, shade below the line (solid line).

Step 2: Graph the second line \( y = \frac{1}{2}x + 2 \).
Slope = 1/2, y-intercept = 2.
Since it’s “≥”, shade above the line (solid line).

Step 3: The solution is where both shadings overlap — a small triangular region between the two lines, above the flatter line and below the steeper one.

Looking at the given graph for #1 — yes, the pink shaded area matches this description. So the sketch is correct.

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Problem 2:

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

Step 1: Graph \( y = \frac{5}{2}x + 2 \) → slope 5/2, y-int 2. Shade above (solid line).

Step 2: Graph \( y = \frac{1}{2}x - 2 \) → slope 1/2, y-int -2. Shade above (solid line).

Step 3: Solution is where both are true — above BOTH lines. That means the region above the higher of the two lines at any x-value. Since the steeper line rises faster, to the right they cross, but to the left, the flatter line is lower. Actually, let’s find intersection:

Set \( \frac{5}{2}x + 2 = \frac{1}{2}x - 2 \)

→ \( \frac{5}{2}x - \frac{1}{2}x = -2 - 2 \)

→ \( 2x = -4 \) → \( x = -2 \), then y = \( \frac{5}{2}(-2)+2 = -5+2 = -3 \)

So they intersect at (-2, -3). For x > -2, the steeper line is on top; for x < -2, the flatter line is on top? Wait no — actually since we’re shading ABOVE both, the solution is everything above the UPPER envelope of the two lines.

But looking at the graph provided for #2 — it shows shading over a large rectangle-like area covering most of the left side and bottom? That doesn’t match.

Wait — let me recheck.

Actually, if you shade above both lines, the overlapping region should be above the maximum of the two lines. At x=0: first line gives y≥2, second gives y≥-2 → so y≥2 dominates. At x=-4: first line y ≥ (5/2)(-4)+2 = -10+2 = -8; second line y ≥ (1/2)(-4)-2 = -2-2 = -4 → so y≥-4 dominates. So the boundary switches at x=-2.

The correct shaded region should be above both lines — which forms a V-shape opening upward with vertex at (-2,-3). But the graph shown in problem 2 has a huge rectangular shaded area including negative y-values and even parts below the lines — that looks wrong.

Wait — maybe I misread the graph. Let me think again.

Actually, looking back at the original image description — for problem 2, the graph shows shading over almost the entire left half-plane and some bottom part — that does NOT match “y ≥ both lines”. It seems like they might have shaded incorrectly.

But hold on — perhaps I made a mistake. Let me double-check the inequalities.

Given:
\( y \geq \frac{5}{2}x + 2 \) — steep line, shade above
\( y \geq \frac{1}{2}x - 2 \) — shallow line, shade above

The solution set is all points that satisfy BOTH — so above both lines. That would be the region that is above the higher of the two lines at every x. As calculated, they cross at (-2, -3). To the right of x=-2, the steep line is higher; to the left, the shallow line is higher? No — wait:

At x = 0:
Line 1: y = 2
Line 2: y = -2 → so line 1 is higher → need y ≥ 2

At x = -4:
Line 1: y = (5/2)(-4)+2 = -10+2 = -8
Line 2: y = (1/2)(-4)-2 = -2-2 = -4 → so line 2 is higher → need y ≥ -4

So the boundary of the solution region follows line 2 for x ≤ -2, and line 1 for x ≥ -2. So the shaded region should look like a "V" pointing down? No — pointing up? Actually, it's the area above both, so it's unbounded above, and bounded below by the piecewise linear function made of the two lines meeting at (-2,-3).

But the graph shown for problem 2 has a solid red box covering from x=-6 to x=0 and y=-6 to y=6 or something — that includes points like (0,0): check if it satisfies both:

For (0,0):
First inequality: 0 ≥ 0 + 2? → 0 ≥ 2? FALSE
Second: 0 ≥ 0 - 2? → 0 ≥ -2? TRUE
But since first is false, (0,0) should NOT be in solution. Yet in the graph it appears shaded — so the graph is incorrect.

Wait — this suggests there might be an error in the provided graphs? Or perhaps I’m misunderstanding.

Alternatively, maybe the student is supposed to verify or complete the sketch? The instruction says “Sketch the solution”, implying the graphs are partially done or examples.

Looking back at the user’s message — it says “Solve the problem accurately.” and the image contains four problems with graphs already drawn. Perhaps the task is to identify which graphs correctly represent the solutions?

That makes more sense. Because otherwise, why include the graphs?

Reinterpreting: The worksheet shows four systems with their proposed solution sketches. Our job is to determine which ones are correctly sketched.

Let’s do that.

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

Graph shows:
- Steep line (slope 5/2) going through (0,-2) — correct. Shaded below? In the graph, the pink region is between the two lines, above the shallow one and below the steep one — yes, that matches “y ≤ steep” and “y ≥ shallow”. And they intersect when:

\( \frac{5}{2}x - 2 = \frac{1}{2}x + 2 \)
→ \( 2x = 4 \) → x=2, y= (1/2)(2)+2=3 → point (2,3)

In the graph, the shaded triangle has vertices around there — looks correct.

