Linear Inequalities

🎯 Key Points

• Multiplying/dividing BOTH sides by a NEGATIVE number flips the inequality sign — the single most common source of errors in this chapter
• Number line: open/hollow circle for strict (<,>), closed/filled circle for non-strict (≤,≥)
• Two-variable inequality graphing: dashed boundary line for strict, solid for non-strict; test the origin (0,0) to decide which half-plane to shade
• System of inequalities solution = the OVERLAP (intersection) of all individual half-planes, not the union

Linear Inequalities

A linear inequality is a statement involving the symbols <, >, ≤, or ≥ between two algebraic expressions where the highest power of the variable is 1. Just like equations, inequalities can be solved, but a few special rules apply.

Rules for Solving Inequalities

• The same number can be added or subtracted from both sides without changing the inequality sign.
• Both sides can be multiplied or divided by the same positive number without changing the sign.
• If both sides are multiplied or divided by a negative number, the inequality sign must be reversed (flipped). For example, if x > 3, then -x < -3.
• Taking reciprocals of both sides (when both sides have the same sign) also reverses the inequality.

Solving Inequalities in One Variable

To solve an inequality such as 3x - 5 < 7, collect the variable terms on one side just like an equation: 3x < 12, so x < 4. The solution is not a single number but an entire range of values, written as an interval such as (-∞, 4).

Representation on a Number Line

• Use an open circle (hollow dot) at the boundary point for strict inequalities (< or >), since that point is not included.
• Use a closed circle (filled dot) at the boundary point for ≤ or ≥, since that point is included.
• Shade or draw an arrow in the direction of all the values that satisfy the inequality.

Graphical Solution in Two Variables

A linear inequality in two variables, such as 2x + 3y ≤ 12, is solved by first drawing the boundary line 2x + 3y = 12 as if it were an equation.

• If the inequality is strict (< or >), draw the boundary as a dashed line (not included). If it is ≤ or ≥, draw a solid line (included).
• Pick a convenient test point not on the line, usually the origin (0,0), and substitute it into the inequality.
• If the test point satisfies the inequality, shade the half-plane containing that point. Otherwise shade the opposite half-plane.
• The shaded region (including the boundary line if solid) represents every point (x, y) that is a solution, called a half-plane.

Solving a System of Two Inequalities

When two linear inequalities in two variables must both be satisfied, graph each one's boundary line and shaded half-plane on the same axes. The common shaded region (the overlap of both half-planes) is the solution set of the system. This common region may be a bounded polygon or an unbounded region depending on the constraints, and forms the basis of the feasible region used later in linear programming.

Word Problems on Linear Inequalities

Many real situations translate into a single inequality that is then solved algebraically.

• Choose a variable for the unknown quantity.
• Convert phrases into symbols: "at least" → ≥, "at most" → ≤, "more than" → >, "less than" → <.
• Form the inequality, solve it, and reject any solutions that make no physical sense (e.g. a negative count, or a non-integer where only whole numbers apply).

Example: The longest side of a triangle is 3 times the shortest side, and the third side is 2 cm shorter than the longest. If the perimeter is at least 61 cm, find the minimum length of the shortest side. Let the shortest side be x; the sides are x, 3x, 3x − 2, so the perimeter 7x − 2 ≥ 61 gives x ≥ 9. The shortest side is at least 9 cm.

Compound (Double) Inequalities and Interval Notation

• A double inequality such as −3 ≤ 2x + 1 < 7 is solved by applying the same operation to all three parts at once: subtract 1 to get −4 ≤ 2x < 6, then divide by 2 to get −2 ≤ x < 3.
• The solution is written as an interval, here [−2, 3). A square bracket [ or ] means the endpoint is included (≤, ≥); a round bracket ( or ) means it is excluded (<, >).
• Unbounded solutions use ∞ or −∞, which always take a round bracket, e.g. (−∞, 4) for x < 4.

Feasible Region of a System of Inequalities

When several linear inequalities in two variables must hold at once (often together with x ≥ 0 and y ≥ 0), each inequality contributes one shaded half-plane. The feasible region is the set of points satisfying every inequality simultaneously — the common overlap of all the half-planes.

• The feasible region may be bounded (a closed polygon) or unbounded (extending to infinity).
• The corner points (vertices) are found by solving the boundary lines two at a time; these vertices are crucial in linear programming, where the optimum of a linear objective always occurs at a corner.
• The non-negativity constraints x ≥ 0, y ≥ 0 restrict the region to the first quadrant, which is standard in resource-allocation problems.

🚀 JEE Advanced Edge

Inequalities involving modulus and reciprocals require sign-case splitting: Solving |x-2| < 3 requires converting to the compound inequality -3 < x-2 < 3 (giving -1 < x < 5), while an inequality like 1/(x-2) > 0 requires checking the sign of (x-2) directly rather than cross-multiplying blindly — cross-multiplying an inequality by an expression of unknown sign (which could be negative) is a common error that silently flips the answer.

Quadratic inequalities via sign chart: To solve x²-5x+6 > 0, factor to (x-2)(x-3) > 0, mark the roots 2 and 3 on a number line, and test the sign of the product in each of the three resulting intervals — the product is positive outside the roots and negative between them (for an upward parabola), giving the solution x<2 or x>3 without needing calculus.

Worked problem: Solve (x-1)/(x+2) ≤ 0. Approach: The expression is undefined at x=-2, so x≠-2. Critical points are x=1 (numerator zero) and x=-2 (denominator zero). Sign chart across intervals (-∞,-2), (-2,1), (1,∞): the expression is positive, negative, positive respectively. We need ≤0, so the solution is the negative interval plus the included zero: -2 < x ≤ 1 (note x=-2 is excluded since the expression is undefined there, but x=1 is included since the inequality is non-strict and numerator=0 there).

Worked Example: Solving a Linear Inequality

Solve 3x − 7 > 5x − 1 and represent the solution on a number line.

3x − 7 > 5x − 1 → −7 + 1 > 5x − 3x → −6 > 2x → x < −3. Solution: x ∈ (−∞, −3), i.e., all real numbers less than −3 (open circle at −3 on the number line, arrow pointing left).

Worked Example: System of Inequalities

Find all integers x satisfying both −5 ≤ 2x − 3 and 2x − 3 ≤ 7 simultaneously.

Left inequality: −5 ≤ 2x − 3 → −2 ≤ 2x → x ≥ −1. Right inequality: 2x − 3 ≤ 7 → 2x ≤ 10 → x ≤ 5. Combined: −1 ≤ x ≤ 5. Integer solutions: −1, 0, 1, 2, 3, 4, 5 (seven values). Always solve both inequalities separately, then take the intersection.