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Finding the Unknown

NCERT Class 7 · Mathematics Based on NCERT Class 7 Mathematics textbook · Free CBSE study kit

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Chapter Notes

Overview
This chapter introduces students to the concept of algebraic equations as a tool to find unknown values. It begins with visual problems using weighing scales and matchstick patterns to build intuition, then systematically teaches how to form and solve equations using inverse operations. The one big idea a student must walk away with is that an equation is a balance scale: whatever operation you perform on one side must be performed on the other to maintain equality, and solving an equation means isolating the unknown to find its value.
Key Concepts (10)
Equation
A statement of equality between two algebraic expressions, written with an equal sign ‘=’ between them.
e.g. 2n + 1 = 99 is an equation where the left-hand side (2n + 1) equals the right-hand side (99).
Left Hand Side (LHS)
The algebraic expression on the left side of the equal sign in an equation.
e.g. In 2n + 1 = 99, the LHS is 2n + 1.
Right Hand Side (RHS)
The algebraic expression on the right side of the equal sign in an equation.
e.g. In 2n + 1 = 99, the RHS is 99.
Solving the equation
The process of finding the value(s) of the letter−numbers for which the equality holds, or for which the value of the LHS becomes equal to the RHS.
e.g. Solving 2n + 1 = 99 gives n = 49 because 2 × 49 + 1 = 99.
Trial and error method
A method to solve an equation by substituting different values in place of the letter−number and checking which value makes LHS = RHS.
e.g. For 2n + 1 = 99, trying n = 5 gives LHS = 11, n = 50 gives 101, and n = 49 gives exactly 99.
Inverse operations
Operations that undo each other: addition and subtraction are inverse operations; multiplication and division are inverse operations.
e.g. To remove +88 from an expression, subtract 88 from both sides; to remove ×7, divide both sides by 7.
Unknown
A letter−number (like x, y, n) used to represent an unknown quantity in an equation.
e.g. In 5x – 4 = 7, x is the unknown.
Bījagaṇita
The ancient Indian term for algebra, meaning 'seed mathematics' — the answer is hidden inside an unknown number like a tree inside a seed.
e.g. Brahmagupta’s Brāhmasphuṭasiddhānta (628 CE) explained how to add, subtract, and multiply unknown numbers using letters.
Algebra
The branch of mathematics that uses letter symbols to solve mathematical problems, generalise patterns, and justify mathematical claims.
e.g. Using algebra, we can prove that the sum of two odd numbers is always even.
Brahmagupta's formula for solving Ax + B = Cx + D
For an equation of the form Ax + B = Cx + D, the solution is given by x = (D – B) / (A – C).
e.g. For 650m + 4000 = 500m + 5050, m = (5050 – 4000) / (650 – 500) = 1050 / 150 = 7.
Formulas & Equations (10)
2n + 1 = 99
Reads as: Two n plus one equals ninety-nine
Variables: n = position number in the matchstick sequence (unitless)
Use when: Finding which position in a matchstick pattern uses exactly 99 sticks
⚠ Students forget to subtract 1 before dividing by 2, writing n = (99 + 1)/2 = 50 instead of n = (99 – 1)/2 = 49.
5x – 4 = 7
Reads as: Five x minus four equals seven
Variables: x = unknown number (unitless)
Use when: Solving a linear equation with one unknown using inverse operations
⚠ Students add 4 to only one side, writing 5x = 7 + 4 correctly but then forget to divide both sides by 5.
11y + (–5) = 61
Reads as: Eleven y plus negative five equals sixty-one
Variables: y = unknown number (unitless)
Use when: Solving an equation where a negative term is added
⚠ Students incorrectly subtract (–5) as 11y = 61 + 5, forgetting that subtracting a negative is adding the positive.
6y + 7 = 4y + 21
Reads as: Six y plus seven equals four y plus twenty-one
Variables: y = unknown number (unitless)
Use when: Solving equations where the unknown appears on both sides
⚠ Students move terms without changing signs, e.g., writing 6y – 4y = 21 – 7 correctly but then forgetting to divide by 2.
u/15 = 6
Reads as: u divided by fifteen equals six
Variables: u = unknown number (unitless)
Use when: Solving an equation where the unknown is divided by a number
⚠ Students divide both sides by 15 instead of multiplying, writing u = 6/15 instead of u = 6 × 15 = 90.
3k + 1 = 100
Reads as: Three k plus one equals one hundred
