Educational Galaxy

Chapter 7: Finding the Unknown — EG's own notes, in simple words

Book: Ganita Prakash — Part II (CBSE, Class 7)  |  Chapter 7  |  Medium: English

Diagram (labels are printed on the image): A balanced scale showing the two sides of an equation. A general diagram; it is not a scene from the book.

Solving for a mystery value. A general picture, not a book scene.

What the chapter is about

An equation is a statement that two expressions are equal, like a balanced weighing scale: the left-hand side (LHS) and the right-hand side (RHS) weigh the same. Finding the unknown means finding the value that makes the scale balance. The chapter's great rule is simple: whatever you do to one side, do the same to the other, and the balance holds.

7.1 Find the unknowns

The chapter starts with balance scales carrying unknown weights — sacks and eggs whose weights you must deduce. The key move is already there: if you remove equal weights from both plates, the scale stays balanced. In one figure, all sacks have the same weight, and removing one sack from each plate leaves a comparison you can read directly. The same idea, written with letters, becomes equation solving: an unknown sack weight e, a known weight, and a balance between them. Before any algebra, the skill is reasoning: what must be true for the scale to balance, and what can you remove from both sides without breaking it?

7.2 Solving equations systematically

Take the book's example: one fried egg weighs e units, and a scale problem gives an equation like 2e + 5 = 13 (two eggs plus 5 units balance 13 units). Solve by undoing, the same operation on both sides at each step:

1. Subtract 5 from both sides: 2e + 5 − 5 = 13 − 5, so 2e = 8.
2. Divide both sides by 2: e = 4.
3. Check: LHS = 2 × 4 + 5 = 13 = RHS. Balanced.

Trial works too — the chapter also tries values, for example testing n = 40 in an equation whose RHS is 99 and noting "the LHS is 81, we are getting closer" — but systematic solving is faster and never depends on luck. The rule behind it: when the same operation is performed on both sides of an equation, equality is maintained. Adding, subtracting, multiplying (by a non-zero number) and dividing (by a non-zero number) all preserve the balance.

7.3 Mind the mistake, mend the mistake

Most equation errors are balance errors: doing something to one side only, flipping a sign when moving a term, or dividing by a number that is not a common factor of the whole side. The chapter's habit is to "mind the mistake, mend the mistake": when a check fails, do not start over blindly — find the step where LHS and RHS stopped agreeing, fix that step, and re-check. A check that fails is information, not failure: it tells you exactly which step to inspect.

7.4 A pinch of history

Equations are ancient. Brahmagupta, in Chapter 18 of his Brāhmasphuṭasiddhānta (628 CE), worked with positive and negative numbers and with equations — his work on integers and fractions appears across this book. His ideas travelled in translations into Arabic and influenced Al-Khwarizmi, who around 825 CE wrote Hisab al-jabr wal-muqabala, meaning "calculation by restoring and balancing" — the book that gave algebra its name (al-jabr). As the mathematician David Mumford put it, Brahmagupta is a key person in the creation of algebra as we know it. When you "restore and balance" an equation today, you are using a method with a 1,400-year history.

What you should learn

• Read an equation as a balanced scale: LHS = RHS.
• Find unknowns on scales by removing equal weights from both plates.
• Solve equations systematically: same operation on both sides, undoing in reverse order.
• Check every solution by substituting it back into the original equation.
• Find and fix the faulty step when a check fails, instead of restarting blindly.

Chart / Table: Solving equations

Practice set — try it yourself

1. Solve x + 9 = 20. Check your answer.
   Hint: What undoes adding 9?
2. Solve 5y = 45.
   Hint: What undoes multiplying by 5?
3. Solve 2x + 3 = 15. Show both steps.
   Hint: Undo in reverse order.
4. If two identical sacks plus 4 kg balance 10 kg, what does one sack weigh?
   Hint: Remove the same weight from both plates first.
5. Someone solves x − 6 = 10 as x = 4. What is the mistake, and what is the correct answer?
   Hint: Check by substitution.

Questions with Answers

1. What is an equation, and what does it mean to solve one?
   An equation states that two expressions are equal (LHS = RHS), like a balanced scale. Solving it means finding the value of the unknown that makes the two sides equal.
2. Solve 2e + 5 = 13 and check the answer.
   Subtract 5 from both sides: 2e = 8. Divide both sides by 2: e = 4. Check: 2 × 4 + 5 = 13. Correct.
3. Solve 4n − 3 = 17 step by step.
   Add 3 to both sides: 4n = 20. Divide both sides by 4: n = 5. Check: 4 × 5 − 3 = 20 − 3 = 17. Correct.
4. A scale balances with 3 identical sacks on one side and 15 kg on the other. What is one sack's weight? Use the remove-from-both-sides idea.
   3 sacks = 15 kg. Divide both sides by 3 (share equally): 1 sack = 5 kg. Check: 3 × 5 = 15. Correct.
5. Why does doing the same thing to both sides keep an equation true?
   Because equality means both sides have the same value. Applying the same operation to equal values gives equal results — just as adding the same weight to both plates of a balanced scale keeps it balanced.

Extension questions with answers — NCERT exercise types

1. Solve 5x + 8 = 3x + 20. Show the balancing steps.
   Subtract 3x from both sides: 2x + 8 = 20. Subtract 8 from both sides: 2x = 12. Divide both sides by 2: x = 6. Check: LHS = 30 + 8 = 38; RHS = 18 + 20 = 38. Correct.
2. Think of a number, multiply it by 2, add 10, divide by 2, subtract the original number, add 3. The book's magic trick predicts 8. Prove it with algebra.
   Let the number be x. Steps: 2x → 2x + 10 → (2x + 10) ÷ 2 = x + 5 → (x + 5) − x = 5 → 5 + 3 = 8. The x cancels, so the result is always 8, whatever number you start with.
3. A number's double plus 9 is 27. Find the number by trial and by systematic solving.
   Trial: try 9 → 2 × 9 + 9 = 27. Correct, so the number is 9. Systematic: 2n + 9 = 27 → subtract 9: 2n = 18 → divide by 2: n = 9. Both methods agree.
4. Solve (x ÷ 4) + 6 = 11, and explain each undo step.
   Subtract 6 from both sides: x ÷ 4 = 5. Multiply both sides by 4: x = 20. Check: 20 ÷ 4 + 6 = 5 + 6 = 11. The undo order reverses the building order (+6 last built, so first undone).
5. A student writes: 3x = 24, so x = 24 − 3 = 21. Identify the error and correct it.
   The error is treating multiplication as addition: 3x means 3 times x, not 3 plus x. The correct undo is division: x = 24 ÷ 3 = 8. Check: 3 × 8 = 24. Correct.
6. Two equations: x + 7 = 15 and 2x = 16. Do they have the same solution? What does that tell you?
   Yes: x = 8 solves both (8 + 7 = 15; 2 × 8 = 16). Different equations can share a solution — equations are not unique to their answers, and checking a solution means testing it in the equation given.

How to use this page: Read the summary, say the Key Terms in your own words, then try the practice set without looking. Use the Questions with Answers to check yourself afterwards. Ask your teacher or parent to check your answers.