Google

A system of linear inequalities can have **no solution** when the individual shaded regions for each inequality do not overlap anywhere on the coordinate plane. This situation can occur when the boundary lines of the inequalities are **parallel** and the inequalities require shading on opposite sides of their respective boundary lines. Because parallel lines never intersect and the shaded half-planes face away from each other, there is no region that satisfies both inequalities simultaneously. When graphing reveals no overlap between the shaded regions, the system has no solution.

This is analogous to an inconsistent system of linear equations, where parallel lines produce no common solution. However, unlike systems of equations — where parallel lines always mean no solution — parallel boundary lines in a system of inequalities do not *always* produce no solution. If both inequalities shade toward the region between the parallel lines, an overlapping strip can exist, and the system would have infinitely many solutions.

A System of Linear Inequalities with No Solution

Solve the system $$\left\{\begin{array}{l} 4x + 3y \geq 12 \\ y < -\frac{4}{3}x + 1 \end{array}\right.$$ by graphing.

**Graph $$4x + 3y \geq 12$$.** The boundary line is $$4x + 3y = 12$$, which has intercepts $$x = 3$$ and $$y = 4$$. Because the inequality uses $$\geq$$ (non-strict), draw a **solid** line. Test $$(0, 0)$$: $$4(0) + 3(0) = 0 \geq 12$$ is **false**. So shade the side that does **not** contain $$(0, 0)$$.

**Graph $$y < -\frac{4}{3}x + 1$$ on the same grid.** The boundary line is $$y = -\frac{4}{3}x + 1$$, which has slope $$m = -\frac{4}{3}$$ and y-intercept $$b = 1$$. Because the inequality uses $$<$$ (strict), draw a **dashed** line. Test $$(0, 0)$$: $$0 < -\frac{4}{3}(0) + 1$$ gives $$0 < 1$$, which is **true**. So shade the side that contains $$(0, 0)$$.

**Identify the solution.** The two boundary lines are **parallel** — both have slope $$-\frac{4}{3}$$ — and the shaded regions face away from each other. Because there is no point that lies in both shaded regions at the same time, the system has **no solution**.

Solving $$\{4x + 3y \geq 12,\; y < -\frac{4}{3}x + 1\}$$ by Graphing

In professional resource management, a system of linear inequalities is often used to identify a feasible region for project variables. If a manager determines that the constraints are mutually exclusive and cannot be satisfied at the same time, state the specific mathematical term used to classify the solution set of that system. Additionally, describe the primary visual characteristic that must be present on a coordinate graph to confirm that the system has this classification.

Defining Infeasible Constraints in Systems of Inequalities

Solve the system $$\left\{\begin{array}{l} 3x - 2y \geq 12 \\ y \geq \frac{3}{2}x + 1 \end{array}\right.$$ by graphing.

**Step 1 — Graph $$3x - 2y \geq 12$$.** The boundary line is $$3x - 2y = 12$$, which has intercepts $$x = 4$$ and $$y = -6$$. Because the inequality uses $$\geq$$ (non-strict), draw a **solid** line. Test $$(0, 0)$$: $$3(0) - 2(0) \geq 12$$ gives $$0 \geq 12$$, which is **false**, so shade the side that does **not** contain the origin.

**Step 2 — Graph $$y \geq \frac{3}{2}x + 1$$ on the same grid.** The boundary line is $$y = \frac{3}{2}x + 1$$. Because the inequality uses $$\geq$$ (non-strict), draw a **solid** line. Test $$(0, 0)$$: $$0 \geq \frac{3}{2}(0) + 1$$ gives $$0 \geq 1$$, which is **false**, so shade the side that does **not** contain the origin.

**Step 3 — Identify the solution.** The two boundary lines are **parallel**; writing $$3x - 2y = 12$$ in slope-intercept form gives $$y = \frac{3}{2}x - 6$$, which has the same slope of $$\frac{3}{2}$$ as the second line. The line $$y = \frac{3}{2}x + 1$$ lies above $$y = \frac{3}{2}x - 6$$. Since the first inequality requires shading below $$y = \frac{3}{2}x - 6$$ and the second requires shading above $$y = \frac{3}{2}x + 1$$, the shaded regions face away from each other. Because there is no region that satisfies both inequalities simultaneously, the system has **no solution**.

Solving $$\left\{3x - 2y \geq 12,\; y \geq \frac{3}{2}x + 1\right\}$$ by Graphing

Solve the system $$\left\{\begin{array}{l} x + 3y > 8 \\ y < -\frac{1}{3}x - 2 \end{array}\right.$$ by graphing.

**Step 1 — Graph $$x + 3y > 8$$.** The boundary line is $$x + 3y = 8$$, which has intercepts $$x = 8$$ and $$y = \frac{8}{3}$$. Because the inequality uses $$>$$ (strict), draw a **dashed** line. Test $$(0, 0)$$: $$0 + 3(0) > 8$$ gives $$0 > 8$$, which is **false**, so shade the side that does **not** contain the origin.

**Step 2 — Graph $$y < -\frac{1}{3}x - 2$$ on the same grid.** The boundary line is $$y = -\frac{1}{3}x - 2$$. Because the inequality uses $$<$$ (strict), draw a **dashed** line. Test $$(0, 0)$$: $$0 < -\frac{1}{3}(0) - 2$$ gives $$0 < -2$$, which is **false**, so shade the side that does **not** contain the origin.

**Step 3 — Identify the solution.** The two boundary lines are **parallel**; writing $$x + 3y = 8$$ in slope-intercept form gives $$y = -\frac{1}{3}x + \frac{8}{3}$$, which has the same slope of $$-\frac{1}{3}$$ as the second line. The line $$y = -\frac{1}{3}x + \frac{8}{3}$$ lies above $$y = -\frac{1}{3}x - 2$$. The first inequality requires shading above $$y = -\frac{1}{3}x + \frac{8}{3}$$, and the second requires shading below $$y = -\frac{1}{3}x - 2$$. The shaded regions face away from each other and do not overlap. Thus, the system has **no solution**.

Solving $$\left\{x + 3y > 8,\; y < -\frac{1}{3}x - 2\right\}$$ by Graphing

If the shaded regions of a system of linear inequalities do not overlap anywhere, the system has _____.

Case context:
A system of two linear inequalities is graphed. The boundary lines are parallel and the shaded regions face away from each other.

Question:
Does this system have a solution? Explain your reasoning based on the graph.

Sample answer:
No, the system has no solution because the shaded regions do not overlap.

Key points:
- The system has no solution.
- The shaded regions do not overlap.

Rubric:
A correct response states that there is no solution and explains that the lack of overlapping shaded regions means there are no points that satisfy both inequalities.

Does this system have a solution? Explain your reasoning based on the graph.

A system of linear inequalities has no solution if the shaded regions do not overlap.

A system of two linear inequalities has parallel boundary lines. What condition results in the system having no solution?

A system of linear inequalities with no overlapping shaded regions is analogous to which type of system of linear equations?

Match each term related to a system of linear inequalities with no solution to its correct description.

Order the steps to graphically determine that a system of linear inequalities has no solution.

What is true if the shaded regions of a system of linear inequalities do not overlap?

Question:
When graphing a system of linear inequalities, how do the shaded regions show that there is no solution?

Sample answer:
The shaded regions do not overlap.

Key points:
- The shaded regions do not overlap
- There is no intersection of shaded areas

Rubric:
A correct answer states that the shaded regions for the inequalities do not overlap or intersect.

Learn Before

Related