Systems of Linear Inequalities

A system of linear inequalities is a set of two or more linear inequalities with the same variables. Instead of looking for a single point where two lines cross, you are looking for an entire region of points that satisfy all the inequalities at once. This region — called the feasible region or solution region — appears on the graph as the overlap of the shaded areas for each individual inequality.

Graphing a Single Linear Inequality (Quick Review)

Before tackling a system, make sure you can graph one inequality:

1. Graph the boundary line ($y = mx + b$). Use a solid line for $\leq$ or $\geq$ (the line itself is included). Use a dashed line for the strict inequalities, where the line is not included.
2. Shade the correct side. For $y \geq mx + b$ or $y > mx + b$, shade above the line. For $y \leq mx + b$ or $y < mx + b$, shade below the line.
3. Test a point (usually the origin if it is not on the line) to confirm you shaded the right side.

Graphing a System of Inequalities

To graph a system, you graph each inequality on the same coordinate plane and identify where the shaded regions overlap.

Example 1: Two Inequalities

Graph the solution region for:

$y \leq 2x + 4$

$y \geq -x + 1$

Step 1 — Graph the first inequality.

Boundary line: $y = 2x + 4$ (solid line, because $\leq$). Slope $= 2$, y-intercept $= 4$.

Shade below the line (because $y \leq$).

Test with $(0, 0)$: Is $0 \leq 2(0) + 4 = 4$? Yes, so the side containing the origin is correct.

Step 2 — Graph the second inequality on the same plane.

Boundary line: $y = -x + 1$ (solid line, because $\geq$). Slope $= -1$, y-intercept $= 1$.

Shade above the line (because $y \geq$).

Test with $(0, 0)$: Is $0 \geq -(0) + 1 = 1$? No, so the origin is not in the shaded region for this inequality. Shade the opposite side from the origin.

Step 3 — Identify the overlap.

The feasible region is the area that is shaded by both inequalities — below the line $y = 2x + 4$ and above the line $y = -x + 1$ at the same time.

Finding the corner point. The two boundary lines intersect where:

$2x + 4 = -x + 1$

$3x = -3$

$x = -1$

$y = 2(-1) + 4 = 2$

The boundary lines meet at $(-1, 2)$. Since both inequalities include $\leq$ and $\geq$, this point is part of the feasible region.

Example 2: System with a Strict Inequality

Graph the solution region for:

$y > x - 2$

$y \leq -2x + 6$

First inequality: Boundary line $y = x - 2$ is dashed (strict $>$). Shade above.

Second inequality: Boundary line $y = -2x + 6$ is solid ($\leq$). Shade below.

Intersection of boundary lines:

$x - 2 = -2x + 6$

$3x = 8$

$x = \frac{8}{3} \approx 2.67$

$y = \frac{8}{3} - 2 = \frac{2}{3} \approx 0.67$

The boundary lines cross at $\left(\frac{8}{3}, \frac{2}{3}\right)$. Since the first inequality is strict ($>$), this intersection point itself is not included in the feasible region, but points arbitrarily close to it (on the correct side) are included.

The feasible region is the area above the dashed line and below the solid line.

Example 3: Three Inequalities (Bounded Region)

Graph the solution region for:

$x \geq 0$

$y \geq 0$

$x + y \leq 6$

First inequality: $x \geq 0$ means everything to the right of (and including) the y-axis.

Second inequality: $y \geq 0$ means everything above (and including) the x-axis.

Third inequality: Boundary line $y = -x + 6$ (solid). Shade below.

The feasible region is a triangle with vertices at:

• $(0, 0)$ — where $x = 0$ meets $y = 0$
• $(6, 0)$ — where $y = 0$ meets $x + y = 6$
• $(0, 6)$ — where $x = 0$ meets $x + y = 6$

Every point inside and on the edges of this triangle satisfies all three inequalities. This is an example of a bounded feasible region — it does not extend to infinity in any direction.

Bounded vs. Unbounded Regions

• A bounded region is enclosed on all sides and has a finite area (like the triangle in Example 3)
• An unbounded region extends infinitely in at least one direction (like the overlap in Example 1, which stretches infinitely to the left)

Whether a region is bounded or unbounded depends on how many constraints (inequalities) you have and how they are oriented.

Introduction to Linear Programming

Linear programming is the practical application of systems of inequalities. It answers the question: given a set of constraints (inequalities), what values of $x$ and $y$ make a certain quantity as large or as small as possible?

The quantity you want to maximize or minimize is called the objective function, and it is always linear (e.g., $P = 5x + 3y$).

The Corner Point Theorem

The key insight of linear programming is this: the maximum and minimum values of a linear objective function always occur at a corner point (vertex) of the feasible region. You do not need to test every point in the region — just evaluate the objective function at each vertex and pick the best one.

Example 4: Maximizing Profit

A small workshop makes two products: chairs and tables. Each chair earns $25 profit and each table earns $40 profit. The workshop has these constraints:

• Assembly time: each chair takes 2 hours, each table takes 4 hours, and there are 40 hours available: $2x + 4y \leq 40$
• Finishing time: each chair takes 3 hours, each table takes 2 hours, and there are 36 hours available: $3x + 2y \leq 36$
• Non-negativity: $x \geq 0$ and $y \geq 0$

Here $x$ = number of chairs, $y$ = number of tables.

