Systems of Inequalities Worksheets with Answer Key - Free Printable

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:
Inequalities:
- 3x + y ≤ 2
- 6x - 2y > -4

Step 1: Graph 3x + y = 2 (solid line because ≤)
→ 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 ≤).

Step 2: Graph 6x - 2y = -4 (dashed line because >)
Simplify: divide by 2 → 3x - y = -2 → y = 3x + 2
→ When x=0, y=2 → (0,2)
→ When x=-1, y=-1 → (-1,-1)
Shade ABOVE the line? Wait — let’s test a point.

Original: 6x - 2y > -4
Test (0,0): 0 - 0 = 0 > -4 → TRUE → so shade side containing (0,0), which is BELOW the line y=3x+2? Let’s check:

Wait — if we rearrange 6x - 2y > -4 → subtract 6x: -2y > -6x -4 → divide by -2 (flip inequality): y < 3x + 2

Ah! So it’s y < 3x + 2 → dashed line, shade BELOW.

So both inequalities are shaded below their lines? But wait — first was 3x+y≤2 → y ≤ -3x + 2 → also shade below.

But let’s double-check with a test point in the overlap.

Try (0,0):
First: 3(0)+0=0 ≤ 2 → yes
Second: 6(0)-2(0)=0 > -4 → yes → so (0,0) is in solution.

Now, do the lines intersect?

Set y = -3x + 2 and y = 3x + 2 → set equal: -3x+2 = 3x+2 → -6x=0 → x=0 → y=2

So they meet at (0,2). The solution is the region below both lines — but since one is solid and one dashed, and they cross at (0,2), the solution is the wedge-shaped area below both, including the solid line but not the dashed.

Actually, since both are “below”, and they cross at (0,2), the solution is the region that is below BOTH lines — which is the area below the lower of the two lines for each x.

For x < 0: y=3x+2 is below y=-3x+2? At x=-1: y=3(-1)+2=-1; y=-3(-1)+2=5 → so y=3x+2 is lower → so for x<0, boundary is y<3x+2
For x>0: y=-3x+2 is lower → so boundary is y≤-3x+2

But actually, since we’re shading below both, the solution is the intersection — which is the region that satisfies both. Since (0,0) works, and it’s below both, the solution is the region bounded by the two lines from (0,2) downward.

We can sketch it as the area below both lines, starting from their intersection at (0,2).

Final for #1: Shade the region that is below or on the line y = -3x + 2 AND below the line y = 3x + 2. It looks like a V-shape opening downward with vertex at (0,2), including the left arm (solid) but not the right arm (dashed)? Wait no — let's clarify:

Line 1: y ≤ -3x + 2 → solid line, shade below
Line 2: y < 3x + 2 → dashed line, shade below

At x=0, both give y≤2 and y<2 → so at x=0, y must be less than 2 (since second is strict).

The two lines cross at (0,2). For x>0, the line y=-3x+2 is below y=3x+2, so the binding constraint is y≤-3x+2. For x<0, y=3x+2 is below y=-3x+2, so binding constraint is y<3x+2.

So the solution region is:
- For x ≥ 0: below or on y = -3x + 2
- For x < 0: below y = 3x + 2

And since at x=0, y<2 (from second inequality), the point (0,2) is NOT included.

So the solution is all points below both lines, not including the dashed line, and including the solid line except at (0,2) where it meets the dashed.

This is getting complex — perhaps better to just say: graph both lines, shade below both, and the overlapping shaded region is the solution. Include the solid line part, exclude the dashed line.

In practice, students would draw:

- Solid line for 3x+y=2, shade below.
- Dashed line for 6x-2y=-4 (or y=3x+2), shade below.
- The solution is where shadings overlap — which is a region that includes points like (0,0), (-1,0), (0.5,0), etc., but not (0,2).

Okay, moving on — I'll summarize each quickly with key points.

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Problem 2:
6x - 2y ≤ 4 → simplify: 3x - y ≤ 2 → y ≥ 3x - 2 (remember to flip when dividing by negative? Wait:

6x - 2y ≤ 4 → -2y ≤ -6x + 4 → divide by -2 → y ≥ 3x - 2 (yes, flipped)

Second: y < x

Graph:
- y ≥ 3x - 2 → solid line, shade above
- y < x → dashed line, shade below

Find intersection: set 3x - 2 = x → 2x=2 → x=1, y=1

Test point (0,0):
First: 0 ≥ 0 - 2 → 0≥-2 true
Second: 0 < 0? false → so (0,0) not in solution.

Test (2,0):
First: 0 ≥ 6-2=4? 0≥4 false

Test (0,-3):
First: -3 ≥ 0 -2 → -3≥-2 false

Test (1,0):
First: 0 ≥ 3-2=1? 0≥1 false

Test (0.5, 0):
First: 0 ≥ 1.5 - 2 = -0.5 → true
Second: 0 < 0.5 → true → so (0.5,0) is in solution.

So solution is region above y=3x-2 and below y=x, between their intersection at (1,1) and extending... actually, since y≥3x-2 and y<x, the region is between the two lines for x < 1? Let's see:

At x=0, y≥-2 and y<0 → so y from -2 to 0 (not including 0)

At x=1, y≥1 and y<1 → impossible → so only for x<1

The lines cross at (1,1), and for x<1, y=3x-2 is below y=x, so the region is above the solid line and below the dashed line, for x < 1.

Include the solid line, exclude the dashed line.

