Isolating the unknown, step by step

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What the video covers, what to say before it, where a class usually goes wrong, and what to set afterwards. An account is free, and it opens every chapter of every book.

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What to assume they know

Doing the same thing to both sides preserves equality — an operation applied to each side leaves the assertion intact; inverse operations
An equation is a claim of equality, not an instruction to compute — LHS, RHS, and what a solution is
Adding, subtracting and multiplying negative numbers, including −(−5)
The additive inverse of a number, from Class 6 integer work
Reading a fraction such as u/15 as a division
Multiplying out a bracket: 4(m + 6) = 4m + 24

What they should be able to do

Solve an equation whose unknown appears on both sides, by first collecting it on one side
Handle a subtracted negative term correctly when isolating the unknown
Set the full form and the shortened form of the same solution side by side and say which line was dropped
State the three printed observations governing the shorthand for a term, a factor and a divisor
Explain why a term changes sign when it moves but a factor does not
Solve an equation containing a bracket by at least two different routes
Produce an equation with no solution, and an equation whose solution is zero
Check a found value and say what the check does and does not settle

Where it usually goes wrong

"Move it across and change the sign." Correct as far as it goes, and the chapter does end up writing exactly that shape — but only after deriving it. Say the derivation out loud every time in the explanation, because the sign flip is a consequence of subtracting the same term from each side, not a rule about crossing a line.
"So a factor changes sign too when it moves." This is the wrong generalisation the shortcut invites, and observation (b) is printed to head it off: 2y = 14 becomes y = 14 ÷ 2, not y = 14 − 2. Error card 3 on p.181 makes exactly this mistake: in 2v − 4 = 6 the 2 leaves the left as a division, since 2v becomes v, and lands on the right as a subtraction, since 6 becomes 6 − 2. Card 1 is the same detachment committed with a term, and in the other direction — there the term crosses the equals sign and keeps its sign instead of flipping. Say which way round each card fails; calling them "matching" invites a teacher to describe card 1 as a sign that wrongly changed, which is the opposite of what it does.
"Subtracting a negative is the same as subtracting." Example 5 turns on 61 − (−5) = 66. Slow the figure down at that step.
"There is a correct order of operations for solving." Example 10 prints three orders side by side and all three work. What matters is that each line follows from the one above by an operation applied to each side.
"Every equation has a solution." Question 2 on p.172 asks for one that does not, and 5s = 3s in question 1 has the solution students most often refuse to write down, which is zero. Both belong in the explanation.
"Simplifying and solving are the same step." Column 3 of Example 10 opens by multiplying out a bracket — that changes the form of one side, and no operation is applied to the other. Distinguish a rewrite of one side from a move applied to both.
"Once I get an answer I am done." The chapter puts a checking prompt after Example 5 and again after Example 6. A check costs one substitution and catches exactly the sign slips this topic is about.

Questions to check understanding

Solve an equation with the unknown on both sides and check the answer
Solve an equation containing a bracket, by two different routes
Given a shortened solution, restore the missing line that justifies each step
Spot and correct a mistake in a worked solution and say what the mistake was — the whole of Part II, §7.3, pp.181–182 is built from these
Build an equation no value of the letter can satisfy
Given 4k + 1 = 13, evaluate related expressions such as 8k + 2 and 4k − 1 (Part II, §7.2, p.181, question 6)
Given 28p − 36 = 98, evaluate 14p − 19 and 28p − 38 (Part II, the exercise block following §7.4, p.187, question 10) — worth flagging, because p itself is not a whole number while both requested values are
Solve a chain of equations to trace a path through the printed maze (Part II, the exercise block following §7.4, pp.188–189, question 19)