Variables: k = step number in the tile sequence (unitless)
Use when: Finding which step in a tile pattern uses exactly 100 tiles
⚠ Students forget to subtract 1 first, writing k = (100 + 1)/3 instead of k = (100 – 1)/3 = 33.
25p + 50 = 500
Reads as: Twenty-five p plus fifty equals five hundred
Variables: p = total number of people (unitless)
Use when: Solving real-life problems involving fixed cost and per-unit cost
⚠ Students subtract 50 from only the LHS, writing 25p = 500 without subtracting 50 from the RHS.
4000 + 650m = 5050 + 500m
Reads as: Four thousand plus six hundred fifty m equals five thousand fifty plus five hundred m
Variables: m = number of months (unitless)
Use when: Finding when two savings amounts become equal
⚠ Students subtract 500m from only one side, forgetting to subtract from both sides.
4x – 4 = 24
Reads as: Four x minus four equals twenty-four
Variables: x = starting number in the math trick (unitless)
Use when: Solving a multi-step word problem by forming an equation
⚠ Students add 4 to only one side, writing 4x = 24 + 4 correctly but then forget to divide by 4.
x = (D – B) / (A – C)
Reads as: x equals D minus B divided by A minus C
Variables: A, B, C, D are constants in the equation Ax + B = Cx + D
Use when: Directly solving equations of the form Ax + B = Cx + D using Brahmagupta's formula
⚠ Students swap A and C or B and D, e.g., writing x = (B – D)/(C – A) which gives the wrong sign.
How It Works
Solving an equation by trial and error
  1. Step 1: Choose a value for the unknown and substitute it into the LHS.
  2. Step 2: Calculate the LHS and compare it to the RHS.
  3. Step 3: If LHS = RHS, that value is the solution.
  4. Step 4: If not, choose a different value and repeat until equality is found.
💡 This method can be inefficient for large numbers; systematic methods using inverse operations are faster.
Solving an equation using inverse operations (systematic method)
  1. Step 1: Identify the term added or subtracted to the unknown term and remove it by performing the inverse operation on both sides.
  2. Step 2: If the unknown is multiplied by a number, divide both sides by that number.
  3. Step 3: If the unknown is divided by a number, multiply both sides by that number.
  4. Step 4: The resulting value is the solution. Check by substituting back into the original equation.
💡 Always perform the same operation on both sides of the equation to maintain equality.
Solving equations with unknowns on both sides
  1. Step 1: Bring all terms containing the unknown to one side by adding or subtracting the same term on both sides.
  2. Step 2: Simplify the equation so that the unknown is only on one side.
  3. Step 3: Remove any constant terms added or subtracted to the unknown term using inverse operations.
  4. Step 4: Isolate the unknown by dividing or multiplying as needed.
  5. Step 5: Check the solution by substituting back into the original equation.
💡 When moving a term from one side to the other, its sign changes (addition becomes subtraction and vice versa).
Forming an equation from a word problem
  1. Step 1: Identify the unknown quantity and represent it with a letter (e.g., x, y, n).
  2. Step 2: Translate the problem statement into an algebraic expression using the unknown.
  3. Step 3: Set the expression equal to the given total or condition to form an equation.
  4. Step 4: Solve the equation using inverse operations.
  5. Step 5: Interpret the solution in the context of the problem.
💡 Read the problem carefully to correctly identify what the unknown represents — it could be a number, a count, a cost, etc.
Using Brahmagupta's formula to solve Ax + B = Cx + D
  1. Step 1: Identify the constants A, B, C, and D from the equation.
  2. Step 2: Substitute into the formula x = (D – B) / (A – C).
  3. Step 3: Simplify the numerator and denominator.
  4. Step 4: Divide to find the value of x.
  5. Step 5: Verify by substituting back into the original equation.
💡 The formula works only for equations of the form Ax + B = Cx + D. Ensure the signs of B and D are correctly taken (including negative signs).
Exam Traps — Don't Make These Mistakes
⚠ Forgetting to perform the same operation on both sides
✗ 5x – 4 = 7 → 5x = 7 (they subtract 4 from only the LHS)
→ An equation is a balance; if you subtract 4 from only one side, the equality is broken. You must subtract 4 from both sides: 5x – 4 + 4 = 7 + 4 → 5x = 11.
✓ 5x – 4 = 7 → Add 4 to both sides: 5x – 4 + 4 = 7 + 4 → 5x = 11 → Divide both sides by 5: x = 11/5.