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Problem 3:
4x + y < 2 → y < -4x + 2 (dashed, shade below)
y > x (dashed, shade above)

Intersection: set -4x+2 = x → 2=5x → x=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

Test (0.5, 0.6):
First: 0.6 < -4*0.5 +2 = -2+2=0 → 0.6<0 false

Test (0,0.5):
First: 0.5 < 2 true
Second: 0.5 > 0 true → in solution

So solution is above y=x and below y=-4x+2, which is a small region near (0.4,0.4), but since both are strict, not including boundaries.

The region is between the two lines for x < 0.4.

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Problem 4:
y < 2x + 1 (dashed, shade below)
y > -1/3 x + 4 (dashed, shade above)

Find 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 (0,0):
First: 0<1 true
Second: 0>4 false → not in solution

Test (0,5):
First: 5<1 false

Test (3,3):
First: 3<6+1=7 true
Second: 3 > -1 +4=3 → 3>3 false

Test (3,3.5):
First: 3.5<7 true
Second: 3.5 > -1 +4=3 → true → in solution

So solution is above y=-1/3x+4 and below y=2x+1, which is a region around the intersection point, extending to the right? Let's see slopes: y=2x+1 has steep positive slope, y=-1/3x+4 has gentle negative slope, so they cross once, and the region between them is where y is greater than the lower line and less than the upper line.

Since at x=0, y>4 and y<1 — impossible, so only for x > 9/7 approximately.

For x > 9/7, y=2x+1 is above y=-1/3x+4, so the region is between them.

Both lines dashed, so exclude boundaries.

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Problem 5:
2x - y ≥ -6 → -y ≥ -2x -6 → y ≤ 2x + 6 (solid, shade below)
x > 2 (dashed vertical 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

Test (2,0): x>2 false

Test (3,10): y≤6+6=12, 10≤12 true, x>2 true → in solution

Test (3,13): 13≤12 false

So solution is to the right of x=2 (not including x=2) and below or on y=2x+6.

It's an infinite region to the right, under the line.

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Problem 6:
y < 1/2 x + 4 (dashed, shade below)
x - 2y ≥ 6 → -2y ≥ -x +6 → y ≤ 1/2 x - 3 (solid, shade below) [divided by -2, flipped]

Wait: x - 2y ≥ 6 → -2y ≥ -x +6 → divide by -2: y ≤ (1/2)x - 3

But first inequality is y < (1/2)x + 4

So we have:
y < (1/2)x + 4
y ≤ (1/2)x - 3

Note that (1/2)x - 3 is always 7 units below (1/2)x + 4, since 4 - (-3)=7.

So y ≤ (1/2)x - 3 implies y < (1/2)x + 4 automatically, because if y is less than or equal to something that is 7 less than the other, it's definitely less than the other.

So the solution is just y ≤ (1/2)x - 3, since it satisfies both.

Is that right? Let me check.

Suppose y = (1/2)x - 3, then first inequality: (1/2)x - 3 < (1/2)x + 4 → -3 < 4 true.

If y < (1/2)x - 3, still true.

But what if y = (1/2)x - 2? Then second inequality: y ≤ (1/2)x - 3? -2 ≤ -3? False. So not satisfied.

So indeed, the second inequality is stricter: y must be ≤ (1/2)x - 3, and that automatically satisfies y < (1/2)x + 4.

So solution is y ≤ (1/2)x - 3, solid line, shade below.

But let's confirm with a point.

Test (0, -4):
First: -4 < 0 +4 =4 true
Second: 0 -2*(-4)=8 ≥6 true → in solution

Test (0, -3):
First: -3<4 true
Second: 0 -2*(-3)=6≥6 true → in solution (since ≥)

Test (0, -2):
First: -2<4 true
Second: 0 -2*(-2)=4≥6? false → not in solution

And y ≤ (1/2)(0) -3 = -3, so at x=0, y≤-3, which matches.

Also, the line y=(1/2)x -3 is solid, and we shade below.

The first inequality is redundant.

So for #6, solution is simply the region below or on the line y = (1/2)x - 3.

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Now, to present final answers clearly for each problem.

Since this is a graphing task, the "answer" is the description of the solution region or how to graph it. But since the user expects a final answer, and it's multiple problems, I'll summarize each.

But the instruction says "provide only the final correct answer" after reasoning. However, for graphing problems, the answer is the graph itself, which we can't draw here. So perhaps describe the solution region for each.

Given the context, I think the expected final answer is to state for each problem what the solution set is, or perhaps list the key features.

But looking back at the user's request: "Sketch the solution for each system" — so the answer should be the sketched graphs, but since we're text-based, we describe them.

Perhaps for the purpose of this exercise, we can say the solution regions as described.

But to follow the format, I'll provide a concise summary for each.

Final Answer:

1. The solution is the region below or on the line \( y = -3x + 2 \) and below the line \( y = 3x + 2 \), excluding the dashed line \( y = 3x + 2 \). The lines intersect at (0,2), and the region includes points like (0,0).

2. The solution is the region above or on the line \( y = 3x - 2 \) and below the line \( y = x \), for \( x < 1 \). The lines intersect at (1,1), and the region includes points like (0.5, 0).

3. The solution is the region above the line \( y = x \) and below the line \( y = -4x + 2 \), for \( x < 0.4 \). The lines intersect at (0.4, 0.4), and neither boundary is included.

4. The solution is the region above the line \( y = -\frac{1}{3}x + 4 \) and below the line \( y = 2x + 1 \), for \( x > \frac{9}{7} \). The lines intersect at approximately (1.29, 3.57), and neither boundary is included.

5. The solution is the region to the right of the line \( x = 2 \) (not including it) and below or on the line \( y = 2x + 6 \).

6. The solution is the region below or on the line \( y = \frac{1}{2}x - 3 \), as it satisfies both inequalities.

Note: In actual graphing, students would shade these regions on the coordinate planes provided.