Substitution: rewriting one unknown so only the other survives

What to assume they know

• Crossing, parallel or lying on top: what each picture says about solutions — the three cases and the words consistent, inconsistent and dependent
• Why a graph stops being trustworthy once the answer is not a whole number — why an exact method is wanted at all
• Solving a linear equation in one unknown, including fractional coefficients
• Making one letter the subject of an equation
• Expanding a bracket with a negative multiplier, and collecting like terms
• Turning a sentence about ages, costs or counts into an equation

What they should be able to do

• Choose which equation to rearrange and which letter to isolate, and justify the choice by the arithmetic it avoids
• Carry out the replacement and reduce the pair to one equation in one unknown
• Recover the second unknown and state the answer as a pair
• Verify a solution in both original equations and say what the check rules out
• Explain why the replacement neither creates nor destroys solutions
• Interpret a variable-free true statement and a variable-free false statement, and match each to its picture
• Model a word problem as a pair and answer the question that was actually asked

Where it usually goes wrong

• "Substitute back into the equation you rearranged, in step 2." Doing so gives a statement that is true for every value, because you have put an equation into itself. The expression has to go into the other equation. This is the single commonest slip in the method, and the empty result looks alarmingly like the infinite-solution case.
• "A statement with no letter in it means the working failed." It is the answer, in a different form. Which answer depends only on whether the statement is true or false.
• "18 = 18 means x and y are both zero." It means nothing further has been pinned down: every point of one line answers the pair.
• "You can stop once you have y." The question asked for a pair, and word problems usually ask for something built from both — the two ages, or the two prices, not one of them.
• "Verification is a formality." It catches sign errors, bracket errors and the wrong-equation slip above, and it is the only step that tests the answer against the problem rather than against your own working.
• "Rounding the fractions before substituting back is fine." In Example 4 the answer is twenty-ninths; rounded values will not satisfy either equation, and the check will fail for a reason that has nothing to do with the method.
• "Substitution is always the best route." When neither letter comes free without fractions — as with 9x – 4y = 2000 alongside 7x – 3y = 2000 in the next section — elimination is much lighter.

Questions to check understanding

• Solve a given pair by substitution and verify the answer in both equations
• Solve a pair with fractional or decimal coefficients, clearing them first
• Recognise, from a variable-free ending, that a pair has no solution or endlessly many, and say which
• Form a pair from a word problem and answer the question in the problem's own units
• Find a parameter value, such as m in Exercise 3.2 Q2, once the pair is solved
• Explain why the substitution goes into the second equation and not the first