✔ Problem 1 graph is CORRECT.

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

As before, solution is above both lines. Intersection at (-2,-3). The region should be above the upper envelope.

But the graph shows a large shaded rectangle covering left side and bottom — including points like (-4,0):

Check (-4,0):
First: 0 ≥ (5/2)(-4)+2 = -10+2 = -8 → TRUE
Second: 0 ≥ (1/2)(-4)-2 = -2-2 = -4 → TRUE
So (-4,0) IS in solution.

What about (0,0)?
First: 0 ≥ 0+2? → 0≥2? FALSE → not in solution.

In the graph for #2, is (0,0) shaded? From description, it seems the shading goes to x=0 and y=0 — probably includes (0,0), which is wrong.

Also, the graph has vertical and horizontal boundaries? Inequalities don't have those — solution should be unbounded.

Moreover, the shading appears to be cut off at x=0 and y=6 etc., which isn't right.

Actually, looking closely — in problem 2's graph, the shaded region is a polygon with corners at approximately (-6,-6), (-6,6), (0,6), (0,-2)? That doesn't make sense.

Perhaps it's shaded incorrectly. Let me calculate a test point.

Take point (-2,0):
First: 0 ≥ (5/2)(-2)+2 = -5+2 = -3 → TRUE
Second: 0 ≥ (1/2)(-2)-2 = -1-2 = -3 → TRUE → should be included.

Point (0,3):
First: 3 ≥ 0+2 → 3≥2 TRUE
Second: 3≥0-2 → 3≥-2 TRUE → should be included.

Point (0,1):
First: 1≥2? FALSE → not included.

In the graph for #2, if it shades up to y=6 at x=0, that might include (0,1) which is invalid.

But also, the graph seems to have a vertical line at x=0? No, probably not.

Another way: the only correct graph should show shading above both lines, forming a wedge starting at (-2,-3) and going up-right and up-left.

The given graph for #2 does not look like that — it looks like a rectangle. So likely INCORRECT.

But let's compare to others.

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

Note: second is strict inequality (>), so dashed line.

Graph shows:
- Shallow line y=(1/2)x+2, solid, shaded below?
- Steep line y=3x-3, dashed, shaded above?
Intersection: set (1/2)x+2 = 3x-3 → 2+3 = 3x - 0.5x → 5 = 2.5x → x=2, y=3(2)-3=3

So intersect at (2,3)

Solution: below shallow line AND above steep line (dashed).

Test point (0,0):
First: 0 ≤ 0+2 → TRUE
Second: 0 > 0-3 → 0>-3 TRUE → should be in solution.

In graph, is (0,0) shaded? Yes, appears so.

Another point (4,4):
First: 4 ≤ (1/2)(4)+2 = 2+2=4 → 4≤4 TRUE
Second: 4 > 3(4)-3=12-3=9? 4>9? FALSE → not in solution.

In graph, at x=4, y=4 is probably not shaded — good.

The shaded region is a triangle extending left-down from (2,3), which matches.

And the steep line is dashed — correct for strict inequality.

✔ Problem 3 graph is CORRECT.

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Problem 4:
Inequalities:
\( y < -3 \)
\( y \geq 4x + 1 \)

First: horizontal line y=-3, dashed (since <), shade below.

Second: line y=4x+1, solid, shade above.

Find intersection: set 4x+1 = -3 → 4x = -4 → x=-1, y=-3

So they meet at (-1,-3)

Solution: below y=-3 AND above y=4x+1.

But below y=-3 and above y=4x+1 — is that possible?

At x=-1, y must be < -3 and ≥ 4(-1)+1 = -3 → so y < -3 and y ≥ -3 → impossible. Only equal at point, but strict inequality.

For x < -1: say x=-2, then y ≥ 4(-2)+1 = -7, and y < -3. So for example y=-5: -5 < -3 TRUE, -5 ≥ -7 TRUE → valid.

For x > -1: say x=0, y ≥ 1, and y < -3 → impossible.

So solution exists only for x < -1, between the line y=4x+1 and y=-3.

Specifically, for each x < -1, y from 4x+1 up to but not including -3.

Now look at graph for #4:
It shows a horizontal dashed line at y=-3, shaded below — good.
Steep line y=4x+1, solid, shaded above — good.
Shaded region is a strip below y=-3 and above the steep line, which for x<-1 is a finite vertical strip? Actually, as x decreases, 4x+1 decreases, so the band gets wider downward.

In the graph, it shows shading in the bottom-left quadrant, bounded on the right by the steep line and on the top by y=-3 — which matches.

Test point (-2,-5):
y < -3? -5 < -3 TRUE
y ≥ 4(-2)+1 = -8+1=-7? -5 ≥ -7 TRUE → should be shaded — and it is.

Test point (0,-4):
y < -3? -4 < -3 TRUE
y ≥ 4(0)+1=1? -4 ≥ 1? FALSE → not in solution — and in graph, at x=0, nothing is shaded below y=-3 except possibly, but the steep line at x=0 is y=1, so above it would be y≥1, which conflicts with y<-3 — so no shading at x=0 — correct.

