Mathematics › 📐 Grade 8 › Graphing Two-Variable Linear Inequalities

Graphing Two-Variable Linear Inequalities

Swap the equals sign for an inequality symbol, and a line stops being a boundary you sit on and becomes an edge between two entire regions.

Grade 8Algebra 1

Graphing Two-Variable Linear Inequalities — illustration

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The big idea: A linear equation like y = mx + b draws one line of solutions. Change the equals sign to an inequality — y > mx + b, or ≥, <, ≤ — and the solution set explodes from a line into an entire half-plane: every point on one side of the line. The line itself becomes the boundary, drawn dashed when its own points don't count and solid when they do, and a single test point tells you which side to shade.

🎯 By the end, you'll be able to

Graph y > mx + b, y ≥ mx + b, y < mx + b, and y ≤ mx + b as a shaded half-plane
Explain when the boundary line is dashed (excluded) versus solid (included)
Use a test point, usually the origin, to decide which side of the line to shade
Check whether a given ordered pair is a solution to a two-variable inequality

📎 You should already know

Graphing linear functions in slope-intercept form (grade8__factory)
Plotting ordered pairs on a coordinate plane

From one line to a whole region

You already know that y = mx + b graphs as a single straight line — every point on that line makes the equation true. A two-variable inequality like y > mx + b asks a looser question: which points make the left side bigger than the right side? The answer is no longer a line — it's an entire half-plane, everything on one side of the boundary line.

🔑 The boundary line: dashed or solid

Graph the boundary as if it were y = mx + b. If the inequality is strict (> or <), draw the line dashed — points exactly on it are not solutions. If the inequality includes equality (≥ or ≤), draw the line solid — points on it are included.

\[ y > mx + b \quad y \ge mx + b \quad y < mx + b \quad y \le mx + b \]

Same boundary line, four different shaded regions depending on the symbol.

Picking a side with a test point

Once the boundary is drawn, half the plane satisfies the inequality and half doesn't — but which half? Pick any point that is clearly not on the line, usually the origin (0, 0) because it's easy to compute with, and substitute its coordinates into the original inequality. If the inequality comes out true, shade the side containing that test point; if false, shade the other side.

🎮 Half-Plane Shader LIVE

Graph the boundary line and shade the solution region for a two-variable inequality.

📝 Worked example: Graph y < 2x − 1.

Graph the boundary y = 2x − 1 (slope 2, y-intercept −1). Since the inequality is strict (<), draw it dashed.
Test the origin (0, 0), which is not on the line: is 0 < 2(0) − 1, i.e. 0 < −1? That's false.
Since the test point fails, shade the side of the line that does not contain (0, 0).

✓ The solution is the half-plane below the dashed line y = 2x − 1 — the side away from the origin.

📝 Worked example: Graph y ≥ −x + 3 and state whether (1, 1) is a solution.

Graph the boundary y = −x + 3. Since the inequality includes equality (≥), draw it solid.
Test the origin: is 0 ≥ −(0) + 3, i.e. 0 ≥ 3? That's false, so shade the side away from the origin (above the line).
Check (1, 1): is 1 ≥ −(1) + 3, i.e. 1 ≥ 2? That's false.

✓ The shaded region is above and including the solid line y = −x + 3, but (1, 1) is not a solution since it fails the inequality.

⚠️ Don't assume 'shade above' — and mind the boundary

It's tempting to guess that > always means “shade above the line,” but that only happens to work when the inequality is already solved for y with a plus sign in front. Always confirm with a test point instead of guessing. And remember what dashed versus solid encodes: a point sitting exactly on the boundary satisfies ≥ or ≤, but it never satisfies a strict > or <.

✨ Rusty on one-variable inequalities?

This lesson assumes you can already isolate a variable in an inequality like px + q > r, flipping the symbol when you multiply or divide by a negative. If that feels rusty, it's worth revisiting the Grade 7 lesson on solving two-step inequalities before tackling these coordinate-plane versions.

A real-world constraint

Two-variable inequalities show up naturally as budgets or limits. Suppose tickets for adults cost $5 and tickets for children cost $3, and a group has at most $60 to spend. If x is the number of adult tickets and y the number of child tickets, the constraint is 5x + 3y ≤ 60. Every point (x, y) in the shaded solution region, restricted to whole numbers of tickets that make sense, represents an affordable combination.

📝 Worked example: A group has at most $60 for tickets: adult tickets cost $5 and child tickets cost $3. Is buying 8 adult tickets and 5 child tickets affordable? The constraint is 5x + 3y ≤ 60.

Substitute x = 8 and y = 5 into the left side: 5(8) + 3(5) = 40 + 15.
That totals 55.
Compare to the budget: is 55 ≤ 60?

✓ Yes — 55 ≤ 60 is true, so (8, 5) lies in the solution region and the group can afford that combination.

Check your understanding

1. For the inequality y > 3x + 2, how should the boundary line be drawn?

Strict inequalities (> or <) exclude the boundary itself, so the line is drawn dashed.

2. To decide which side of the boundary line to shade, you should:

There's no universal 'shade above/below' rule; a test point (commonly the origin) tells you which side is correct.

3. Does the point (2, 3), which lies exactly on the boundary line y = x + 1, satisfy y ≥ x + 1?

Substituting gives 3 ≥ 3, which is true, so a point exactly on the boundary does satisfy a ≥ (or ≤) inequality — that's what 'solid line, included' means.

4. For y < −x + 4, testing the origin (0, 0) gives 0 < 4, which is true. What does this tell you?

When the test point makes the inequality true, you shade the region containing that test point.

5. A budget constraint is 5x + 3y ≤ 60. Is the point (10, 5) a solution?

5(10) + 3(5) = 50 + 15 = 65, which is greater than 60, so (10, 5) fails the constraint and is not a solution.

✅ Key takeaways

A two-variable linear inequality's solution set is an entire half-plane, not just a line.
The boundary is graphed like y = mx + b, then drawn dashed for strict > or < and solid for ≥ or ≤.
A test point not on the line — usually the origin — tells you which side to shade: true means shade that side.
A point on a dashed boundary is never a solution; a point on a solid boundary always is.
Real-world limits like budgets translate directly into inequalities such as 5x + 3y ≤ 60, with every point in the shaded region a feasible combination.