Worksheet for practicing systems of linear inequalities with graphing exercises.

Let’s solve each problem one by one, step by step.

---

Problem 1: Is (0, 0) a solution to the system of linear inequalities?

Looking at the graph:

- There are two lines:
- One is a solid diagonal line going through the origin (y = x).
- The other is a dashed vertical line at x = -2.
- The shaded region is to the right of x = -2 and above or on y = x.

Check point (0, 0):

- Is it to the right of x = -2? → Yes, because 0 > -2.
- Is it above or on y = x? → At x=0, y=x means y=0. So (0,0) is ON the line → which is included since the line is solid.

✔ So yes, (0,0) is in the shaded region → it IS a solution.

---

Problem 2: Which ordered pair is a solution to the graphed system?

Graph shows:

- Two dashed lines forming an “X” shape.
- Shaded region is the bottom triangle, below both lines.
- Lines appear to be:
- One with positive slope: maybe y = x + 1? (passes through (-1,0), (0,1))
- One with negative slope: maybe y = -x -1? (passes through (-1,0), (0,-1))

But we don’t need equations — just test the points in the shaded area.

Shaded region is below both lines, around the bottom center.

Test each option:

a. (2, 4) → too high up → not in shaded region
b. (-1, -3) → let’s see: x=-1, y=-3. That’s low down. Check if under both lines.
- For line going up: at x=-1, line is at y=0 → -3 < 0 → OK
- For line going down: at x=-1, line is at y=0 → -3 < 0 → OK
→ This looks like it’s in the shaded region.

c. (-2, 3) → too high → no
d. (-3, -1) → x=-3, y=-1. Let’s check:
- Upward line at x=-3: y = -3 + 1 = -2 → so y=-1 is ABOVE that line → NOT in shaded region (shaded is BELOW both)

So only b. (-1, -3) is in the shaded region.

✔ Answer: b

---

Problem 3: Is (-2, 4) a solution to the system?

Graph has:

- Two solid lines crossing.
- Shaded region is to the right of both lines? Actually, looking carefully:

One line goes from top-left to bottom-right (negative slope).
Other goes from bottom-left to top-right (positive slope).
Shaded region is the area that is below the downward-sloping line and above the upward-sloping line? Wait — actually, the shading is on the side where both conditions are met.

Point (-2, 4): plot it — x=-2, y=4 → that’s left and high up.

Look at graph: the shaded region is mostly on the right side, starting around x=0 or so. Point (-2,4) is far left and high — definitely outside the shaded region.

To confirm: suppose the lines are approximately:

- Line 1 (downward): passes through (0,2) and (2,0) → equation y = -x + 2
- Line 2 (upward): passes through (0,1) and (2,3) → equation y = x + 1

At x = -2:

- On line 1: y = -(-2)+2 = 4 → so point (-2,4) is ON this line → but is the inequality ≤ or ≥? Since line is solid, and shading is below it? In the graph, for the downward line, shading is BELOW it → so y ≤ -x+2 → at (-2,4): 4 ≤ 4 → OK.

On line 2: y = x+1 → at x=-2, y=-1 → point (-2,4) is WAY above → and shading is ABOVE the upward line? Looking at graph: for the upward line, shading is above it → so y ≥ x+1 → 4 ≥ -1 → true.

Wait — so according to this, (-2,4) satisfies both?

But visually, in the graph, the shaded region does NOT include (-2,4). Why?

Because the two lines intersect somewhere around (0.5, 1.5), and the shaded region is the "wedge" to the RIGHT of the intersection.

Actually, re-examining the graph: the shaded region is bounded between the two lines, but only for x greater than about 0. So even though mathematically (-2,4) might satisfy the inequalities if extended, in the context of the graph shown, the shading stops at the intersection and doesn't go left.

But wait — the problem says “is (-2,4) a solution to the system?” — meaning, does it satisfy BOTH inequalities represented by the graph?

We need to deduce the inequalities from the graph.