Objective function: Maximize $P = 25x + 40y$

Step 1 — Find the feasible region. Graph all four inequalities. The feasible region is the area satisfying all constraints.

Step 2 — Find the corner points. Each corner is where two boundary lines intersect.

Corner 1: $x = 0$ and $y = 0$: the origin $(0, 0)$

Corner 2: $x = 0$ and $2(0) + 4y = 40$: $y = 10$, giving $(0, 10)$

Corner 3: $y = 0$ and $3x + 2(0) = 36$: $x = 12$, giving $(12, 0)$

Corner 4: Intersection of $2x + 4y = 40$ and $3x + 2y = 36$.

Simplify the first equation: $x + 2y = 20$, so $x = 20 - 2y$

Substitute into the second: $3(20 - 2y) + 2y = 36$

$60 - 6y + 2y = 36$

$-4y = -24$

$y = 6$

$x = 20 - 2(6) = 8$

Corner 4: $(8, 6)$

But we need to verify that $(0, 10)$ satisfies all constraints:

Check $3(0) + 2(10) = 20 \leq 36$. Yes.

And verify $(12, 0)$: Check $2(12) + 4(0) = 24 \leq 40$. Yes.

Step 3 — Evaluate the objective function at each corner.

Answer: Maximum profit is $440, achieved by making 8 chairs and 6 tables.

This is a fundamentally different kind of answer than systems of equations give you. Instead of “where do the lines meet?”, linear programming asks “where in this region is the best outcome?” — and the Corner Point Theorem guarantees the answer is always at a vertex.

Real-World Application: Carpentry — Material and Time Constraints

A carpenter builds bookshelves and desks for a furniture shop. Each bookshelf requires 20 board-feet of lumber and 5 hours of labor. Each desk requires 30 board-feet and 8 hours. The shop has 600 board-feet of lumber and 160 hours of labor available this month. Bookshelves sell for $120 profit and desks for $200 profit.

Define variables: Let $b$ = bookshelves, $d$ = desks.

Constraints:

$20b + 30d \leq 600 \quad \text{(lumber)}$

$5b + 8d \leq 160 \quad \text{(labor)}$

$b \geq 0, \quad d \geq 0$

Objective: Maximize $P = 120b + 200d$

Find the corner points:

• $(0, 0)$: $P = 0$
• $(0, 20)$: From $30d = 600$, $d = 20$. Check labor: $8(20) = 160 \leq 160$. Works. $P = 200(20) = 4000$
• $(32, 0)$: From $5b = 160$, $b = 32$. Check lumber: $20(32) = 640$. That exceeds 600, so this is NOT feasible. Instead, from $20b = 600$, $b = 30$. Check labor: $5(30) = 150 \leq 160$. Works. Corner is $(30, 0)$. $P = 120(30) = 3600$
• Intersection of $20b + 30d = 600$ and $5b + 8d = 160$:

From the second equation: $b = 32 - 1.6d$

Substitute: $20(32 - 1.6d) + 30d = 600$

$640 - 32d + 30d = 600$

$-2d = -40$

$d = 20$

Then $b = 32 - 1.6(20) = 32 - 32 = 0$

This gives the point $(0, 20)$, which we already found. The lumber and labor constraints intersect at the same corner here, meaning the carpenter cannot produce a mix that uses all of both resources. The optimal strategy is either all desks or all bookshelves.

Answer: The carpenter should make 0 bookshelves and 20 desks for a maximum profit of $4,000.

Graphing Tips for Systems of Inequalities

1. Use different shading patterns (or colors) for each inequality so you can clearly see the overlap
2. Label each boundary line with its equation
3. Mark all corner points with their coordinates
4. Shade lightly — heavy shading makes it hard to see where regions overlap
5. Always test a point in your final feasible region to confirm it satisfies every inequality

Common Mistakes to Avoid

1. Using the wrong line style. Solid lines for $\leq$ and $\geq$; dashed lines for strict inequalities (where the boundary is not included). Mixing these up changes the solution set.
2. Shading the wrong side. Always test a point (like the origin) to confirm which side to shade. Do not rely on memorization alone.
3. Missing a constraint. In word problems, $x \geq 0$ and $y \geq 0$ are often implied but must be included. You cannot produce a negative number of products.
4. Testing interior points instead of corners in linear programming. The maximum or minimum of a linear objective function is always at a corner point, never in the interior of the feasible region.
5. Forgetting to verify corner points satisfy ALL constraints. A corner formed by two boundary lines might violate a third constraint — always check every inequality.

Practice Problems

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

Key Takeaways

• A system of inequalities defines a feasible region — the set of all points satisfying every inequality simultaneously
• Graph each inequality on the same coordinate plane and look for where the shaded areas overlap
• Use solid lines for $\leq$ and $\geq$; use dashed lines for strict inequalities where the boundary is not included
• Linear programming finds the maximum or minimum value of an objective function within the feasible region
• The Corner Point Theorem guarantees that optimal values occur at vertices of the feasible region — evaluate the objective function at each corner and choose the best
• Always include non-negativity constraints ($x \geq 0$, $y \geq 0$) in real-world problems

Return to Algebra for more topics in this section.