Examples worth working on the board

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

• Example 4 (§3.3.1, p. 30): 7x – 15y = 2 together with x + 2y = 3. The second equation is the one to open, because x is already alone but for its sign, giving x = 3 – 2y. Verified: replacing x in the first gives 21 – 14y – 15y = 2, so 29y = 19, so y = 19/29 and then x = 3 – 38/29 = 49/29. Verified in both original equations: 7(49/29) – 15(19/29) = 58/29 = 2, and 49/29 + 38/29 = 87/29 = 3. The answer is printed on p. 30; the instruction to substitute it back into both equations is on p. 31, under its own Verification heading, so the example runs across the page turn.
• Why that choice of equation. Verified as a contrast worth showing: opening the first equation instead gives x = (2 + 15y)/7, and the substitution then runs through sevenths for two lines before they cancel. Same answer, more arithmetic. The chapter's own advice is to take whichever is convenient; this is what convenient means in practice.
• The three steps, as printed (§3.3.1, p. 31). Free one letter from either equation; put that expression into the other equation and solve what is left; return the value to the rearranged equation to get the second unknown. The page also warns, ahead of Examples 9 and 10, that step 2 can end with no letter at all, and says what a true and a false ending each mean. A remark below the steps explains where the method's name comes from.
• Example 5 (§3.3.1, pp. 31–32), Aftab and his daughter. Seven years back his age was seven times hers; three years ahead it will be three times hers. With s and t for the two ages in years, the chapter forms s – 7 = 7(t – 7), tidied to s – 7t + 42 = 0, and s + 3 = 3(t + 3), tidied to s – 3t = 6. It substitutes s = 3t + 6 and reaches t = 12 and then s = 42. Verified: seven years back the two ages were 35 and 5, and 35 is seven times 5; three years ahead they are 45 and 15, and 45 is three times 15.
• Example 6 (§3.3.1, p. 32): two pencils with three erasers cost 9, and four pencils with six erasers cost 18. The pair is 2x + 3y = 9 with 4x + 6y = 18. The chapter frees x as (9 – 3y)/2 and the substitution collapses to a statement with both sides equal and no letter left, so no single cost can be named for either item. Verified: the second equation is the first doubled, so the two conditions are one condition; verified as a sample of the endless answers, a pencil at 3 with an eraser at 1 fits, and so does a pencil at 1.5 with an eraser at 2.
• Example 7 (§3.3.1, pp. 32–33): two rails, one carrying x + 2y – 4 = 0 and the other 2x + 4y – 12 = 0, asked whether they will ever cross. Freeing x as 4 – 2y and substituting leaves a false numerical statement, so they never meet. Verified: 2(4 – 2y) + 4y – 12 collapses to –4 for every y, and –4 is not 0.
• The callback worth building the explanation's ending on. Verified against Table 3.1 on p. 26: the pair in Example 6 is that table's second row and the pair in Example 7 is its third row. The chapter is re-solving, by algebra alone, the two pairs it had already classified by ratios — so the true ending belongs to coincident lines and the false ending to parallel ones, and that is not a coincidence but the chapter's design.
• Exercise 3.2 material (pp. 33–34). Q1 gives six pairs: x + y = 14 with x – y = 4; s – t = 3 with s/3 + t/2 = 6; 3x – y = 3 with 9x – 3y = 9; 0.2x + 0.3y = 1.3 with 0.4x + 0.5y = 2.3; √2x + √3y = 0 with √3x – √8y = 0; and 3x/2 – 5y/3 = –2 with x/3 + y/2 = 13/6. Q2 asks for 2x + 3y = 11 with 2x – 4y = –24 and then the value of m making y = mx + 3 pass through the answer. Q3 sets six word problems: two numbers differing by 26 where one is three times the other; two supplementary angles differing by 18 degrees; seven bats with six balls at 3800 and three bats with five balls at 1750; a taxi charging a fixed amount plus a rate per km, 105 for 10 km and 155 for 15 km, with the fare for 25 km also asked; a fraction that becomes 9/11 when 2 is added to both parts and 5/6 when 3 is added to both; and Jacob, three times his son's age in five years and seven times it five years ago. Verified, as answers the explanation may check itself against: (14 → 9 and 5); (s = 9, t = 6); the third pair is one line twice; (2, 3); the fifth pair meets only at the origin; (2, 3); m = –1 with x = –2 and y = 5; 13 and 39; 99° and 81°; a bat 500 and a ball 50; 5 fixed with 10 per km and ` 255 for 25 km; the fraction 7/9; Jacob 40 and his son 10.

Figures to have open

• A replacement movement: the expression from one equation lifted and dropped into every place the letter stands in the other. Standard schematic, and the image the thesis rests on.
• Two small line drawings for sections 9 and 10 — one line carrying all the answers of Example 6, and the two parallel rails of Example 7. The chapter draws neither; both are needed here because the whole point is that the algebra reproduces the geometry.
• Table 3.1 rows 2 and 3 (§3.2, p. 26) recalled for section 11.
• No figure is required for Example 5. The chapter asks for a graphical representation there but prints none, so the explanation either draws its own line pair for the two age equations or says plainly that the page leaves it to the reader.

Where this sits in the book

• NCERT Class 10 Mathematics, Chapter 3, §3.3.1, the substitution section, pp. 30–33: Example 4 on p. 30; the stepwise statement and the remark on the method's name on p. 31; Example 5 on pp. 31–32; Example 6 on p. 32; Example 7 on pp. 32–33.
• Exercise 3.2, pp. 33–34.
• Backward pointer inside the chapter: Table 3.1 on p. 26 already classified the pairs that reappear as Examples 6 and 7.
• Forward pointer inside the chapter: p. 31 refers ahead to Examples 9 and 10 when it warns that the unknown may vanish; Example 9 is on p. 35 and belongs to Elimination: scaling the equations so a variable cancels on addition.