✔ Problem 4 graph is CORRECT.

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Now back to Problem 2.

We said solution is y ≥ max( (5/2)x+2 , (1/2)x-2 )

Which is equivalent to being above both lines.

The graph shown for #2 has a shaded region that includes, for example, the point (-4, -5):

Check:
First: y ≥ (5/2)(-4)+2 = -10+2 = -8 → -5 ≥ -8 TRUE
Second: y ≥ (1/2)(-4)-2 = -2-2 = -4 → -5 ≥ -4? FALSE → so (-4,-5) should NOT be in solution.

But in the graph, if it's shaded down to y=-6 at x=-4, it might include (-4,-5), which is invalid.

Moreover, the graph appears to have straight vertical and horizontal edges, which shouldn't be there.

Whereas for other problems, the shading follows the lines appropriately.

Therefore, Problem 2's graph is likely INCORRECT.

But let's confirm with another point.

Take point (-2, -3): intersection point.
y ≥ (5/2)(-2)+2 = -5+2 = -3 → -3 ≥ -3 TRUE
y ≥ (1/2)(-2)-2 = -1-2 = -3 → -3 ≥ -3 TRUE → should be included.

Point (-2, -4):
First: -4 ≥ -3? FALSE → not in solution.

In graph for #2, if it shades below y=-3 at x=-2, that would be wrong.

From the initial description, the graph for #2 has shading that goes down to y=-6 and across to x=0, which probably includes invalid points.

Thus, only Problems 1, 3, and 4 have correct sketches. Problem 2 does not.

But the problem says "Sketch the solution", and provides graphs — perhaps the student is to recognize which are correct, or maybe the graphs are meant to be completed.

Given that, and since the user said "solve the problem accurately", and the image has four items, I think the expected answer is to state which graphs are correct.

Perhaps in context, all are supposed to be evaluated.

Maybe for Problem 2, the graph is intended to show the correct region, but described poorly.

Another thought: in Problem 2, if we consider the inequalities again:

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

The solution is indeed the intersection of the two half-planes.

The boundary is the two rays from (-2,-3): one along y=(5/2)x+2 for x≥-2, and one along y=(1/2)x-2 for x≤-2, and above them.

The graph should show shading above this V-shaped boundary.

If the provided graph for #2 shows a different shape, it's wrong.

Given that, and since in my analysis Problems 1,3,4 are correct, and 2 is not, but let's see the answer format.

Perhaps the task is simply to accept the graphs as given and move on, but that doesn't make sense.

Another idea: maybe "solve" means to describe the solution region for each, but the graphs are provided as aids.

I think the safest approach is to assume that the graphs are part of the problem, and we need to verify them.

Based on careful checking:

- Problem 1: Correct
- Problem 2: Incorrect (shading includes points not satisfying both inequalities)
- Problem 3: Correct
- Problem 4: Correct

But since the user might expect the final answer to be the identification, or perhaps just to acknowledge.

Looking back at the response format, it says "Final Answer: [Provide only the final correct answer]"

Perhaps for this type of problem, the final answer is to list which ones are correctly sketched.

Maybe the problem is to solve each system, meaning to describe the solution set.

But that would be verbose.

Given the context, and since it's a worksheet with graphs, I believe the intent is to evaluate the sketches.

So I'll go with that.

Final Answer: Problems 1, 3, and 4 are correctly sketched. Problem 2 is not correctly sketched.

To be precise, let's box that.

But the instruction says "final correct answer", so perhaps list the numbers.

Since it's multiple parts, maybe state for each.

I think for clarity:

The correct sketches are for problems 1, 3, and 4. Problem 2's sketch is incorrect.

But to fit the format, perhaps:

Final Answer: 1, 3, 4

Or specify.

Another way: perhaps the problem expects us to say that all are correct except 2, but let's see.

I recall that in some worksheets, the graphs are pre-drawn and students select or something, but here it's "sketch", so maybe the graphs are the answers, and we need to verify.

I think I've spent enough time. Based on analysis:

Problem 1: correct
Problem 2: incorrect
Problem 3: correct
Problem 4: correct

So the final answer is that problems 1, 3, and 4 have accurate sketches, while problem 2 does not.

For the boxed answer, perhaps list the correct ones.

Since the user might expect a specific format, and given that in many such contexts, they want the evaluation, I'll put:

Final Answer: Problems 1, 3, and 4 are correctly solved; Problem 2 is not.

To make it concise:

Final Answer: 1, 3, 4

But let's confirm Problem 2 once more.

Suppose in Problem 2's graph, the shading is only where both are satisfied. If it's a large area, but perhaps it's correct if we interpret.

Take point (-6, -6) in Problem 2's graph:
y ≥ (5/2)(-6)+2 = -15+2 = -13 → -6 ≥ -13 TRUE
y ≥ (1/2)(-6)-2 = -3-2 = -5 → -6 ≥ -5? FALSE → so not in solution.

If the graph shades (-6,-6), it's wrong.

Similarly, point (0,0) as before is not in solution, but if shaded, wrong.

Whereas in Problem 1,3,4, the shading matches the inequalities.

So yes.

Final Answer: 1, 3, 4