Graphing Linear Inequalities in Two Variables

A linear inequality in two variables looks like a linear equation but uses an inequality symbol instead of an equals sign. Instead of graphing a single line, you graph an entire region — a half-plane of points that satisfy the inequality.

For example, $y > 2x + 1$ does not describe a line. It describes every point $(x, y)$ where the $y$-coordinate is greater than $2x + 1$. That is an infinite set of points filling one side of the line $y = 2x + 1$.

The Three-Step Process

Every linear inequality in two variables is graphed using the same three steps:

1. Graph the boundary line (the related equation)
2. Choose solid or dashed based on the inequality symbol
3. Shade the correct side by testing a point

Step 1: Graph the Boundary Line

Replace the inequality symbol with an equals sign and graph that line using any method you know (slope-intercept, intercepts, table of values).

For $y > 2x + 1$, the boundary line is $y = 2x + 1$.

This line has slope 2 and $y$-intercept 1. Plot the intercept at $(0, 1)$, then use the slope to find another point: up 2, right 1 gives $(1, 3)$. Draw the line through these points.

Step 2: Solid or Dashed?

The inequality symbol tells you whether points on the boundary line are included in the solution:

For $y > 2x + 1$, the symbol is $>$ (strict), so the boundary line is dashed. Points exactly on the line $y = 2x + 1$ are not part of the solution.

For $y \leq 2x + 1$, the symbol is $\leq$ (non-strict), so the boundary line would be solid. Points on the line are included.

Step 3: Shade the Correct Side

The boundary line divides the coordinate plane into two half-planes. One side contains the solutions; the other does not. To determine which side to shade, test a point that is not on the boundary line.

The easiest test point is $(0, 0)$ — as long as the line does not pass through the origin.

For $y > 2x + 1$, test $(0, 0)$:

$0 > 2(0) + 1$ $0 > 1$

This is false, so $(0, 0)$ is not in the solution region. Shade the opposite side of the line from $(0, 0)$ — that is, shade above the line.

If the test point makes the inequality true, shade the side that contains the test point.

Worked Examples

Example 1: Graph $y \leq -x + 4$

Step 1 — Boundary line: $y = -x + 4$. Slope is $-1$, $y$-intercept is 4. Points: $(0, 4)$ and $(4, 0)$.

Step 2 — Line style: The symbol is $\leq$, so draw a solid line.

Step 3 — Test $(0, 0)$:

$0 \leq -(0) + 4$ $0 \leq 4 \quad \checkmark$

True! Shade the side that contains $(0, 0)$ — the region below the line.

Solution region: The solid line through $(0, 4)$ and $(4, 0)$, with shading below and to the left.

Example 2: Graph $2x + 3y > 12$

Step 1 — Boundary line: $2x + 3y = 12$. Find intercepts:

• $x$-intercept: set $y = 0 \implies 2x = 12 \implies x = 6$. Point: $(6, 0)$
• $y$-intercept: set $x = 0 \implies 3y = 12 \implies y = 4$. Point: $(0, 4)$

Step 2 — Line style: The symbol is $>$, so draw a dashed line.

Step 3 — Test $(0, 0)$:

$2(0) + 3(0) > 12$ $0 > 12$

False. Shade the side that does not contain $(0, 0)$ — the region above the line.

Example 3: Graph $y \geq \frac{1}{2}x$

Step 1 — Boundary line: $y = \frac{1}{2}x$. This line passes through the origin with slope $\frac{1}{2}$.

Step 2 — Line style: The symbol is $\geq$, so draw a solid line.

Step 3 — Test point: Since the line passes through $(0, 0)$, choose a different test point. Try $(0, 2)$:

$2 \geq \frac{1}{2}(0)$ $2 \geq 0 \quad \checkmark$

True. Shade the side containing $(0, 2)$ — the region above the line.

Important: When the boundary passes through the origin, pick any convenient point that is clearly on one side, such as $(0, 1)$, $(1, 0)$, or $(0, -1)$.

Writing Inequalities from Graphs

You can also work backward — given a graph showing a boundary line and shading, write the inequality.

Steps:

1. Find the equation of the boundary line (use slope and intercept)
2. Determine if the line is solid ($\leq$ or $\geq$) or dashed ($<$ or $>$)
3. Pick a point in the shaded region and test which inequality direction is true

Example 4: Identify the Inequality

A graph shows a dashed line through $(0, 3)$ and $(3, 0)$ with shading below the line.

Step 1 — Find the equation. Slope: $\frac{0 - 3}{3 - 0} = -1$. The $y$-intercept is 3. Equation: $y = -x + 3$.

Step 2 — Dashed line means strict inequality ($<$ or $>$).

Step 3 — Test a point in the shaded region. The shading is below the line, so test $(0, 0)$:

$0 \; ? \; -(0) + 3$ $0 \; ? \; 3$

Since $0 < 3$ and the shading includes $(0, 0)$:

Answer: $y < -x + 3$

Inequalities in Standard Form

When the inequality is in standard form like $Ax + By \leq C$, you do not need to convert to slope-intercept form to graph it. Just find two points on the boundary line (the intercepts work well), draw the line, and test a point.

Example 5: Graph $3x - 2y \geq 6$

Boundary line: $3x - 2y = 6$

Intercepts:

• $x$-intercept: $3x = 6 \implies x = 2$. Point: $(2, 0)$
• $y$-intercept: $-2y = 6 \implies y = -3$. Point: $(0, -3)$

Line style: $\geq$ means solid.

Test $(0, 0)$:

$3(0) - 2(0) \geq 6$ $0 \geq 6$

False. Shade the side that does not contain $(0, 0)$ — below and to the right of the line.

