Solving Linear Inequalities in One Variable

The Two Rules — Almost Like Equations

For equations you learned two moves: add/subtract the same number on both sides, and multiply/divide both sides by the same non-zero number. Inequalities keep both moves, with one crucial amendment.

Key Point (Rule 1): Equal numbers may be added to or subtracted from both sides of an inequality without affecting its sign.

Key Point (Rule 2): Both sides may be multiplied or divided by the same positive number freely. But when both sides are multiplied or divided by a negative number, the inequality sign reverses: $<$ becomes $>$, $\le$ becomes $\ge$, and vice versa.

Why the flip? Watch the number line: $3 > 2$, yet $-3 < -2$ — multiplying by $-1$ mirrors every number through 0 and reverses order. A quick check: $-8 < -7$, but $(-8)(-2) = 16 > 14 = (-7)(-2)$.

[Board Tip] The single most-lost mark in this chapter: dividing by a negative and forgetting the flip. Habit to build — the moment you write $\div(-3)$ or $\times(-1)$, flip the sign in the same line, then continue.

Writing Answers: Interval Notation

Over the reals, solution sets of linear inequalities are intervals — unbroken stretches of the number line. Compact notation:

• $(a, b)$ — open: $a < x < b$, endpoints excluded.
• $[a, b]$ — closed: $a \le x \le b$, endpoints included.
• $[a, b)$ and $(a, b]$ — half-open: one end in, one end out.
• $(a, \infty)$ means $x > a$; $[a, \infty)$ means $x \ge a$; $(-\infty, b)$ means $x < b$; $(-\infty, b]$ means $x \le b$.

Key Point: Round bracket = endpoint excluded; square bracket = endpoint included. The symbols $\infty$ and $-\infty$ are not numbers, so they always take a round bracket.

Unless a question says otherwise, this chapter solves inequalities over the real numbers — so expect interval answers. When the question restricts to naturals or integers, list the values instead.

[JEE Tip] Interval notation is the working language of Sets (Chapter 1) revisited: solution sets combine by intersection ($\cap$) when conditions must hold together and union ($\cup$) when either suffices. JEE answers are almost always reported as intervals or unions of intervals.

Solved Examples

Example 1: Same inequality, different worlds

Solve $30x < 200$ when (i) $x$ is a natural number (ii) $x$ is an integer.

Solution:

Step 1 — Divide by the positive coefficient (Rule 2, no flip). $x < \frac{200}{30} = \frac{20}{3} \approx 6.67$.

Step 2 — (i) Collect the naturals. Naturals below $\frac{20}{3}$: solution set $\{1, 2, 3, 4, 5, 6\}$.

Step 3 — (ii) Collect the integers. $\{\ldots, -3, -2, -1, 0, 1, 2, 3, 4, 5, 6\}$ — the domain changes the answer, not the algebra.

Takeaway: Solve once over the reals, then filter by the allowed number system.

Example 2: Collect, then divide

Solve $5x - 3 < 3x + 1$ when (i) $x$ is an integer (ii) $x$ is a real number.

Solution:

Step 1 — Add 3 to both sides (Rule 1). $5x < 3x + 4$.

Step 2 — Subtract 3x (Rule 1). $2x < 4$.

Step 3 — Divide by the positive 2 (Rule 2). $x < 2$.

Step 4 — Report per domain. (i) Integers: $\{\ldots, -2, -1, 0, 1\}$. (ii) Reals: $x \in (-\infty, 2)$.

Takeaway: Rule 1 moves terms across; only the final division touches the coefficient.

Example 3: The flip appears

Solve $4x + 3 < 6x + 7$.

Solution:

Step 1 — Collect x on one side. Subtract $6x$ and 3 from both sides: $-2x < 4$.

Step 2 — Divide by $-2$ — negative, so FLIP. $x > -2$.

Step 3 — Write the interval and check. Solution set $(-2, \infty)$. Check x = 0: $3 < 7$ ✓ and $0 > -2$ ✓.

Takeaway: The flip happens in the same line as the negative division — never a line later.

Example 4: Fractions first

Solve $\frac{5 - 2x}{3} \le \frac{x}{6} - 5$.

Solution:

Step 1 — Clear denominators with the positive 6. $2(5 - 2x) \le x - 30$ — no flip, 6 is positive.