Assume:

Line A (downward sloping, solid): passes through (0,2) and (2,0) → y = -x + 2
Since shading is below it → y ≤ -x + 2

Line B (upward sloping, solid): passes through (0,1) and (1,2) → y = x + 1
Shading is above it → y ≥ x + 1

Now test (-2,4):

First inequality: y ≤ -x + 2 → 4 ≤ -(-2) + 2 → 4 ≤ 2 + 2 → 4 ≤ 4 → TRUE

Second inequality: y ≥ x + 1 → 4 ≥ -2 + 1 → 4 ≥ -1 → TRUE

So mathematically, it satisfies both.

But why isn’t it shaded in the graph? Because the graph only shows a portion — the actual solution set extends infinitely, but the drawing cuts off. The point (-2,4) is part of the solution set, even if not visibly shaded in the small window.

However, looking again at the graph: the shaded region is only the area that is simultaneously below the first line AND above the second line — which forms a wedge opening to the right. But at x=-2, the two lines are at y=4 and y=-1, so any y between -1 and 4 would satisfy? No — wait:

For a given x, y must be ≥ (x+1) and ≤ (-x+2)

At x=-2: y ≥ -1 and y ≤ 4 → so y=4 is allowed.

But in the graph, at x=-2, is there any shading? The graph's x-axis starts around -3 or -4, and at x=-2, the shaded region should be between y=-1 and y=4 — but in the image, the shading seems to start only after x=0 or so. That might be due to the scale or clipping.

Actually, looking closely at Problem 3’s graph: the shaded region is ONLY where both inequalities overlap, and it appears as a triangular-like region on the right. But mathematically, the solution set is all points where y ≥ x+1 and y ≤ -x+2 — which is a band between the two lines, extending infinitely left and right? No — wait, when you have y ≥ x+1 and y ≤ -x+2, the region is bounded on the left and right? Let’s find where the lines intersect:

Set x+1 = -x+2 → 2x = 1 → x=0.5, y=1.5

For x < 0.5, say x=0: y≥1 and y≤2 → possible
For x=-2: y≥-1 and y≤4 → still possible
But as x decreases, the lower bound y≥x+1 decreases, upper bound y≤-x+2 increases — so the region actually widens to the left.

In the graph provided, however, the shading is only shown for x ≥ some value — probably because the grid is limited. But the point (-2,4) is on the boundary of the upper line, and since the line is solid, it should be included.

Moreover, in the graph, at x=-2, if you go up to y=4, it’s on the line, and the shading includes the line — so technically, it should be considered a solution.

But let me double-check the visual: in the image for Problem 3, the shaded region does NOT extend to x=-2; it starts around x=0. However, that might be an artifact of how much was drawn. The mathematical solution includes (-2,4).

Perhaps I made a mistake in identifying the lines.

Alternative approach: pick a point clearly in the shaded region, say (2,2):

- Is it below the downward line? Downward line at x=2 is y=0? No, earlier I said y=-x+2, at x=2, y=0 — but (2,2) is above that — contradiction.

I think I misidentified the lines.

Look again at Problem 3 graph:

- One line goes from top-left to bottom-right: passes through (0,3) and (3,0)? Or (0,2) and (2,0)? Let's count grids.

Assuming each grid is 1 unit.

Downward line: when x=0, y=2; when x=2, y=0 → so y = -x + 2

Upward line: when x=0, y=1; when x=2, y=3 → y = x + 1

At x=1: downward line y=1, upward line y=2 — so for x=1, y must be ≥2 and ≤1? Impossible. That can't be.

Ah! I see the mistake. If the upward line is y=x+1 and downward is y=-x+2, then for them to have a overlapping region, we need y ≥ x+1 and y ≤ -x+2, but at x=1, x+1=2, -x+2=1, so 2 ≤ y ≤ 1 — impossible. So my assumption about which side is shaded is wrong.

Looking at the graph: the shaded region is where the two half-planes overlap. From the arrows and shading, it seems that for the downward-sloping line, the shading is BELOW it, and for the upward-sloping line, the shading is ALSO BELOW it? But that wouldn't make sense with the picture.

Perhaps for the upward line, shading is ABOVE, and for downward, shading is BELOW, but they only overlap for x ≤ 0.5.

At x=0: y ≥ 1 (from upward line) and y ≤ 2 (from downward line) — so y between 1 and 2 is shaded.

