The one rule that differs from equation-solving, and the reason it differs

What to assume they know

• Why testing values one by one is not a method, and what replaces it — solution and solution set, and why substituting candidates cannot settle a real-number answer
• The two rules for solving linear equations: adding or subtracting equally on both sides, and multiplying or dividing both sides by a number that is not zero
• Arithmetic with signed numbers, including the product of two negatives
• Reading the number line left to right as an order, and locating negative numbers on it
• Clearing a fraction by multiplying through by a common denominator

What they should be able to do

• State the two rules the chapter gives for rewriting an inequality, and say which one differs from its equation counterpart
• Justify the difference by describing what multiplication does to positions on the number line
• Explain why the corresponding rule for equations needs only that the multiplier is not zero
• Solve a linear inequality in which the reversal is required, and one in which it can be sidestepped
• Check a reversal by solving the same statement a second way and comparing answers
• Apply the reversal to a double inequality and report the result with its ends in the right order
• Identify statements in which the reversal must not be applied, and say what would go wrong

Where it usually goes wrong

• "Flip whenever a minus sign shows up." The turn is triggered by scaling both sides by a negative quantity, nothing else. Miscellaneous Exercise item 5 puts a minus inside a denominator on one side and requires no turn at all.
• "Flip when the answer comes out negative." The sign of the answer is irrelevant. Example 4 uses a turn and ends at x ≥ 8.
• "Subtracting a negative number flips it." Rule 1 never turns the symbol, whatever is added or subtracted.
• "Rule 2 for equations only asked for a non-zero multiplier, so non-zero is enough here." Not enough. Here the rule splits at the sign, and a negative multiplier is legal but changes the statement's direction.
• "Multiplying by zero is harmless, just unhelpful." For an equation it gives something true; for a strict inequality it gives something false. Zero is barred here for a stronger reason.
• "Dividing both sides by x is the same kind of move." You do not know whether x is positive or negative, so neither half of Rule 2 licenses it. This chapter never does it.
• "The reversal is a convention you just have to remember." It is checkable: every one of the chapter's worked cases can be run a second way that avoids the turn, and the two routes agree.

Questions to check understanding

• Solve a linear inequality for real x where a negative divisor is unavoidable, and state the answer as an interval
• Solve one that clears fractions first, and name the step at which the symbol turns
• Solve a double inequality that requires division by a negative quantity, and report the bounds in the correct order
• Say whether a given step is legitimate, and if not, which rule it breaks
• Given a solved inequality, verify the answer by substituting one interior value and both bounds
• Explain in one or two lines why the equation rules need no sign condition

Examples worth working on the board

Items marked verified are worked out here from the chapter's stated data. The book's separate answers file was not read, checked or extracted at any point.

