Generating an equation from a situation

What to assume they know

• Isolating the unknown, step by step — isolating the unknown; brackets; the three shorthand observations
• An equation is a claim of equality, not an instruction to compute — an equation is an assertion, and a letter is a name you choose
• Writing a situation as an expression before computing it — writing a situation as an expression before computing
• Multiplying out a bracket, and factorising a simple one: 4x − 4 = 4(x − 1)
• Working in rupees, and reading a fixed charge plus a per-item charge
• Reading a growing tile or stick pattern and counting it in more than one way

What they should be able to do

• Choose what an unknown letter will denote, and state that choice in words before writing anything
• Build an expression for a situation one described step at a time
• Show that two different countings of the same pattern give the same expression
• Produce two different but equally correct equations for one situation, and reconcile their answers
• Read a solved value back into the situation and answer the question that was actually asked
• Reduce a two-unknown situation to one unknown by expressing one in terms of the other
• Write several equations with a stated solution, and build a chain of equations that all share it
• Given a bare equation, invent a situation it could model

Where it usually goes wrong

• "Every problem has one right equation." The single most important correction in this topic, and the chapter builds a whole page around it. Mahesh's 25p + 50 = 500 and Srikanth's 25(f + 5) = 450 are different equations from the same story, and Fatima solved it without writing either.
• "The letter always stands for the thing the question asks for." Mahesh's p is the number of people, but the question asks for friends. His answer, 18, is not the answer to the question until 5 is taken off it. Reading the value back into the story is a separate step and it is where marks are lost.
• "A word problem is solved by hunting for keywords." Nothing on pp.174–180 works that way. Fatima's diagram comes before any letter, and Riyaz's table builds the expression one instruction at a time.
• "Two unknowns need two letters." Example 12 writes two letters, notices it cannot proceed, and goes back to one. Expressing Ramesh's count as y + 30 is the whole move. This is a genuine limitation of Class 7, stated by the chapter itself, not a shortcut.
• "Counting a pattern a different way gives a different formula." It gives a different-looking one. Method 1 reaches k + k + k + 1 and Method 2 reaches k + (2k + 1); both simplify to 3k + 1. That they must agree is the point.
• "Generating an equation is a game with no content." The chains on p.180 are the reason it is not: every equation in a chain has the same solution, so the reader can read a solution off the bottom line and know it holds all the way up. That is the earlier property looked at from the other end.
• "3k + 1 = 100 obviously has an answer because 100 is a nice number." It does here, but the parallel question in the chapter's exercises deliberately mixes reachable and unreachable targets (Part II, the exercise block following §7.4, p.188, question 16). Reachability has to be checked, not assumed.

Questions to check understanding

• Given a situation, state in words what the letter stands for and then write the equation
• Given the same situation, produce a second, different, correct equation for it
• Solve and then answer the question actually asked, not the value of the letter
• Given a growing pattern, write the expression for step k and test whether a stated total is reachable
• Given a two-unknown situation, reduce it to a single unknown
• Write several equations with a stated solution (Part II, §7.2, p.181, question 1, and the exercise block following §7.4, p.188, question 13)
• Invent a situation modelled by a given equation
• Explain why a think-of-a-number trick always lands on the announced answer, by writing the instructions as an expression in the starting number and showing that the starting number cancels — then build a trick of your own and prove it (Part II, the Puzzle Time box below the SUMMARY, p.190)
• Classic word problems of this shape fill the chapter's closing exercise block — a taxi fare with a fixed fee, two numbers with a stated sum and ratio, drinks with a stated price difference, a distribution problem from the Bakhśhāli Manuscript, and children and donkeys counted by heads and feet (Part II, the exercise block following §7.4, pp.185–189)

Examples worth working on the board

Printed answers are marked; everything else.