At x=1: y ≥ 2 and y ≤ 1 — no overlap.

So the solution region is only for x ≤ 0.5.

At x=-2: y ≥ -1 and y ≤ 4 — so y=4 is on the boundary, and since the line is solid, it should be included.

And in the graph, although the shading may not be drawn at x=-2 due to space, mathematically it is part of the solution.

Moreover, the point (-2,4) is on the downward line, and the shading includes the line, so yes.

But let's verify with another point. Take (0,1.5): should be in shaded region.

y ≥ 0+1 =1, y ≤ -0+2=2 → 1≤1.5≤2 → yes.

Take (-2,4): y ≥ -2+1 = -1, y ≤ -(-2)+2=4 → 4≤4 and 4≥-1 → yes.

So it is a solution.

However, in the actual image, the shaded region might not include it because the graph is cut off, but based on the inequalities implied by the lines and shading direction, it should be included.

Perhaps the upward line is different. Let's assume the upward line passes through (0,0) and (2,2) — but in the graph, at x=0, the upward line is at y=1, not 0.

Another idea: perhaps the upward line is y = (1/2)x + 1 or something. But to simplify, since the problem is multiple choice or true/false, and based on standard interpretation, (-2,4) satisfies the inequalities if we take the lines as y = -x + 2 (solid, shade below) and y = x + 1 (solid, shade above).

And since the point is on the boundary of the first inequality and inside the second, it should be a solution.

But let's look back at the user's image description — in Problem 3, the shaded region is only on the right side, and (-2,4) is on the left, so perhaps in the context of the graph as drawn, it's not considered, but that doesn't make sense mathematically.

I recall that in some textbooks, if the graph shows a bounded region, they mean only that part, but here the lines have arrows, indicating they extend infinitely, so the solution set is unbounded.

Given that, I think (-2,4) is a solution.

But let's calculate numerically.

Suppose the two inequalities are:

1. y ≤ -x + 2 (downward line, solid, shade below)
2. y ≥ x + 1 (upward line, solid, shade above)

Then for (-2,4):

1. 4 ≤ -(-2) + 2 = 2 + 2 = 4 → 4 ≤ 4 → true
2. 4 ≥ -2 + 1 = -1 → true

So yes.

Perhaps the upward line is different. Let's say it passes through (-2,0) and (0,2) — then slope = (2-0)/(0-(-2)) = 2/2 = 1, so y = x + 2? At x=0, y=2, but in the graph at x=0, the upward line is at y=1, not 2.

In the graph for Problem 3, at x=0, the upward line is at y=1, downward at y=2.

So y = x + 1 and y = -x + 2 are correct.

Therefore, (-2,4) is a solution.

But I have a feeling that the intended answer is no, because in the graph, the shading doesn't reach there. Perhaps the shading is only for the region where both are satisfied and x is greater than the intersection.

Let's find the intersection: x+1 = -x+2 -> 2x=1 -> x=0.5, y=1.5

For x < 0.5, the region y ≥ x+1 and y ≤ -x+2 is valid, and for x > 0.5, it's empty because x+1 > -x+2 when x>0.5.

At x=0.6, x+1=1.6, -x+2=1.4, so 1.6 > 1.4, no overlap.

So the solution region is for x ≤ 0.5, and y between x+1 and -x+2.

At x= -2, y between -1 and 4, so (-2,4) is on the upper boundary.

And since the line is solid, it is included.

So I think the answer is yes.

But let's move on and come back if needed.

---

Problem 4: Solve by graphing: y < 3x - 4 and y ≥ -2x + 1

Step 1: Graph y = 3x - 4

- Slope 3, y-intercept -4.
- When x=0, y=-4
- When x=2, y=6-4=2
- Since it's y < , draw dashed line, shade below.

Step 2: Graph y = -2x + 1

- Slope -2, y-intercept 1.
- When x=0, y=1
- When x=1, y=-1
- Since y ≥ , draw solid line, shade above.

Step 3: Find overlap of shaded regions.

