Worksheet on graphing two-variable systems of inequalities with solutions.

Let’s go through each problem one by one. For each, we’ll check if the given point satisfies both inequalities in the system. If it does, then it’s a solution to the system. If not, then it’s not.

We’ll plug the x and y values of the point into both inequalities and see if they make true statements.

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Problem 1:
Inequalities:
- y > -¾x + 4
- y > ½x - 1
Point: (4, 1)

Check first inequality:
y > -¾(4) + 4 → 1 > -3 + 4 → 1 > 1? ✘ No, because 1 is NOT greater than 1 — it’s equal. So this fails.

Even though the second might work, since one fails, the point is NOT a solution.

Wait — let’s double-check the graph too. The shaded region is where BOTH are true. Point (4,1) is on the boundary of the first line (dashed), so it’s not included. Correct — not a solution.

But wait — the answer key says (4,1). Let me recheck.

Actually, looking at the graph for #1: the pink shaded area includes (4,1)? Let’s test again.

First inequality: y > -¾x + 4
At x=4: -¾*4 = -3 → -3 + 4 = 1 → y > 1 → but y=1 → 1 > 1? False.

Second inequality: y > ½x - 1 → ½*4 = 2 → 2 - 1 = 1 → y > 1 → again, 1 > 1? False.

So (4,1) does NOT satisfy either inequality strictly. But maybe the graph shows it as part of the solution? Wait — no, both lines are dashed, meaning strict inequality — boundaries not included.

Hmm… perhaps there’s a mistake? Or maybe I misread the point?

Wait — look at the graph: the intersection point of the two lines is around (4,1), and the shading is above both lines. At (4,1), you’re ON both lines — so not in the solution set.

But the answer written is (4,1). That must be wrong? Or maybe the problem is asking for something else?

Wait — rereading the title: “Two Variable System of Inequalities - Graphing” — and under each graph, a point is listed. Maybe the task is to verify if that point is in the solution region?

Looking at graph #1: the pink region is above both lines. Point (4,1) is exactly at the intersection — which is NOT included because both lines are dashed. So (4,1) should NOT be a solution.

But perhaps the question is just listing the point shown? Or maybe it’s a trick?

Wait — let’s check problem #2.

Problem 2:
Inequalities:
- y ≥ -5/2 x + 10
- y ≤ ½x + 4
Point: (2,5)

Check first: y ≥ -5/2*(2) + 10 → -5 + 10 = 5 → 5 ≥ 5? ✔ Yes.

Second: y ≤ ½*(2) + 4 → 1 + 4 = 5 → 5 ≤ 5? ✔ Yes.

So (2,5) works — and it’s on both boundaries, which are solid lines — so included. Good.

Back to #1 — maybe the point is meant to be tested, and the answer is that it’s NOT a solution? But the sheet lists (4,1) as if it’s correct.

Wait — perhaps I made a calculation error.

For #1: y > -¾x + 4 at (4,1):
-¾ * 4 = -3; -3 + 4 = 1 → y > 1 → 1 > 1? False.

Similarly, y > ½x -1 → ½*4=2; 2-1=1 → y>1 → 1>1? False.

So definitely not satisfied.

But looking at the graph: the green line is y = ½x -1, blue is y = -¾x +4. They intersect when:

½x -1 = -¾x +4
Multiply both sides by 4: 2x -4 = -3x +16
2x +3x = 16+4 → 5x=20 → x=4
Then y= ½*4 -1 = 2-1=1 → so (4,1) is intersection.

Since both inequalities are strict (>), the solution region does NOT include the boundary — so (4,1) is not in the solution set.

Therefore, if the task is to find which points are solutions, (4,1) is not.

But the worksheet has (4,1) written below — perhaps it’s indicating the intersection point, not necessarily a solution? Or maybe the student is supposed to identify the solution region, and the point is just labeled?

This is confusing.

Wait — looking at all problems, each has a point listed below the graph. And in some cases, like #3, point (-2,-2) — let’s check that.

Problem 3:
Inequalities:
- y < 2x + 2
- y < -2
Point: (-2,-2)

First: y < 2*(-2)+2 → -4+2=-2 → y < -2 → but y=-2 → -2 < -2? ✘ False.

Second: y < -2 → -2 < -2? ✘ False.

Again, not satisfied. But the lines: y=2x+2 is dashed, y=-2 is dashed. Point (-2,-2) is on both lines? Let’s see:

If x=-2, y=2*(-2)+2=-4+2=-2 → yes, on first line.

