Graphing Inequalities in Two Variables Worksheet for 8th - 12th ... - Free Printable

Let’s solve each inequality step by step. We’ll graph them on a coordinate plane. Remember:
- If the inequality is < or >, draw a dashed line.
- If it’s ≤ or ≥, draw a solid line.
- Then shade the region that satisfies the inequality.

---

Problem 1: \( y > 3x \)

Step 1: Graph the line \( y = 3x \).
This line passes through (0,0) and has slope 3 → go up 3, right 1 from origin.

Step 2: Since it’s “>”, use a dashed line.

Step 3: Test a point not on the line — like (0,1).
Is \( 1 > 3(0) \)? → \( 1 > 0 \) → TRUE. So shade the side containing (0,1) — which is above the line.

✔ Final graph: Dashed line through origin with slope 3, shaded above.

---

Problem 2: \( y < -2 \)

Step 1: This is a horizontal line at \( y = -2 \).

Step 2: Since it’s “<”, use a dashed line.

Step 3: Shade below the line (because y-values less than -2 are below).

✔ Final graph: Horizontal dashed line at y = -2, shaded below.

---

Problem 3: \( y - x < -2 \)

First, rewrite in slope-intercept form:
\( y < x - 2 \)

Step 1: Graph \( y = x - 2 \).
Y-intercept: (0, -2), slope = 1 → go up 1, right 1.

Step 2: “<” means dashed line.

Step 3: Test (0,0): Is \( 0 < 0 - 2 \)? → \( 0 < -2 \)? FALSE.
So shade the opposite side — below the line.

✔ Final graph: Dashed line with slope 1, y-intercept -2, shaded below.

---

Problem 4: \( y - x \leq 6 \)

Rewrite: \( y \leq x + 6 \)

Step 1: Graph \( y = x + 6 \).
Y-intercept: (0,6), slope = 1.

Step 2: “≤” means solid line.

Step 3: Test (0,0): Is \( 0 ≤ 0 + 6 \)? → \( 0 ≤ 6 \)? TRUE.
Shade the side containing (0,0) — below the line.

✔ Final graph: Solid line with slope 1, y-intercept 6, shaded below.

---

Problem 5: \( 3x - y > -8 \)

Rewrite to solve for y:
Subtract 3x: \( -y > -3x - 8 \)
Multiply by -1 (flip inequality sign!): \( y < 3x + 8 \)

Step 1: Graph \( y = 3x + 8 \).
Y-intercept: (0,8), slope = 3.

Step 2: “<” → dashed line.

Step 3: Test (0,0): Is \( 0 < 3(0) + 8 \)? → \( 0 < 8 \)? TRUE.
Shade the side containing (0,0) — below the line.

✔ Final graph: Dashed line with slope 3, y-intercept 8, shaded below.

---

Problem 6: \( 3x - y > 4 \)

Rewrite:
Subtract 3x: \( -y > -3x + 4 \)
Multiply by -1 (flip sign): \( y < 3x - 4 \)

Step 1: Graph \( y = 3x - 4 \).
Y-intercept: (0,-4), slope = 3.

Step 2: “<” → dashed line.

Step 3: Test (0,0): Is \( 0 < 3(0) - 4 \)? → \( 0 < -4 \)? FALSE.
Shade the opposite side — below the line? Wait — let’s double-check.

Actually, since (0,0) gives false, we shade the side NOT containing (0,0).
The line \( y = 3x - 4 \) goes through (0,-4) and (1,-1).
Point (0,0) is ABOVE the line. Since test fails, shade BELOW the line.

Wait — let’s pick another test point to be sure. Try (2,0):
Is \( 0 < 3(2) - 4 \)? → \( 0 < 6 - 4 = 2 \)? TRUE.
(2,0) is below the line? Let’s see: At x=2, line y = 3(2)-4 = 2. So (2,0) is below y=2 → yes, below the line.

So shading below is correct.

✔ Final graph: Dashed line with slope 3, y-intercept -4, shaded below.

---

Final Answer:

For each problem, you should have drawn:

1. \( y > 3x \): Dashed line through origin, slope 3, shaded above.
2. \( y < -2 \): Horizontal dashed line at y = -2, shaded below.
3. \( y - x < -2 \) → \( y < x - 2 \): Dashed line, slope 1, y-int -2, shaded below.
4. \( y - x \leq 6 \) → \( y \leq x + 6 \): Solid line, slope 1, y-int 6, shaded below.
5. \( 3x - y > -8 \) → \( y < 3x + 8 \): Dashed line, slope 3, y-int 8, shaded below.
6. \( 3x - y > 4 \) → \( y < 3x - 4 \): Dashed line, slope 3, y-int -4, shaded below.

Make sure your graphs match these descriptions exactly — including dashed/solid lines and correct shading direction.