Doing the same thing to both sides preserves equality

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What to assume they know

An equation is a claim of equality, not an instruction to compute — an equation asserts that two expressions name one number; LHS and RHS
Finding an unknown by reasoning about what must balance — equal removals from both sides of a scale leave it balanced
Adding and subtracting negative numbers, including subtracting a negative
Reciprocals, and Brahmagupta's rule for dividing fractions — reciprocals, and dividing by a fraction by multiplying by its reciprocal (Part I, §8.2)
A fraction of a fraction, and why the numerators and denominators multiply — multiplying two fractions (Part I, §8.1)
Multiplication and division as operations that undo each other, from Class 6
Reading a long arithmetic expression without evaluating it

What they should be able to do

Carry out a trial-and-error search on a given equation and narrate the closing-in
State two distinct weaknesses of the trial-and-error method
Justify, from the meaning of the equals sign, why an operation applied to both sides leaves the assertion true
Shorten a known arithmetic statement by using the fact that adding and subtracting undo each other
Do the same for a known product, using multiplication and division as the undoing pair, including with a fraction
Remove a subtracted negative term by adding that negative to both sides
Solve a two-step equation such as 5x − 4 = 7 by two applications of the move
Verify a solution by substitution, and say what the verification does and does not establish

Where it usually goes wrong

"Trial and error is just the lazy method." Not the chapter's complaint. It succeeds on p.168 and gets the right answer. Its two real weaknesses are that it can take arbitrarily long, and that it can never certify it has found everything — which is why the question about other solutions is printed straight after.
"You can only do this to equations with an unknown in them." 15 + 8 = 23 has no unknown at all, and Examples 1 to 4 have none either. The property belongs to the equals sign. Establishing it on letterless statements first is the chapter's strategy.
"Whatever you do to one side, do the opposite to the other." A garbled half-memory of a later shorthand. The rule at this stage is genuinely symmetrical: the same operation, on each side. The sign flip a student half remembers only appears once the middle line is skipped, and that comes later (Part II, §7.2, p.173).
"You have to work out the left-hand side before you can do anything." The four arithmetic examples exist to break this. 14593 − 1459 + 145 − 14 is never evaluated anywhere in Example 1.
"Adding is always the undo for a minus sign." Example 3 subtracts a negative, and the undo is adding that negative. Slow down on the sign; this is where the arithmetic actually bites.
"Dividing both sides by a fraction is a different rule." Example 4 is placed precisely to show it is not. Dividing by 8/9 is multiplying by 9/8, and the move itself has not changed.
"Checking proves the answer is the only one." It proves the value found does satisfy the equation. Whether anything else does is a separate question, and it is the one trial and error could not settle.

Questions to check understanding

Solve a given equation by trial and error, then by the systematic route, and compare the effort
Given a known long arithmetic statement, read off the value of a shortened version of it without evaluating anything
State the operation that must be applied to both sides to remove a named term or factor
Solve a two-step equation and check the answer by substitution
Explain why applying the same operation to both sides cannot change what the equation asserts
Given an equation with a fractional solution, produce it exactly rather than rounding it
Items of the solve-and-check shape above open the first "Figure it Out" block of the section (Part II, §7.2, p.172) — that block is five equations to solve and check, and holds no machine
Given a machine diagram and its output, recover the input. Two different shapes hide under that one sentence, and only one of them inverts: p.185 q4(a) and p.186 q5 are straight chains, run back by undoing each operation in turn; p.186 q4(b) is not a chain — the input forks into two arms, one scaled and one added to, and the arms are subtracted. Its input cannot be recovered by inverting anything and has to be turned into an equation first (Part II, the exercise block following §7.4, pp.185–186, questions 4 and 5)