Find intersection point: set 3x - 4 = -2x + 1

3x +2x = 1 +4 → 5x=5 → x=1, y=3(1)-4= -1

So lines intersect at (1, -1)

Now, the solution is the region that is below the first line and above the second line.

For example, test point (2,0):

- y < 3x-4? 0 < 6-4=2 → 0<2 true
- y ≥ -2x+1? 0 ≥ -4+1= -3 → 0≥-3 true

So (2,0) is in solution.

The solution region is to the right of x=1, between the two lines.

Since we're to solve by graphing, we describe the region or sketch it.

But for the answer, since it's "solve", we can say the solution is all points (x,y) such that y < 3x-4 and y ≥ -2x+1, which is the region below the line y=3x-4 and above y=-2x+1, for x > 1.

At x=1, y must be ≥ -1 and < -1? 3(1)-4= -1, so y < -1 and y ≥ -1 — no solution at x=1.

For x>1, it works.

For example, x=2, y<2 and y≥-3, so y in [-3,2)

But typically, we leave it as the graphical region.

Since the problem asks to solve by graphing, and provides a blank grid, the expected answer is to graph it, but since we're text-based, we can describe.

But for final answer, perhaps they want the description or the intersection.

In many contexts, "solve" means find the solution set, which is the region.

But for this format, since it's practice, and other problems have specific answers, perhaps for this one, we just note that the solution is the region satisfying both.

But let's see the other problems.

Perhaps for Problems 4,5,6, since they say "solve by graphing", and provide grids, the answer is the graph, but since we can't draw, we might skip or describe.

But the user expects a final answer for each.

Looking back, for Problem 1,2,3, we have yes/no or choice, for 4,5,6, it's to graph, so perhaps no numerical answer, but the instruction is to provide final answer.

Perhaps for those, the final answer is the description or the key point.

To save time, let's do Problems 5 and 6 similarly.

---

Problem 5: Solve by graphing: 2x + y > -4 and x - 2y ≤ 4

First, rewrite in slope-intercept form.

1. 2x + y > -4 → y > -2x -4

2. x - 2y ≤ 4 → -2y ≤ -x +4 → divide by -2, reverse inequality: y ≥ (1/2)x - 2

So:

- y > -2x -4 : dashed line, slope -2, y-intercept -4, shade above
- y ≥ (1/2)x - 2 : solid line, slope 1/2, y-intercept -2, shade above

Find intersection: set -2x -4 = (1/2)x - 2

Multiply both sides by 2: -4x -8 = x -4

-4x - x = -4 +8 → -5x =4 → x= -4/5 = -0.8

y = -2(-0.8) -4 = 1.6 -4 = -2.4

Or from other: (1/2)(-0.8) -2 = -0.4 -2 = -2.4

So intersect at (-0.8, -2.4)

Solution region is above both lines, which is most of the plane except the wedge below.

Test point (0,0):

- y > -2x-4? 0 > 0 -4 = -4 → true
- y ≥ 0.5x -2? 0 ≥ 0 -2 = -2 → true

So (0,0) is in solution.

The region is above both lines.

---

Problem 6: Solve by graphing: y ≤ (2/3)x + 2 and y > -x -3

- First line: y = (2/3)x + 2, solid (since ≤), shade below
- Second line: y = -x -3, dashed (since >), shade above

Find intersection: (2/3)x + 2 = -x -3

(2/3)x + x = -3 -2 → (5/3)x = -5 → x = -5 * 3/5 = -3

y = -(-3) -3 = 3-3=0

Or from first: (2/3)(-3) +2 = -2+2=0

So intersect at (-3,0)

Solution region is below the first line and above the second line.

Test point (0,0):

- y ≤ (2/3)(0)+2 =2 → 0≤2 true
- y > -0 -3 = -3 → 0>-3 true

So (0,0) is in solution.

Region is between the two lines, for x > -3 or something.

At x=0, y ≤2 and y>-3, so yes.

Now, back to Problem 3.

After re-thinking, in the graph for Problem 3, the shaded region is only the area that is below the downward-sloping line and above the upward-sloping line, and since at x= -2, the upward line is at y= -1, and downward at y=4, and the point (-2,4) is on the downward line, and the shading includes the line, it should be included.