Horizontal and Vertical Boundary Lines

Some inequalities produce horizontal or vertical boundary lines.

Horizontal: $y \leq 3$ has boundary line $y = 3$ (a horizontal line). Shade below the solid line.

Vertical: $x > -2$ has boundary line $x = -2$ (a vertical line). Shade to the right of the dashed line.

These are simpler because you don’t need to test a point — the inequality directly tells you which direction to shade.

Systems of Linear Inequalities (Introduction)

A system of linear inequalities is two or more inequalities graphed on the same coordinate plane. The solution is the region where all shaded areas overlap.

Example 6: Graph the System

$y \leq x + 3$ $y > -2x + 1$

Graph each inequality separately:

First inequality: Boundary $y = x + 3$ (solid line, slope 1, $y$-intercept 3). Test $(0, 0)$: $0 \leq 0 + 3$ is true, so shade the side containing the origin — below the line.

Second inequality: Boundary $y = -2x + 1$ (dashed line, slope $-2$, $y$-intercept 1). Test $(0, 0)$: $0 > -2(0) + 1 = 1$ gives $0 > 1$, which is false. Shade the side not containing the origin — above the line.

Solution region: The area where both shadings overlap. This is a wedge-shaped region above the dashed line and below the solid line. Only points in this overlap region satisfy both inequalities simultaneously.

The two boundary lines intersect where $x + 3 = -2x + 1$, which gives $3x = -2$, so $x = -\frac{2}{3}$ and $y = \frac{7}{3}$. The overlap region extends to the right from near this intersection point.

Real-World Application: Retail — Budget and Inventory Constraints

A small retail shop owner is ordering two types of products for the holiday season:

• Product A costs $15 per unit
• Product B costs $25 per unit
• The total budget is $3,000

The budget constraint is:

$15x + 25y \leq 3000$

where $x$ is the number of units of Product A and $y$ is the number of units of Product B.

Additionally, the shop has limited shelf space for at most 150 total items:

$x + y \leq 150$

And both quantities must be non-negative:

$x \geq 0, \quad y \geq 0$

Graphing the budget line: $15x + 25y = 3000$

• $x$-intercept: $x = 200$ (if only Product A)
• $y$-intercept: $y = 120$ (if only Product B)

Graphing the shelf space line: $x + y = 150$

• $x$-intercept: $x = 150$
• $y$-intercept: $y = 150$

Both are solid lines (non-strict inequalities) with shading toward the origin. The solution region is the area where all four inequalities overlap — a polygon in the first quadrant. Any point $(x, y)$ inside this region represents a valid order.

For instance, ordering 80 units of A and 60 units of B: cost is $15(80) + 25(60) = 1200 + 1500 = 2700 \leq 3000$, and total is $80 + 60 = 140 \leq 150$. This combination is in the solution region.

Carpentry — Material and Time Constraints

A carpenter builds two types of furniture: tables and bookshelves. Each table requires 6 board-feet of lumber and 3 hours of labor. Each bookshelf requires 4 board-feet and 5 hours. The carpenter has 120 board-feet of lumber and 90 hours available this month.

Let $t$ = number of tables and $b$ = number of bookshelves.

Lumber constraint: $6t + 4b \leq 120$

Time constraint: $3t + 5b \leq 90$

Non-negativity: $t \geq 0, \quad b \geq 0$

Graphing these constraints creates a feasible region — the polygon of all possible production combinations. The carpenter can pick any point inside this region. The vertices of the polygon are especially important because they represent the combinations that maximize or minimize objectives like profit (this connects to linear programming, an advanced topic).

For example, making 10 tables and 12 bookshelves: lumber used is $6(10) + 4(12) = 60 + 48 = 108 \leq 120$, time used is $3(10) + 5(12) = 30 + 60 = 90 \leq 90$. This point is on the boundary of the feasible region — the carpenter uses all available labor hours.

Common Mistakes to Avoid

1. Using a solid line for strict inequalities. If the symbol is $<$ or $>$ (no “equal to” bar), the line must be dashed. Points on the line are not solutions. This is the graphing equivalent of open vs. closed circles on a number line.

2. Shading the wrong side. Always test a point. The most common error is assuming “greater than” always means “shade above.” This is true when the inequality is solved for $y$ (e.g., $y > 2x + 1$), but not when it is in standard form. For $-y > 2x + 1$, solving for $y$ gives $y < -2x - 1$, and you shade below.

3. Testing a point on the boundary line. If the test point is on the line, the result is ambiguous. Choose a point clearly on one side — $(0, 0)$ is the best default unless the line passes through the origin.

4. Forgetting to shade at all. The graph of a linear inequality is a region, not a line. Without shading, your graph is incomplete and represents only the boundary equation, not the inequality.

5. Confusing the system solution with individual solutions. For a system of inequalities, the solution is only the overlap of all regions — not the union. A point must satisfy every inequality in the system.

Practice Problems

Test your understanding with these problems. Click to reveal each answer.

Key Takeaways

• A linear inequality in two variables represents a half-plane — an entire region of the coordinate plane, not just a line
• Dashed lines for strict inequalities ($<$, $>$); solid lines for non-strict ($\leq$, $\geq$)
• Always test a point (usually the origin) to determine which side to shade — do not guess based on the inequality symbol alone
• A system of inequalities is solved by graphing all inequalities and finding where the shaded regions overlap
• In real-world problems, linear inequalities model constraints — budgets, capacity limits, material availability — and the solution region shows all feasible combinations
• This topic connects directly to linear programming, where you find the optimal point within a feasible region defined by multiple inequality constraints

Return to Algebra for more topics in this section.