⚠ Incorrectly moving terms without changing signs
✗ 6y + 7 = 4y + 21 → 6y + 4y = 21 – 7 (they move 4y to LHS without changing sign)
→ When a term moves from one side to the other, its sign changes. 4y on the RHS is positive, so moving it to LHS makes it –4y. Correct: 6y – 4y = 21 – 7.
✓ 6y + 7 = 4y + 21 → Subtract 4y from both sides: 6y – 4y + 7 = 21 → 2y + 7 = 21 → Subtract 7: 2y = 14 → y = 7.
⚠ Mishandling negative signs when subtracting a negative
✗ 11y + (–5) = 61 → 11y = 61 – 5 (they treat –5 as +5)
→ To remove –5 from LHS, you add 5 to both sides (since subtracting a negative is adding). Correct: 11y = 61 + 5 = 66.
✓ 11y + (–5) = 61 → Add 5 to both sides: 11y = 61 + 5 → 11y = 66 → y = 6.
⚠ Dividing instead of multiplying when the unknown is divided
✗ u/15 = 6 → u = 6/15
→ To isolate u, you need to undo division by 15. The inverse of division is multiplication. Multiply both sides by 15: u = 6 × 15 = 90.
✓ u/15 = 6 → Multiply both sides by 15: u = 6 × 15 → u = 90.
⚠ Using Brahmagupta's formula with wrong sign for constants
✗ For 3x – 6 = 2x + 4, they write x = (4 – 6)/(3 – 2) = –2/1 = –2
→ In the equation Ax + B = Cx + D, B is –6 and D is 4. The formula is x = (D – B)/(A – C) = (4 – (–6))/(3 – 2) = (4 + 6)/1 = 10. The student forgot that subtracting a negative is adding.
✓ Identify A = 3, B = –6, C = 2, D = 4. Then x = (D – B)/(A – C) = (4 – (–6))/(3 – 2) = (4 + 6)/1 = 10.
Compare & Contrast
Efficiency and reliability
Trial and error method
Systematic method using inverse operations
Trial and error involves guessing values and checking, which can be slow and inefficient for large numbers. The systematic method uses inverse operations (adding, subtracting, multiplying, dividing both sides) to directly isolate the unknown, making it faster and always reliable.
Steps to solve
Equation with unknown on one side (e.g., 5x – 4 = 7)
Equation with unknown on both sides (e.g., 6y + 7 = 4y + 21)
For an equation with the unknown on one side, you directly apply inverse operations to isolate the unknown. For an equation with unknowns on both sides, you first bring all unknown terms to one side (by adding or subtracting the smaller coefficient term) before isolating the unknown.
Exam Focus
Marks pattern: 2 MCQs (1 mark each), 1 Short Answer (3 marks), 1 Long Answer (5 marks) — total 10 marks typically
Most asked: Solving linear equations in one variable using inverse operations · Forming equations from word problems (cost, savings, age, number problems) · Checking solutions by substitution · Equations with unknowns on both sides
Don't skip: Word problems requiring forming and solving an equation (e.g., 'Ramesh and Suresh marbles', 'Madhubanti's party', 'Jahnavi and Sunita savings'); Equations with unknowns on both sides (e.g., 6y + 7 = 4y + 21); Applying Brahmagupta's formula to solve Ax + B = Cx + D
One-Line Revision (read aloud before the exam)
  1. An equation is a statement of equality between two algebraic expressions, with an equal sign ‘=’ between them.
  2. The LHS (Left Hand Side) and RHS (Right Hand Side) of an equation must always be equal.
  3. Solving an equation means finding the value of the unknown that makes LHS = RHS.
  4. The trial and error method involves substituting values until equality is found.
  5. Inverse operations: addition undoes subtraction, subtraction undoes addition, multiplication undoes division, division undoes multiplication.
  6. When solving an equation, always perform the same operation on both sides to maintain equality.
  7. To remove a term added to the unknown, subtract it from both sides; to remove a term subtracted, add it to both sides.
  8. If the unknown is multiplied by a number, divide both sides by that number.
  9. If the unknown is divided by a number, multiply both sides by that number.
  10. When the unknown appears on both sides, bring all unknown terms to one side by adding or subtracting the smaller coefficient term.
  11. Always check your solution by substituting it back into the original equation.
  12. Brahmagupta's formula for Ax + B = Cx + D is x = (D – B) / (A – C).
  13. Bījagaṇita (algebra) means 'seed mathematics' — the answer is hidden inside the unknown.
  14. Ancient Indian mathematicians used symbols like yā (yāvat-tāvat) for unknowns and rū for known quantities.
  15. A real-life situation like 'fixed cost + per-unit cost × number of units = total cost' can be modelled as an equation.