Moreover, in many similar problems, boundary points on solid lines are included.

So I'll go with yes.

But let's confirm with the initial quick assessment.

Perhaps the upward line is y = -x + c or something else.

Another way: in the graph, the two lines intersect at approximately (0.5,1.5), and the shaded region is the triangle-like shape to the right, but since the lines have arrows, it should extend.

I think it's safe to say yes.

For the sake of completing, let's list the answers.

Final Answers:

1. Yes
2. b. (-1, -3)
3. Yes
4. The solution is the region where y < 3x - 4 and y ≥ -2x + 1, which is below the line y=3x-4 and above y= -2x+1, for x > 1.
5. The solution is the region where y > -2x -4 and y ≥ (1/2)x - 2, which is above both lines.
6. The solution is the region where y ≤ (2/3)x + 2 and y > -x -3, which is below the first line and above the second line.

But for Problems 4,5,6, since they ask to solve by graphing, and no specific answer is expected beyond the graph, perhaps in this context, we can omit or state the region.

Since the user might expect a boxed answer for each, and for 4,5,6, it's graphical, perhaps for those, the final answer is "see graph" but that's not satisfactory.

Looking at the worksheet, for Problems 4,5,6, there are blank grids, so the student is to graph, so no numerical answer.

But the instruction is to provide final answer, so for those, we can say the solution set is described by the inequalities.

To comply, I'll provide the key point or something.

For Problem 4, the lines intersect at (1,-1), and the solution is for x>1, etc.

But let's see the format.

Perhaps for this response, since ExplainAnswer is false, we only provide final answer, but the rules say to work through first.

I think for clarity, I'll provide the answers as follows.

After careful consideration:

Problem 1: Yes

Problem 2: b

Problem 3: Yes (based on mathematical satisfaction)

Problem 4: The solution is all points (x,y) such that y < 3x - 4 and y ≥ -2x + 1. Graphically, it is the region below the dashed line y=3x-4 and above the solid line y= -2x+1, which occurs for x > 1.

Problem 5: The solution is all points (x,y) such that 2x + y > -4 and x - 2y ≤ 4, or equivalently y > -2x -4 and y ≥ (1/2)x - 2. Graphically, it is the region above both lines.

Problem 6: The solution is all points (x,y) such that y ≤ (2/3)x + 2 and y > -x -3. Graphically, it is the region below the solid line y=(2/3)x+2 and above the dashed line y= -x -3.

For the Final Answer section, since it's multiple parts, I'll list them.

But the instruction is "provide only the final correct answer", so perhaps for each problem.

Since it's a single response, I'll box the answers for 1,2,3, and for 4,5,6, describe briefly.

To match the format, let's do:

Final Answer:
1. Yes
2. b
3. Yes
4. Region below y=3x-4 and above y=-2x+1 for x>1
5. Region above y=-2x-4 and above y=0.5x-2
6. Region below y=(2/3)x+2 and above y=-x-3

But for 5, "above both" is fine.

For Problem 3, I recall that in some interpretations, if the graph shows a bounded region, but here it's not bounded, so yes.

Upon second thought, in the graph for Problem 3, the shaded region is only where the two half-planes overlap, and at x= -2, it is included, so yes.

So I'll go with that.

Final Answer:

1. Yes
2. b
3. Yes
4. The solution is the region where \( y < 3x - 4 \) and \( y \geq -2x + 1 \), which is below the line \( y = 3x - 4 \) and above the line \( y = -2x + 1 \), occurring for \( x > 1 \).
5. The solution is the region where \( 2x + y > -4 \) and \( x - 2y \leq 4 \), or equivalently \( y > -2x - 4 \) and \( y \geq \frac{1}{2}x - 2 \), which is above both lines.
6. The solution is the region where \( y \leq \frac{2}{3}x + 2 \) and \( y > -x - 3 \), which is below the line \( y = \frac{2}{3}x + 2 \) and above the line \( y = -x - 3 \).

For the boxed format, since it's multiple, perhaps list them.

Since the user might expect concise answers, and for 4,5,6, the graph is the answer, but in text, this is fine.

I think this is acceptable.