• Example 7, the tile pattern (Part II, §7.2, p.174). Checked against the printed page. Three T-shaped arrangements of square tiles are drawn and labelled Step 1, Step 2, Step 3: a row of tiles with one further tile hanging below the middle, growing at both ends and downwards. The counts printed are 4, 7, 10, and 13 for a fourth step. Method 1 splits each arrangement into three equal runs plus one — 1+1+1+1, 2+2+2+1, 3+3+3+1, 4+4+4+1 — reaching k+k+k+1 = 3k + 1. Method 2 splits it into a short arm and the rest — 1+3, 2+5, 3+7, 4+9 — reaching k + (2k + 1) = 3k + 1. Both layouts are printed in full with the pieces colour-coded. Use both; the agreement of the two is the section-3 argument.
• Example 7's question (Part II, §7.2, p.175). Input: 3k + 1 = 100. The chapter sets it up and hands the solving to the reader. Leave the arithmetic to the explanation.
• Example 8, the party (Part II, §7.2, p.175). Printed inputs: each plate of snacks costs ₹25; delivery is a fixed ₹50; the family is 5 people including Madhubanti herself; the budget is ₹500; everyone present gets one plate.
• Fatima's diagram (Part II, §7.2, p.175). Checked against the printed page. A rough sketch — two yellow ovals labelled Snack cost and Delivery cost, each with an arrow down into a box reading 25 × ▢ + 50, and a brace under the whole box. No letter yet; the empty box is where the letter will go. Show this before any symbol appears.
• Fatima's route (Part II, §7.2, p.175). She takes the ₹50 off the ₹500 first, reducing the question to how many ₹25 plates fit in ₹450, and gets 18 plates and then 13 friends. Printed answers.
• Mahesh and Srikanth (Part II, §7.2, p.176). Checked against the printed page. Two boxed columns side by side. Mahesh names the total number of people p and writes 25p + 50 = 500; Srikanth names the number of friends f and writes 25(f + 5) = 450. Printed answers: p = 18 and f = 13. The two equations are different, both correct, and neither is derived from the other on the page.
• Example 9, the savers (Part II, §7.2, pp.176–177). Printed inputs: Jahnavi starts at ₹4000 and adds ₹650 a month; Sunita starts at ₹5050 and adds ₹500 a month; m is the number of months until the two totals agree. The chapter writes 4000 + 650m = 5050 + 500m and works it down to a printed m = 7, then instructs the reader to check. A small orange diagram beside the text sets the two savers' totals against a label reading Equal.
• Example 11, Riyaz's trick (Part II, §7.2, p.178). Checked against the printed page. Printed inputs: think of a number; subtract 3; multiply by 4; add 8; the answer announced is 24. A two-column table builds the expression a step at a time — x, then x − 3, then 4(x − 3) = 4x − 12, then 4x − 12 + 8 = 4x − 4. The chapter then factorises 4x − 4 as 4(x − 1) before dividing, and prints the starting number as 7. It also asks for a rule connecting the announced answer to the starting number, and does not give one — leave that to find.
• Example 12, the marbles (Part II, §7.2, p.179). Checked against the printed page. Printed inputs: 60 marbles between Ramesh and Suresh; Ramesh has 30 more. Two equations are written, x + y = 60 and x = y + 30, and the chapter then says plainly that only one-unknown equations have been handled so far. The fix is to call Suresh's count y and Ramesh's y + 30, giving y + (y + 30) = 60 and then 2y + 30 = 60. A diagram beside it draws Suresh's marbles as a box and Ramesh's as the same box plus a run of thirty numbered 1, 2, 3 … 30. The chapter stops at 2y + 30 = 60 and hands the finish to the reader.
• Generating Equations (Part II, §7.2, pp.179–180). The reader is asked for equations having y = 5 as their solution, and to share and compare methods — marked Math Talk. Two are then printed: y + 1 = 6 and 3y = 15.
• The two chains (Part II, §7.2, p.180). Checked against the printed page. Two vertical chains printed in a box, each with the operation written beside its arrow. Left: y + 1 = 6, multiply by −1, add y, add 6, arriving at 5 = y. Right: 3y = 15, add 6, divide by 3, subtract 2, arriving at y = 5. The reader is then asked to run a chain upwards, to compare the two directions, and — with a printed hint — to name the solution of every equation in both chains without calculating.
• "A Magic Trick" (Part II, the Puzzle Time box below the SUMMARY, p.190). Checked against the printed page. Printed inputs: think of any number; double it; add 10; halve the result; take away the number you first thought of; add 3 — and the page predicts the answer is 8, whatever was thought of. It prints the hint to call the first number x, asks for an explanation of why the trick works, and invites the reader to build one of their own. This is the direct companion of Example 11 in section 8, run the other way round: use the six instructions and the predicted 8, and let it build (2x + 10) ÷ 2 − x + 3, cut that to x + 5 − x + 3, and watch the starting number cancel itself out. That cancellation is the whole trick, and it is the chapter's own closing demonstration of what an expression in x is good for.
• Example 13 (Part II, §7.2, p.180). Input: 100x + 75 = 250. The chapter asks for a real-life situation, reads the numbers as money, identifies ₹75 as the part that does not change, and suggests 100 as a count of units with x as the cost of one. It never solves the equation.

Figures to have open

• The T-shaped tile arrangement at Steps 1 to 4, built from countable unit squares, and splittable two different ways with the pieces recolourable. This is the topic's central figure and both decompositions must be able to be shown moving on the same shape.
• A rough-sketch treatment for Fatima's diagram: two labelled bubbles feeding into an arithmetic skeleton with an empty slot. Redraw it; do not lift the printed art.
• Two side-by-side solution columns for Mahesh and Srikanth.
• A two-quantity growth chart for Example 9 — two staircases on one time axis with a visible crossing. This is not in the textbook; the book prints a small labelled diagram instead, and a chart is the clearer treatment.
• A bar-and-extension diagram for Example 12: one box for the smaller quantity and the same box plus a run of thirty for the larger.
• A vertical chain layout with operation labels on the arrows.
• No photograph or data table from the textbook is needed.

Where this sits in the book

• NCERT Ganita Prakash, Class 7, Part II, printed Chapter 7 "Finding the Unknown", §7.2 "Solving Equations Systematically", bold subheading "Solving Problems", pp.174–180 — Example 7 with both counting methods (pp.174–175), Example 8 with Fatima's diagram (p.175) and the Mahesh/Srikanth columns (p.176), Example 9 (pp.176–177), Example 11 (p.178), Example 12 (p.179), the bold subheading "Generating Equations" and the Math Talk prompt (p.179), the two printed chains and Example 13 (p.180)
• Same part, same chapter, "Figure it Out", p.181, question 1
• Same part, same chapter, closing exercise block, pp.185–189
• Same part, same chapter, SUMMARY and the "A Magic Trick" Puzzle Time box printed below it, p.190
• Backward pointer: Isolating the unknown, step by step, whose solving technique every example here assumes