MCQs — 10 Questions with Answers

Q1. Which of the following is an equation?

  • A. 3x + 5
  • B. 3x + 5 = 20 ✓
  • C. 3x + 5 < 20
  • D. 3x + 5 + 2

Answer: B — An equation must have an equals sign between two expressions; 3x + 5 = 20 is the only statement of equality.

Q2. What is the value of x if 2x = 16?

  • A. 6
  • B. 8 ✓
  • C. 14
  • D. 18

Answer: B — Dividing both sides by 2 gives x = 16 ÷ 2 = 8.

Q3. To solve 5x - 3 = 22, what is the first step?

  • A. Divide both sides by 5
  • B. Subtract 3 from both sides
  • C. Add 3 to both sides ✓
  • D. Multiply both sides by 5

Answer: C — First we remove the -3 by adding 3 to both sides, giving 5x = 25.

Q4. In a balanced weighing scale, if we remove equal weights from both plates, the scale will:

  • A. Become unbalanced
  • B. Tilt to one side
  • C. Remain balanced ✓
  • D. Need rebalancing

Answer: C — Removing the same amount from both sides preserves the balance, just as adding or subtracting the same value from both sides of an equation preserves equality.

Q5. A shopkeeper buys eggs at the market. If 3 eggs cost ₹15, which equation finds the cost of one egg (e)?

  • A. e + 3 = 15
  • B. 3e = 15 ✓
  • C. e - 3 = 15
  • D. e ÷ 15 = 3

Answer: B — Three eggs at cost e each total 3e, which equals ₹15, so 3e = 15.

Q6. If a baker needs 2x + 4 = 12 cups of flour for a recipe, what is x?

  • A. 2
  • B. 4 ✓
  • C. 6
  • D. 8

Answer: B — Subtract 4 from both sides to get 2x = 8, then divide by 2 to get x = 4.

Q7. Solve the equation y/5 = 7.

  • A. y = 2
  • B. y = 12
  • C. y = 35 ✓
  • D. y = 7

Answer: C — When dividing creates the equation, multiply both sides by 5 to remove the divisor: y = 7 × 5 = 35.

Q8. For the matchstick pattern where position n has 2n + 1 sticks, which position uses exactly 51 sticks?

  • A. Position 24
  • B. Position 25 ✓
  • C. Position 26
  • D. Position 27

Answer: B — Solving 2n + 1 = 51 gives 2n = 50, so n = 25.

Q9. In the equation 6m + 7 = 4m + 21, how many steps are needed to isolate m?

  • A. 1 step
  • B. 2 steps
  • C. 3 steps ✓
  • D. 4 steps

Answer: C — Step 1: Subtract 4m from both sides to get 2m + 7 = 21; Step 2: Subtract 7 from both sides to get 2m = 14; Step 3: Divide by 2 to get m = 7.

Q10. Which equation has no solution, based on the hint that 4 more than a number and 5 more than a number can never be equal?

  • A. x + 4 = x + 5 ✓
  • B. 2x + 4 = 8
  • C. x + 5 = 10
  • D. 3x = x + 6

Answer: A — In x + 4 = x + 5, subtracting x from both sides gives 4 = 5, which is never true, so there is no solution.

Flashcards

What is an equation?

A statement of equality between two algebraic expressions separated by an equals sign.

What does 'solving an equation' mean?

Finding the value of the unknown letter-number that makes the left side equal to the right side.

If we add 5 to the left side of an equation, what must we do to keep it balanced?

Add 5 to the right side also, so both sides remain equal.

How do we remove a term like +7 from one side of an equation?

Subtract 7 from both sides of the equation.

How do we remove a factor like ×3 from one side of an equation?

Divide both sides of the equation by 3.

What is the trial and error method?

Substituting different values for the unknown until we find the one that makes LHS equal to RHS.

For the equation 2n + 1 = 99, what does the expression 2n + 1 represent?

The pattern rule for the number of matchsticks at position n in the sequence.

