Linear Inequalities

Two real numbers or two algebraic expressions related by the symbols $<$, $>$, $\le$, or $\ge$ form an inequality. While equations express exact balance, inequalities define quantitative boundaries, limits, and tolerances.

• Strict Inequalities: Involve strictly less than ($<$) or strictly greater than ($>$), e.g., $ax + b < 0$ or $ax + b > 0$. The boundary values are strictly excluded.
• Slack (Non-Strict) Inequalities: Involve less than or equal to ($\le$) or greater than or equal to ($\ge$), e.g., $ax + b \le 0$ or $ax + b \ge 0$. The boundary values are included in the solution set.
• Linear Inequality in One Variable: An inequality that can be written in the form $ax + b < 0$, $ax + b \le 0$, $ax + b > 0$, or $ax + b \ge 0$, where $a \neq 0$ and $a, b \in \mathbb{R}$.
• Double (Compound) Inequalities: Involve two inequality signs bounding an expression between two values, e.g., $a < f(x) \le b$.

1. Reciprocal Rule: If $a$ and $b$ have the same sign (both positive or both negative) and $a < b$, then $\frac{1}{a} > \frac{1}{b}$. (e.g., $2 < 5 \implies \frac{1}{2} > \frac{1}{5}$; $-5 < -2 \implies -\frac{1}{5} > -\frac{1}{2}$).
2. Cross-Multiplication Caution: In an inequality like $\frac{x-1}{x-2} > 0$, you cannot cross-multiply by $(x - 2)$ because the sign of $(x - 2)$ depends on $x$. If $x < 2$, $(x - 2)$ is negative, which would reverse the inequality! Always multiply by $[(x-2)^2 > 0]$ or use the wavy curve method.

Any value of the unknown variable that converts an inequality into a true mathematical statement is termed a solution. The set of all such values is called the solution set, which critically depends on the specified replacement set (domain).

Consider the inequality $4x - 2 < 14 \implies 4x < 16 \implies x < 4$:

• If replacement set is $\mathbb{N}$ (Natural numbers $\{1, 2, 3, \dots\}$): Solution set is finite: $\{1, 2, 3\}$.
• If replacement set is $\mathbb{Z}$ (Integers $\{\dots, -2, -1, 0, 1, 2, \dots\}$): Solution set is infinite discrete: $\{\dots, -2, -1, 0, 1, 2, 3\}$.
• If replacement set is $\mathbb{R}$ (Real numbers): Solution set is a continuous interval: $(-\infty, 4)$.

1. Clear fractions by multiplying all terms by the positive least common multiple (LCM) of the denominators.
2. Expand brackets and group all terms containing the variable on the left-hand side (LHS) and all constant numbers on the right-hand side (RHS).
3. Combine like terms to reduce the inequality to the form $Ax < B$, $Ax \le B$, $Ax > B$, or $Ax \ge B$.
4. Divide both sides by the coefficient $A$. If $A < 0$, remember to reverse the inequality sign!
5. Express the final solution set in standard interval notation and plot it on the real number line.

Complex boundary conditions in physical problems frequently require variables to satisfy simultaneous double constraints or absolute magnitude bounds. Modulus inequalities represent one of the highest-weightage topics in WBCHSE algebra.

An inequality of the form $a \le f(x) \le b$ is a simultaneous conjunction of two inequalities:

The solution is the intersection of the solution sets of both inequalities: $S = S_1 \cap S_2$. Linear double inequalities can often be solved simultaneously by performing identical operations on all three parts.

Recall the definition of the absolute value function: $|x| = x$ if $x \ge 0$, and $|x| = -x$ if $x < 0$. Geometrically, $|x|$ denotes the distance of $x$ from the origin $0$ on the real number line.

For any positive real constant $a > 0$:

To solve rational inequalities of the form $\frac{P(x)}{Q(x)} \ge 0$ where $P(x)$ and $Q(x)$ are factored polynomials:

1. Ensure the right-hand side is zero, and factorize numerator and denominator completely into linear factors $(x - c_i)$. Make sure the coefficient of $x$ in each factor is $+1$.
2. Find all critical points (roots of numerator and denominator) and plot them on the number line in increasing order.
3. Start from the rightmost interval above the axis ($+$), and alternate signs across each critical point of odd multiplicity. For critical points of even multiplicity $(x - c)^{2k}$, the curve stays on the same side (does not cross).
4. Select intervals marked $(+)$ for $\ge 0$ or $(-)$ for $\le 0$. Exclude roots of the denominator $Q(x) = 0$ unconditionally to prevent division by zero!

While an inequality in one variable corresponds to an interval on a 1D line, an inequality in two variables represents an entire half-plane (semi-infinite 2D region) in the Cartesian plane $\mathbb{R}^2$.

Consider the linear equation $ax + by = c$ with $a, b \neq 0$. This straight line divides the Cartesian plane into two distinct regions called half-planes:

• Open Half-Plane: Consists of all points $(x, y)$ strictly satisfying $ax + by < c$ or $ax + by > c$. The boundary line itself is excluded from the solution set and is drawn as a dashed (broken) line.
• Closed Half-Plane: Consists of all points $(x, y)$ satisfying $ax + by \le c$ or $ax + by \ge c$. The boundary line is included in the solution set and is drawn as a solid (continuous) line.

To determine which half-plane corresponds to the solution region of $ax + by \le c$ (or $<, \ge, >$):

A collection of two or more linear inequalities in the same variables forms a system of linear inequalities. The solution of such a system is the region of the plane that satisfies all the inequalities simultaneously.

Geometrically, the solution set of a system of linear inequalities is the intersection (common overlap) of the half-planes determined by each individual inequality:

• Bounded Region: The solution region can be completely enclosed inside a sufficiently large circle. A bounded region forms a convex polygon.
• Unbounded Region: The solution region extends infinitely in at least one direction.
• Infeasible System: If the individual half-planes have no common intersection ($\mathcal{R} = \emptyset$), the system has no solution.

In almost all physical and economic modeling problems, quantities such as units produced, hours worked, or ingredients mixed cannot be negative. These conditions are represented by:

Geometrically, $x \ge 0$ represents the half-plane to the right of the $Y$-axis, and $y \ge 0$ represents the half-plane above the $X$-axis. Their simultaneous intersection restricts the feasible region strictly to the First Quadrant (Quadrant I).

The vertices or corner points of the feasible region are the points of intersection of the boundary lines forming the perimeter of the polygon. They are determined algebraically by solving the equations of intersecting pairs of lines simultaneously using substitution or elimination.

Translating complex verbal descriptions into formal systems of linear inequalities is an essential mathematical modeling competency tested extensively in WBCHSE examinations.

A classic WBCHSE long-answer problem involves blending solutions of different concentrations. If $x$ litres of a $p_1\%$ solution are added to $V$ litres of a $p_2\%$ solution to yield a resulting concentration between $C_1\%$ and $C_2\%$, the total amount of pure solute is:

This generates the compound inequality:

In Class 12, students will encounter Linear Programming, where a linear objective function $Z = ax + by$ (representing profit or cost) is maximized or minimized subject to linear inequality constraints. The Corner Point Theorem guarantees that the optimal value of $Z$, if it exists, occurs at one of the vertices (corner points) of the feasible region defined by the inequalities studied in Class 11.