And y=-2 → on second line.

So again, on boundary, not included.

But in the graph, the shaded region is below both lines — so (-2,-2) is not in it.

Yet the answer is listed as (-2,-2).

This suggests that perhaps the point listed is the intersection point of the two lines, not necessarily a solution to the system.

Maybe the task is to find the intersection point of the two boundary lines? That would make sense, because in every case, the point given is where the two lines cross.

Let’s verify with #4.

Problem 4:
Inequalities:
- -x + 2y ≤ 6
- y ≥ 2
Point: (-2,2)

Find intersection of boundaries:
Boundary 1: -x + 2y = 6
Boundary 2: y = 2

Plug y=2 into first: -x + 4 = 6 → -x=2 → x=-2 → so (-2,2) — matches.

And in the graph, the lines intersect at (-2,2).

Similarly, #5: point (-4,-3)

Boundaries: y = 3/2 x + 3 and y = -3

Set equal: 3/2 x + 3 = -3 → 3/2 x = -6 → x = -4 → yes, (-4,-3)

#6: (-3,-1)

Boundaries: 2x+3y=-9 and x=-3

Plug x=-3: 2*(-3)+3y=-9 → -6+3y=-9 → 3y=-3 → y=-1 → yes.

#7: (4,1)

Boundaries: -5x+4y=-16 and x+4y=8

Solve:

From second: x = 8 - 4y

Plug into first: -5(8-4y) +4y = -16 → -40 +20y +4y = -16 → 24y = 24 → y=1

Then x=8-4*1=4 → (4,1) — yes.

#8: (3,-1)

Boundaries: 5x+3y=12 and -x+3y=-6

Subtract second from first: (5x+3y) - (-x+3y) = 12 - (-6) → 6x = 18 → x=3

Then from second: -3 +3y = -6 → 3y=-3 → y=-1 → (3,-1)

So in every case, the point listed is the intersection point of the two boundary lines.

Therefore, the task is likely: "Find the point of intersection of the two boundary lines for each system."

Not whether it's in the solution region, but where the lines cross.

That makes sense now.

So for each problem, we solve the system of equations formed by replacing the inequality signs with equals signs.

Let’s do that for each.

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Problem 1:
Equations:
y = -¾x + 4
y = ½x - 1

Set equal:
-¾x + 4 = ½x - 1
Multiply both sides by 4 to eliminate denominators:
-3x + 16 = 2x - 4
16 + 4 = 2x + 3x
20 = 5x → x=4
Then y= ½*4 -1 = 2-1=1 → (4,1)

✔ Matches.

Problem 2:
y = -5/2 x + 10
y = ½x + 4

Set equal:
-5/2 x + 10 = ½x + 4
Multiply by 2: -5x + 20 = x + 8
20 - 8 = x + 5x → 12 = 6x → x=2
y= ½*2 +4 =1+4=5 → (2,5)

✔

Problem 3:
y = 2x + 2
y = -2

Set equal: 2x + 2 = -2 → 2x = -4 → x=-2 → y=-2 → (-2,-2)

✔

Problem 4:
-x + 2y = 6
y = 2

Substitute y=2: -x + 4 = 6 → -x=2 → x=-2 → (-2,2)

✔

Problem 5:
y = 3/2 x + 3
y = -3

Set equal: 3/2 x + 3 = -3 → 3/2 x = -6 → x = -6 * 2/3 = -4 → y=-3 → (-4,-3)

✔

Problem 6:
2x + 3y = -9
x = -3

Substitute x=-3: 2*(-3) + 3y = -9 → -6 + 3y = -9 → 3y = -3 → y=-1 → (-3,-1)

✔

Problem 7:
-5x + 4y = -16
x + 4y = 8

Subtract second equation from first:
(-5x + 4y) - (x + 4y) = -16 - 8
-6x = -24 → x=4

Then from second: 4 + 4y = 8 → 4y=4 → y=1 → (4,1)

✔

Problem 8:
5x + 3y = 12
-x + 3y = -6

Subtract second from first:
(5x + 3y) - (-x + 3y) = 12 - (-6)
6x = 18 → x=3

Then from second: -3 + 3y = -6 → 3y = -3 → y=-1 → (3,-1)

✔

All match.

So the task was to find the intersection point of the two boundary lines for each system.

Final Answer:
The points of intersection for each system are:
1) (4,1)
2) (2,5)
3) (-2,-2)
4) (-2,2)
5) (-4,-3)
6) (-3,-1)
7) (4,1)
8) (3,-1)