Examples worth working on the board

Example 5 (Part II, §7.2, p.171). Input: 11y + (−5) = 61. The chapter subtracts (−5) from each side to reach 11y = 66, then notes the value can be seen directly from 11 × 6, and also divides each side by 11. Printed answer: y = 6. Show both finishes — the seen-directly one and the divided one — because the two-column comparison later leans on the division.
Example 6 (Part II, §7.2, pp.171–172). Input: 6y + 7 = 4y + 21. The chapter subtracts 4y from each side to reach 2y + 7 = 21, then subtracts 7 to reach 2y = 14, then divides by 2. Printed answer: y = 7. This is the first equation in the chapter with the unknown on both sides.
The two-column comparison (Part II, §7.2, p.173). Checked against the printed page. Both boxed tables sit on this page, after a lead-up on p.172. Each is five or seven lines tall and shows the same equation solved twice: the left column keeps the operation written on both sides, the right column writes only the result. The first box is Example 5's equation, the second is Example 6's. This layout is the lesson — reproduce the side-by-side structure exactly, and show the middle line of the left column fading out to produce the right.
The divided case (Part II, §7.2, pp.172–173). Input: u/15 = 6, which appears first as an exercise item on p.172 and is then worked on p.173. Multiplying each side by 15 gives the printed answer u = 90.
The three observations (Part II, §7.2, p.173). Printed with their own worked instances: for a term, 2y + 7 = 21 becomes 2y = 21 − 7; for a factor, 2y = 14 becomes y = 14 ÷ 2; for a divisor, u/15 = 6 becomes u = 6 × 15.
Example 10 (Part II, §7.2, p.177). Checked against the printed page. Input: 28(x + 4) + 300 = 1000, printed with three parallel solutions in three columns. Column 1 subtracts 300, divides by 28, then subtracts 4. Column 2 notices that 28, 300 and 1000 all divide by 4, divides the whole equation by 4 to reach 7(x + 4) + 75 = 250, and continues from there. Column 3 multiplies out the bracket first to reach 28x + 412 = 1000. Printed answer in all three: x = 21. Do not merge the columns; three routes landing together is the argument that no single route is the right one.
Figure it Out, question 1 (Part II, §7.2, p.172). Five equations to solve and check: 3x − 10 = 35; 5s = 3s; 3u − 7 = 2u + 3; 4(m + 6) − 8 = 2m − 4; u/15 = 6. Note that (b) is the one whose solution is zero, and (d) is the one with a bracket and a negative answer.
Figure it Out, question 2 (Part II, §7.2, p.172). The reader is asked to build an equation that no value can satisfy, and the printed hint points at a number four greater than something set equal to that thing five greater. Marked Math Talk. This is the first admission in the chapter that solving can come back empty.
The owl's note (Part II, §7.2, p.172). Checked against the printed page. A tinted band with a cartoon owl carrying a standing instruction to check the answer. It sits between Example 6 and the exercises. A plainer checking prompt, set as a question rather than in a band, follows Example 5 on p.171.
The error catalogue (Part II, §7.3, pp.181–182). Checked against the printed page. Nine short solutions in orange cards, each with a mistake to be found and mended, numbered 1 to 9. Do not work them here — they belong to the exercise topic — but point at cards 1, 3, 4, 6 and 7, which are all the same kind of failure: a piece of the equation shifted across, or an operation applied to only part of a side, without the both-sides step that would have licensed it. Card 5 is a different fault and should be named separately if it is shown at all — its first step, 15w − 4w = 26 becoming 15w = 26 + 4w, is a legitimate move, and the damage is done on the next line, where 4w is quietly replaced by 4; the lesson there is that 15w − 4w should have been collected into 11w first. The Mind the Mistake, Mend the Mistake topic covers all nine and is marked as an exercise set rather than an explanation.

Figures to have open

A two-column solution layout in which the same equation is solved twice and the middle line of the left column can be faded out to yield the right column. This is the central figure of the topic and must be built, not narrated.
A three-track parallel-solution layout for Example 10, with all three tracks visible at once and converging.
An annotated-equation treatment where an operation can be written beneath both sides and a term or factor shown cancelling — reused from Doing the same thing to both sides preserves equality.
No photograph, table or data figure from the textbook is needed. The book's own boxed tables carry the argument, but redraw them; do not reproduce the printed pages.

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.170–173 — Examples 5 and 6 (pp.171–172), the "Figure it Out" block and the observation that the procedure can be shortened (p.172), the two boxed column comparisons, the divided case and the three lettered observations (p.173)
Same part, same chapter, §7.2, bold subheading "Solving Problems", p.177 — Example 10 and its three parallel solutions
Same part, same chapter, §7.3 "Mind the Mistake, Mend the Mistake", pp.181–182 — the nine error cards this topic points forward to
Same part, same chapter, SUMMARY, p.190, bullet 4
Sibling topic: Mind the Mistake, Mend the Mistake, the error-spotting set, marked video: no