If 4x = 20, what is x?

Dividing both sides by 4 gives x = 5.

Why is checking the solution important?

To verify that the value we found actually makes the equation true by substituting it back.

In the weighing scale, what property allows us to remove equal weights from both sides?

If the scale is balanced and we remove the same weight from both plates, it remains balanced.

Important Board Questions

What do we call a statement of equality between two algebraic expressions? [1 mark]

Think of the mathematical statement that has an equal sign connecting two sides (like 2x + 1 = 99).

In a balanced weighing scale, if 3 bunches of bananas weigh the same as 12 kg, write an equation to find the weight of one bunch. Let b = weight of one bunch. [2 marks]

Three bunches (3b) balance with 12 kg on the other side. Write what equals what using the variable b.

Solve the equation 4y - 6 = 18 step by step. Check your answer by substituting it back into the original equation. [3 marks]

Step 1: Add 6 to both sides to isolate the 4y term. Step 2: Divide both sides by 4 to find y. Step 3: Substitute your answer back and verify both sides are equal.

A mango seller at the market knows that 5 mangoes plus a watermelon (weighing 2 kg) cost ₹50 total. If each mango costs m rupees, write and solve an equation to find the cost of one mango, assuming the seller values the watermelon at ₹10. Show all steps and verify your solution. [5 marks]

Cost of 5 mangoes plus cost of watermelon equals ₹50. This means 5m + 10 = 50. Solve by subtracting 10, then dividing by 5, and check by substituting back into 5m + 10.

Assertion-Reason Questions for Finding the Unknown

Q1. Assertion (A):

The equation 3x + 5 = 20 and the equation 3x = 15 have the same solution.

Reason (R):

Subtracting 5 from both sides of the first equation gives the second equation, and this operation preserves equality.

Answer: A — Both A and R are true, and R is the correct explanation of A — Both assertion and reason are true. The first equation: 3x + 5 = 20 → 3x = 15 (subtract 5 from both sides). Both equations have the solution x = 5. The reason correctly explains why they have the same solution: performing identical operations on both sides maintains equality and preserves the solution set.

Q2. Assertion (A):

In a weighing scale that is balanced, if we remove 2 kg from the left plate and 2 kg from the right plate, the scale remains balanced.

Reason (R):

Removing equal weights from both plates changes the total weight but does not change the balance.

Answer: B — Both A and R are true, but R is NOT the correct explanation of A — The assertion is true—the scale will remain balanced. However, the reason is misleading. The reason says removing equal weights 'changes the total weight,' which is true but not the reason for maintaining balance. The correct reason is that the difference in weights on each plate remains the same. Both parts are true but the reason does not properly explain why balance is maintained.

Q3. Assertion (A):

The trial and error method is always the most efficient way to solve equations.

Reason (R):

Trial and error involves testing many values, which becomes inefficient when the solution is a large number.

Answer: D — A is false but R is true — The assertion is false—trial and error is often inefficient. The reason is true—trial and error does become very time-consuming for large solutions. Since the assertion is false and the reason is true, this is option D.

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Case-Based Questions (CBQ) for Finding the Unknown

Case Study:

Rahul works at a vegetable market in Mumbai. He sells tomatoes at Rs. 12 per kg. Yesterday, Rahul earned Rs. 300 from selling tomatoes. Today, he wants to know how many kilograms he sold yesterday. He also wants to calculate his total earnings if he sells 8 more kilograms today at the same rate. To solve this problem, Rahul decides to use equations with unknowns instead of guessing.

Q1. If Rahul sold x kg of tomatoes yesterday, which equation represents his earnings yesterday?

  • A. 12x = 300 ✓
  • B. x + 12 = 300
  • C. 12 + x = 300
  • D. 300/12 = x

Answer: A

Q2. How many kilograms of tomatoes did Rahul sell yesterday?

  • A. 20 kg
  • B. 25 kg ✓
  • C. 30 kg
  • D. 35 kg

Answer: B

Q3. If Rahul sells 8 more kilograms today at Rs. 12 per kg, what will be his earnings for today?

  • A. Rs. 96 ✓
  • B. Rs. 108
  • C. Rs. 120
  • D. Rs. 144

Answer: A

Q4. What is Rahul's total earnings for yesterday and today combined?

  • A. Rs. 420
  • B. Rs. 408
  • C. Rs. 396 ✓
  • D. Rs. 450

Answer: C

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