• The rules as the chapter recalls them for equations (§5.3, p. 91). Adding or subtracting equally on both sides; multiplying or dividing both sides by a number that is not zero. Then the chapter says the second one behaves differently once the statement is an inequality, and gives its evidence.
• The chapter's numerical evidence (§5.3, p. 91). First pair: 3 is above 2, while −3 is below −2. Second pair: −8 is below −7, while (−8)(−2) is 16 and (−7)(−2) is 14, so after the multiplication the first is above the second. Put all six numbers on one line; the exchange of left and right is the whole argument in one picture.
• Verified contrast for Rule 1, built to sit beside it: take −8 < −7 and add 10 to each side, giving 2 < 3. Both points move ten steps right and the order is untouched. Add −10 instead and they become −18 and −17, still in the same order. A shift never reorders anything, however large or however negative.
• Verified contrast for the positive half of Rule 2: take −8 < −7 and multiply both sides by 2, giving −16 < −14. The gap has doubled, the order has not moved. Multiply by ½ instead and it becomes −4 < −3.5. Stretching about zero keeps left on the left.
• Rule 1 and Rule 2 as the chapter finally states them for inequalities (§5.3, p. 92). Rule 1 is the same move as for equations and does not disturb the symbol. Rule 2 splits: a positive multiplier leaves the symbol alone, a negative one turns it round. State both in the wording used here — they are printed as formal rules and must not be carried across.
• Verified, section 7, and flag it as an added inference: multiplying by zero costs more here than in an equation. An equation multiplied by zero becomes 0 = 0, which is true and useless; the strict inequality 3 < 5 multiplied by zero becomes 0 < 0, which is false outright. So zero does not merely lose information here, it destroys a true statement. Note where the two rules stand on it, because they do not stand in the same place: the equation rule recalled on p. 91 names a non-zero multiplier outright, while the rule stated for inequalities on p. 92 names a positive multiplier and a negative one and says nothing about zero at all. Zero is left out there rather than ruled out.
• Example 3 (§5.3, p. 93): 4x + 3 < 6x + 7. The chapter's route collects the x terms on the left, reaching −2x < 4, then divides by −2 and turns the symbol round to get x > −2, recorded as the interval from −2 upward.
• Verified second route, and this is section 8's payoff: collect the x terms on the right instead. From 3 − 7 < 6x − 4x comes −4 < 2x, then dividing by the positive 2 gives −2 < x. Same answer, no reversal used. The reversal is therefore checkable — it is not something a student has to take on trust.
• Example 4 (§5.3, p. 93): (5 − 2x)/3 ≤ x/6 − 5. Multiplying through by 6, which is positive and so needs no turn, gives 2(5 − 2x) ≤ x − 30, then 10 − 4x ≤ x − 30, then −5x ≤ −40, and dividing by −5 turns the symbol to give x ≥ 8, recorded as the interval from 8 upward with 8 included.
• Verified second route: from 10 − 4x ≤ x − 30 move to 40 ≤ 5x and then 8 ≤ x, again with no reversal anywhere. Both routes land on the same set.
• Miscellaneous Example 10 (p. 96): −5 ≤ (5 − 3x)/2 ≤ 8. Multiplying by 2 gives −10 ≤ 5 − 3x ≤ 16, and subtracting 5 throughout gives −15 ≤ −3x ≤ 11. Dividing by −3 turns both symbols and swaps which end is which, and the chapter writes the result with the smaller bound restored to the left: x runs from −11/3 up to 5, both ends included.
• Verified, and this is the strongest check in the brief: the two ends map to each other in exchanged order. Put x = 5 into (5 − 3x)/2 and it gives −5, the lower bound of the original chain. Put x = −11/3 in and it gives 8, the upper bound. An interior value confirms the same: x = 0 gives 5/2, which sits between −5 and 8, and 0 sits between −11/3 and 5. The end-for-end turn is not a bookkeeping quirk; it is what the numbers actually do.
• Miscellaneous Exercise item 5 (p. 98): −12 < 4 − 3x/(−5) ≤ 2. Hand this to the explanation as the trap. Verified: the negative sits in a denominator on one side, not as a multiplier of both sides, so 4 − 3x/(−5) is 4 + 3x/5 and no reversal is called for at any step. The chain runs to −16 < 3x/5 ≤ −2, then −80 < 3x ≤ −10, then x between −80/3 and −10/3, the lower end open and the upper end closed. A student flipping on sight of a minus sign gets this wrong twice.
• Exercise 5.1 item 2 (p. 95): −12x > 30. Verified: dividing by −12 turns the symbol, giving x < −5/2.
• Example 11's second constraint (p. 96): 11 − 5x ≤ 1. Verified: it reduces to −5x ≤ −10 and then, with the turn, x ≥ 2. This is the constraint that puts the filled endpoint into Fig 5.3, which Drawing the answer as a piece of the number line, hollow circle or solid reads.
• Where the rules are silent. No worked example or exercise in this chapter multiplies or divides both sides by an expression containing the letter — checked across all eleven page images. Every clearing step scales by a constant. That is not an accident: neither half of Rule 2 applies when the multiplier's sign is unknown.

Figures to have open

• The paired number line of section 4, with a multiplier control. Two labelled points, their images, and the arrow between them reversing when the multiplier goes negative. This is an added figure and it is the one the topic cannot be taught without; the chapter argues the point from two numerical instances only.
• A two-column solution panel for Example 3, the reversal route beside the reversal-free route, meeting at the same answer. Standard schematic.
• An end-swap diagram for the double chain: the bounds −15 and 11 above, the bounds 5 and −11/3 below, with crossing arrows showing which came from which. Standard schematic.
• A verification strip for the double chain: x = −11/3, x = 0 and x = 5 each fed into (5 − 3x)/2, producing 8, 5/2 and −5 against the original bounds. Standard schematic, built from working added here.
• No textbook artwork is needed for this topic. Pages 91, 92, 96 and 98 carry no figures — checked on all four page images. The chapter's three number-line drawings are on pp. 93, 94 and 97 and belong to Drawing the answer as a piece of the number line, hollow circle or solid.

Where this sits in the book

• NCERT Class XI Mathematics, Chapter 5 Linear Inequalities, §5.3, p. 91, for the recollection of the two equation rules and for the two numerical instances motivating the difference.
• §5.3, p. 92, for Rule 1 and Rule 2 as stated for inequalities.
• §5.3, p. 93, for Example 3 and Example 4.
• Miscellaneous Examples, p. 96, for Example 10 and for the second constraint of Example 11.
• Exercise 5.1 item 2, p. 95; Miscellaneous Exercise on Chapter 5 item 5, p. 98.
• Summary, p. 99, for the compact restatement of both rules.