Elimination: scaling the equations so a variable cancels on addition

What to assume they know

• Substitution: rewriting one unknown so only the other survives — substitution, and the meaning of a variable-free ending
• Predicting which of the three you will get by comparing coefficients — the coefficient comparison, which predicts what elimination is about to reveal
• The lowest common multiple of two numbers, used to match coefficients cheaply
• Adding and subtracting equations term by term, with signs kept straight
• Ratio notation such as 9 : 7, and the trick of writing the two quantities as 9x and 7x
• Place value: a two-digit number as ten times its first digit plus its second

What they should be able to do

• Justify multiplying an equation by a non-zero constant, and adding one equation to another, in terms of the solutions each step preserves
• Pick multipliers that match the coefficients of a chosen unknown, using the lowest common multiple
• Decide whether to add or to subtract, from the signs the matched coefficients carry
• Eliminate either unknown from the same pair and check that the answers agree
• Interpret the two variable-free endings and connect them to parallel and coincident lines
• Return from the solved letters to the quantities the question actually asked for
• Split a condition that is not linear into the linear cases it covers, and report every answer that results

Where it usually goes wrong

• "Adding equations is a trick that happens to work." It is a consequence of what an equation says. If two quantities are equal and two others are equal, the totals are equal. Saying that once removes the mystery.
• "You may multiply by anything." Anything except zero. Multiplying an equation by zero turns it into a statement true everywhere, and the information in it is gone for good — which is why the step cannot then be undone.
• "Always subtract." Subtract when the matched coefficients share a sign, add when they oppose. Getting this backwards doubles the unknown you meant to remove and is the commonest arithmetic failure in the method.
• "x is the first person's income." In Example 8 the incomes are 9x and 7x. Reporting ` 2000 as an income is the standard error on ratio-modelled problems, and the verification against both ratios is what catches it.
• "A problem has one answer." Example 10 has two, because the stated condition covers two linear cases. Reporting only 42 is a wrong answer, not half a right one.
• "Digits differing by 2 is a linear equation." It is not — a difference of size two is two possible signed statements, and the chapter quietly splits them. Naming that split is the most transferable idea in the section.
• "Elimination is for hard pairs and substitution for easy ones." They answer the same pairs. Exercise 3.3 Q1 deliberately asks for both on the same four pairs so the class can feel which is lighter where.

Questions to check understanding

• Solve a given pair by elimination, showing the multipliers chosen
• Solve the same pair by both algebraic methods and comment on which was lighter
• Conclude from a variable-free ending that a pair is inconsistent, or dependent
• Model a ratio problem by writing the quantities as multiples of one unknown, and report the quantities rather than the unknown
• Model a two-digit number problem using place value and solve it
• Handle a stated difference of size by splitting into cases, and give every answer
• Find a parameter for which a pair has no solution, using the same coefficient reasoning

Examples worth working on the board

Inputs. Values marked verified are worked out here on the chapter's data.