Examples worth working on the board

The trial-and-error run (Part II, §7.2, p.168). Checked against the printed page. The chapter tries n = 5 (giving 11), n = 10 (21), n = 30 (61), n = 40 (81), n = 50 (101, overshooting) and finally n = 49 (99). All six trials and all six LHS values are printed. Show the search as a walk that closes in from below, then overshoots, then corrects — that shape is the argument for why it is unsatisfying.
The question left hanging (Part II, §7.2, p.168). Immediately after the search the reader is asked whether this equation could have another solution. The chapter does not answer it there. Leave it unanswered too; it is the hinge into the rest of the section.
The second trial task (Part II, §7.2, p.168). Input: 5x − 4 = 7, offered for trial and error. Its solution is not a whole number, which is exactly why it is set here — but do not announce that in section 1. Let the search fail on camera.
The letterless test case (Part II, §7.2, p.169). Input: 15 + 8 = 23, with the suggestion of adding 10 to both sides. Nothing is unknown; the point is that the property is about equality itself, not about unknowns.
Example 1 (Part II, §7.2, p.169). Given: 14593 − 1459 + 145 − 14 + 88 = 13353. Asked: the value of 14593 − 1459 + 145 − 14..
Example 2 (Part II, §7.2, p.169). Given: 23 × 41 × 11 × 8 × 7 = 5,80,888. Asked: the value of 23 × 41 × 11 × 8. The printed route is to divide by 7, and the chapter then asks, as a Math Talk prompt, whether that amounts to dividing each side by 7. It does.
Example 3 (Part II, §7.2, pp.169–170). Given on p.169: 12345 − 5432 + 135 − 24 − (−67) = 7091. Asked: the value of 12345 − 5432 + 135 − 24. The move is to add (−67) to each side. Typesetting warning: the working printed on p.170 shows 132 where p.169's statement shows 135, three times over. 135 is the consistent figure — with 135 the shortened expression comes to 7024, which is the answer the book itself prints, and the original statement then checks out against 7091. With 132 neither number works. Checked against both pages.
Example 4 (Part II, §7.2, p.170). Given: (35/113) × 24 × 14 × (8/9) = 94080/1017. Asked: the value of (35/113) × 24 × 14. The printed route divides both sides by 8/9, which the chapter carries out as multiplication by 9/8. The answer printed is 11760/113.
Solving 5x − 4 = 7 (Part II, §7.2, p.170). The chapter adds 4 to both sides to reach 5x = 11, then divides both sides by 5. The printed solution is 11/5. A non-integer answer, deliberately, on the very first equation solved this way.
The check (Part II, §7.2, p.171). The chapter puts 11/5 back into the LHS, cancels the 5 against the denominator, and lands on 7, which is the RHS. Show the cancellation; it is the reason the fractional answer is not frightening.

Figures to have open

A two-pan scale that can be re-labelled LHS and RHS, reused from Finding an unknown by reasoning about what must balance so the continuity is visible. Standard schematic.
An annotated-equation treatment where an operation can be written beneath both sides simultaneously and a term or factor can be shown cancelling. This is the workhorse for sections 5 to 11 and needs to be built once and reused.
A closing-in visual for the trial-and-error search — a value climbing towards a fixed target line and overshooting it.
No photograph, table or data figure 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", pp.168–171 — the trial-and-error search and the bold naming of the method (p.168), the look back at the scales, the 15 + 8 = 23 probe, Examples 1 and 2 (p.169), Example 3 and Example 4 (pp.169–170), the solving of 5x − 4 = 7 (p.170) and the substitution check (p.171)
Same part, same chapter, the exercise block following §7.4, pp.185–186 — questions 4 and 5, the machine diagrams. q4(a) and q5 are chains, recovered by running the printed operations backwards; q4(b) branches and must be solved as an equation instead
Same part, same chapter, SUMMARY, p.190, bullets 2 and 4
Backward pointer: Finding an unknown by reasoning about what must balance, whose removal move is what p.169 looks back at
Forward pointer: Isolating the unknown, step by step, where the two-line shorthand for this move is derived