Which System Of Inequalities Has No Solution

A system of inequalities has no solution when the solution sets of the individual inequalities do not overlap, meaning there is no coordinate pair $(x, y)$ that satisfies every condition simultaneously. Graphically, this appears as shaded regions that share no common area. In real terms, algebraically, it often results in a logical contradiction, such as requiring a variable to be simultaneously greater than 5 and less than 3. Understanding which system of inequalities has no solution is a fundamental skill in algebra, linear programming, and optimization, as it identifies infeasible constraints in real-world modeling scenarios.

Understanding the Basics of Inequality Systems

Before diving into the specific conditions that create empty solution sets, it is helpful to review what a system of inequalities represents. A system consists of two or more inequalities involving the same variables. The solution to the system is the intersection of the solution sets of each individual inequality.

For linear inequalities in two variables ($x$ and $y$), each inequality divides the coordinate plane into two half-planes: one that satisfies the inequality (the shaded region) and one that does not. But the boundary line is solid if the inequality includes equality ($\le$ or $\ge$) and dashed if it is strict (${content}lt;$ or ${content}gt;$). The solution to the system is the region where all shading overlaps. If that overlap is empty, the system has no solution.

The official docs gloss over this. That's a mistake.

Graphical Indicators: Parallel Lines and Opposite Half-Planes

The most common and visually intuitive example of a system with no solution involves parallel boundary lines shading in opposite directions.

Consider the system:

1. $y > 2x + 1$
2. $y < 2x - 3$

Both inequalities have the same slope ($m = 2$), meaning their boundary lines are parallel.

• The first inequality shades the region above the line $y = 2x + 1$.
• The second inequality shades the region below the line $y = 2x - 3$.

Because the lines are parallel, the region "above the top line" and the region "below the bottom line" never meet. There is a band of empty space between them. Here's the thing — no point $(x, y)$ can exist in both shaded regions at the same time. This is the classic textbook example of an inconsistent system.

Key Takeaway: If the boundary lines are parallel (same slope) and the inequality symbols point in opposite directions (one "greater than," one "less than") such that the shaded regions face away from each other, the system has no solution That alone is useful..

Conversely, if parallel lines shade toward each other (e.g., $y < 2x + 1$ and $y > 2x - 3$), the solution is the infinite band between the lines, which contains infinitely many solutions.

Non-Linear Systems: Curves That Never Touch

Systems involving quadratic, exponential, or absolute value inequalities can also yield no solution. The principle remains identical: the intersection of the shaded regions is empty And it works..

Quadratic vs. Linear

Imagine a system with a parabola opening upward and a line strictly below it:

1. $y > x^2 + 2$ (Region strictly above the parabola, vertex at $(0,2)$)
2. $y < 1$ (Region strictly below the horizontal line $y=1$)

The minimum value of the first inequality's region is just above $y=2$. That's why the maximum value of the second inequality's region is just below $y=1$. Since the regions are separated by a gap (from $y=1$ to $y=2$), they never intersect. This system has no solution.

Absolute Value Systems

Absolute value inequalities create V-shaped regions. System:

1. $|x| + |y| < 1$ (Interior of a diamond centered at origin)
2. $|x| + |y| > 3$ (Exterior of a larger diamond centered at origin)

The first inequality defines a bounded region close to the origin. There is a "moat" between the diamond of radius 1 and the diamond of radius 3 where neither condition is met, but crucially, no point satisfies both being inside the small diamond AND outside the large diamond simultaneously. Because of that, the second defines the region far away from the origin. The intersection is empty.

Algebraic Verification: Deriving Contradictions

While graphing provides visual proof, algebraic manipulation confirms the absence of a solution definitively. This is especially useful for systems in one variable or when graphing is impractical.

One-Variable Systems

Consider the system:

1. $3x - 5 > 10$
2. $2x + 4 \le 0$

Solve each independently:

1. $3x > 15 \implies x > 5$
2. $2x \le -4 \implies x \le -2$

The first requires $x$ to be strictly greater than 5. On a number line, these intervals $(5, \infty)$ and $(-\infty, -2]$ do not touch. The second requires $x$ to be less than or equal to -2. There is no real number that is both ${content}gt;5$ and $\le -2$. The intersection is the empty set, denoted $\emptyset$ And it works..

Two-Variable Systems (Substitution/Elimination)

For linear systems, attempting to find an intersection point algebraically often leads to a false statement (a contradiction) That alone is useful..

System:

1. $x + y > 5$
2. $x + y < 2$

If we treat the boundaries as equations ($x+y=5$ and $x+y=2$), subtracting the second from the first gives $0 = 3$, which is false. This confirms the lines are parallel and distinct Practical, not theoretical..

Even with non-parallel lines, a contradiction can arise if the feasible regions are "pushed" apart by other constraints. Because of that, $x \ge 0$ 2. System:

1. $y \ge 0$

The first two inequalities restrict the solution to the first quadrant (including axes). Since $x \ge 0$ and $y \ge 0$, their sum must be $\ge 0$. The third inequality requires the sum of two non-negative numbers to be negative ($x+y \le -1$). It is mathematically impossible for a non-negative number to be $\le -1$. The algebraic logic creates a direct contradiction: $0 \le x+y \le -1 \implies 0 \le -1$ (False).

The Role of Boundary Types: Strict vs. Non-Strict Inequalities

The presence of equality ($\le, \ge$) versus strict inequality (${content}lt;, >$) can determine whether a system has no solution or exactly one solution (a single point).

Case A: Parallel Lines Touching at a "Point at Infinity" (No Solution)

Lines: $y \ge 2x + 1$ and $y \le 2x + 1$. These are the same line. The solution is the line itself (infinite solutions).

Case B: Parallel Lines with a Gap (No Solution)

Lines: $y \ge 2x + 1$ and $y \le 2x - 1$. Gap exists. No solution.

Case C: Parallel Lines "Kissing" (Strict vs Non-Strict)

System:

1. $y > 2x + 1$ (Dashed line, shade above)
2. $y < 2x + 1$ (Dashed line, shade below)

The boundary line is $y = 2x + 1$. The regions are mutually exclusive. The second excludes the line (strict ${content}lt;$). The first inequality excludes the line (strict ${content}gt;$). **No solution.

System:

1. $y \ge 2x + 1$ (Solid line, shade above)
2. $y \