Step 2 — Expand and collect. $10 - 4x \le x - 30$ gives $-5x \le -40$.

Step 3 — Divide by $-5$ and FLIP. $x \ge 8$.

Step 4 — Interval. $[8, \infty)$ — square bracket, since the slack sign includes 8.

Takeaway: Clear fractions with a POSITIVE multiplier first; save any negative division for one deliberate, flipped step.

Example 5: Negative coefficient from the start

Solve $-12x > 30$ when (i) $x$ is a natural number (ii) $x$ is an integer.

Solution:

Step 1 — Divide by $-12$ and FLIP. $x < -\frac{30}{12} = -\frac{5}{2}$.

Step 2 — (i) Filter by naturals. No natural number is negative: solution set $\varnothing$ (empty).

Step 3 — (ii) Filter by integers. Integers below $-2.5$: $\{\ldots, -5, -4, -3\}$.

Takeaway: An empty solution set is a legitimate answer — say so explicitly rather than forcing values.

Example 6: Brackets both sides

Solve $3(x - 1) \le 2(x - 3)$ for real $x$.

Solution:

Step 1 — Expand both sides. $3x - 3 \le 2x - 6$.

Step 2 — Collect. Subtract 2x and add 3: $x \le -3$.

Step 3 — Interval. $(-\infty, -3]$ — square bracket because the sign is slack.

Takeaway: Expansion first, collection second; the bracket type in the answer mirrors the sign type.

Example 7: When everything cancels

Solve (i) $2(2x + 3) - 10 < 6(x - 2)$ (ii) $37 - (3x + 5) \ge 9x - 8(x - 3)$.

Solution:

Step 1 — (i) Expand. $4x + 6 - 10 < 6x - 12$, i.e. $4x - 4 < 6x - 12$.

Step 2 — Collect. $-4 + 12 < 6x - 4x$ gives $8 < 2x$, so $x > 4$: solution $(4, \infty)$.

Step 3 — (ii) Simplify each side fully first. LHS $= 37 - 3x - 5 = 32 - 3x$; RHS $= 9x - 8x + 24 = x + 24$.

Step 4 — Collect. $32 - 24 \ge x + 3x$ gives $8 \ge 4x$, so $x \le 2$: solution $(-\infty, 2]$.

Takeaway: Simplify each side COMPLETELY before moving terms across — most sign errors happen mid-shuffle.

Example 8: Fractional two-sided

Solve $\frac{x}{3} > \frac{x}{2} + 1$.

Solution:

Step 1 — Multiply by the positive 6. $2x > 3x + 6$.

Step 2 — Collect. $-x > 6$.

Step 3 — Multiply by $-1$ and FLIP. $x < -6$: solution set $(-\infty, -6)$.

Takeaway: $-x > 6$ is NOT $x > -6$ — the flip converts it to $x < -6$; check with x = $-7$: $-\frac{7}{3} > -\frac{7}{2} + 1 = -\frac{5}{2}$ ✓.

Example 9: Checking a claimed answer

A student solves $7 - 4x < -5$ and reports $x < 3$. Verify and correct.

Solution:

Step 1 — Redo the algebra. Subtract 7: $-4x < -12$; divide by $-4$ and FLIP: $x > 3$.

Step 2 — Diagnose. The student forgot the flip on dividing by $-4$.

Step 3 — Refute the claim with a test value. x = 0 satisfies "$x < 3$", but $7 - 0 = 7 < -5$ is FALSE — so $x < 3$ cannot be the solution set.

Step 4 — State the correction. $(3, \infty)$.

Takeaway: One test value can demolish a wrong answer — build the checking habit.

Example 10: Strictness survives the algebra

Solve $\frac{3(x - 2)}{5} \ge \frac{5(2 - x)}{3}$.

Solution:

Step 1 — Clear denominators with the positive 15. $9(x - 2) \ge 25(2 - x)$.

Step 2 — Expand. $9x - 18 \ge 50 - 25x$.

Step 3 — Collect. $34x \ge 68$, so $x \ge 2$.

Step 4 — Interval. $[2, \infty)$.

Takeaway: The $\ge$ stayed $\ge$ throughout — only negative multipliers change it, and we used only positive ones.