• Example 8 (§3.3.2, p. 34). Two people earn in the ratio 9 : 7 and spend in the ratio 4 : 3, and each puts by 2000 a month. The chapter writes the two incomes as 9x and 7x and the two expenditures as 4y and 3y, so the savings give 9x – 4y = 2000 and 7x – 3y = 2000. It scales the first by 3 and the second by 4, reaching 27x – 12y = 6000 and 28x – 12y = 8000, subtracts to leave x = 2000, and returns to get y = 4000, so the monthly incomes are 18,000 and ` 14,000. The page verifies both ratios.
• Why those two multipliers. Verified: the y-coefficients are 4 and 3, whose lowest common multiple is 12, which is what 3 and 4 produce. Matching the x-coefficients instead needs 7 and 9, giving 63 — heavier, but no harder in principle.
• Eliminating the other unknown (the chapter invites this and does not do it, §3.3.2, p. 35). Verified: scaling by 7 and 9 gives 63x – 28y = 14000 alongside 63x – 27y = 18000; subtracting the first from the second leaves y = 4000 directly, after which x = 2000. Same pair of values by an independent route, which is the most convincing single demonstration in the topic.
• What the letters are not. Verified and worth a slow beat: x = 2000 is not an income and y = 4000 is not an expenditure. The incomes are 9x and 7x, that is 18,000 and 14,000; the expenditures are 4y and 3y, that is 16,000 and 12,000. Savings check: 18000 – 16000 = 2000 and 14000 – 12000 = 2000. Ratio check: 16000 : 12000 reduces to 4 : 3, and 18000 : 14000 reduces to 9 : 7.
• The four steps, as printed (§3.3.2, p. 35). Scale both equations by suitable non-zero constants until one unknown's coefficients match numerically; add or subtract so that unknown goes; solve what is left; put the value back into either original equation for the other unknown. The page also says what to conclude when step 2 leaves a statement with no unknown in it — true means endlessly many answers, false means none — and two remarks above the steps name the method and invite comparison with substitution and with drawing.
• Example 9 (§3.3.2, p. 35): 2x + 3y = 8 with 4x + 6y = 7. Scaling the first by 2 makes both left sides 4x + 6y, and subtracting leaves a false numerical statement, so nothing satisfies the pair. Verified by the coefficient test of the earlier topic: 2/4 and 3/6 both give 1/2 while –8/–7 gives 8/7, which is the parallel pattern — the two methods agree, as they must.
• Example 10 (§3.3.2, pp. 35–36). A two-digit number added to the number its digits reversed gives 66, and the two digits differ by 2. Writing the digits as x and y, the chapter sets the number as 10x + y and its reverse as 10y + x, adds them to get 11(x + y) = 66, so x + y = 6; then it takes x – y = 2 and y – x = 2 as separate cases, solving each with the sum equation, and reports 42 and 24. Verified: the sum of any two-digit number and its reverse is eleven times the digit sum, which is why 66 collapses so cleanly; 42 + 24 = 66 with digits differing by 2, and the same holds read the other way. Verified as the reason there are exactly two: the difference condition is a size, not a signed quantity, so it splits into two equations rather than one — and both of those are perfectly good linear equations, which is why the split is legitimate in the first place. What produces exactly two answers is not the split by itself but what survives it: each branch yields a digit pair that a two-digit number can actually use, with both digits under ten and the leading one not zero. Two branches, two survivors, two numbers.
• Exercise 3.3 (pp. 36–37). Q1 gives four pairs and asks for each to be done by elimination and by substitution: x + y = 5 with 2x – 3y = 4; 3x + 4y = 10 with 2x – 2y = 2; 3x – 5y – 4 = 0 with 9x = 2y + 7; and x/2 + 2y/3 = –1 with x – y/3 = 3. Q2 sets five word problems: a fraction that becomes 1 when the numerator gains 1 and the denominator loses 1, and 1/2 when only the denominator gains 1; Nuri thrice Sonu's age five years back and twice it ten years on; a two-digit number whose digits sum to 9 and whose ninefold equals twice its reverse; Meena drawing 2000 as 25 notes of 50 and 100 only; and a library whose fixed charge covers three days, with a daily rate after that, 27 being paid for a seven-day loan and 21 for a five-day one. *Verified, as answers the explanation may check itself against:* Q1 gives (19/5, 6/5), then (2, 1), then (9/13, –5/13), then (2, –3); Q2 gives the fraction 3/5, Nuri 50 with Sonu 20, the number 18, ten fifties with fifteen hundreds, and 15 fixed with ` 3 for each day past the third.
• A detail to show. Verified by solving all four: the fractional answers in Q1 are the first and the third, not the first two — Q1(ii) comes out in whole numbers, as does Q1(iv). The two fractional pairs are nineteen-fifths with six-fifths, and nine-thirteenths with minus five-thirteenths. These are exactly the readings the previous topic said a graph could not give, arriving in the very first exercise after the method is taught.

Figures to have open

• A balance-scale movement for section 1: two balanced scales tipped together into one that still balances. Standard schematic; nothing like it is printed, and it is what makes the permission believable rather than merely stated.
• A two-column layout for Example 8 showing the pair scaled on the left and the matched coefficients aligned on the right, so the cancellation is visible rather than asserted.
• A place-value strip for a two-digit number and its reverse, with 10x + y above and 10y + x below, and the 11(x + y) falling out of the addition. Standard schematic, and the key to Example 10.
• A branch diagram for the case split, ending in the two numbers 42 and 24.
• Recall panels for rows 2 and 3 of Table 3.1 (§3.2, p. 26) in section 10.

Where this sits in the book

• NCERT Class 10 Mathematics, Chapter 3, §3.3.2, the elimination section, pp. 34–37: Example 8 with its verification on p. 34; the two remarks and the four-step statement on p. 35; Example 9 on p. 35; Example 10 on pp. 35–36.
• Exercise 3.3, pp. 36–37.
• Backward pointers inside the chapter: Table 3.1 on p. 26 for the two odd endings, and §3.3 on p. 30 for why an exact method was wanted.
• §3.4, the chapter summary, p. 37, lists both algebraic methods by name.