Algebra | ICSE Class 7 Maths Notes

This note covers variables and constants, terms and coefficients, like and unlike terms, forming and evaluating algebraic expressions, adding and subtracting expressions, simple linear equations in one variable, contextual problems, and simple inequalities.

What do letters represent in algebra?

Algebra uses letters to express relationships between numbers and quantities. A variable is a symbol that can represent different numerical values. A constant has a fixed value. An unknown is a quantity whose value is to be found in a particular problem.

An algebraic expression combines numbers and letters through operations such as addition, subtraction and multiplication. Letters used to represent numbers are also called letter-numbers. Define what a letter represents before using it, including the unit when it represents a measured quantity.

How can a relationship stay the same while values change?

Shabnam is 3 years older than Aftab. Let a represent Aftab’s age in years and s represent Shabnam’s age in years. Then s = a + 3. The symbol = means “is equal to”, and + means addition.

The expression a + 3 tells us to add 3 to Aftab’s age. The age represented by a can change, but the age difference remains 3 years. We can also write a = s − 3, where − means subtraction.

Worked example 1. Shabnam is 3 years older than Aftab. Find Shabnam’s age when Aftab is 23 years old.

Answer: Let a and s represent their ages as defined above. Put a = 23 into s = a + 3. Then s = 23 + 3 = 26. Shabnam is 26 years old.

A letter represents a numerical quantity; it is not an abbreviation for the object itself. Here a represents an age, not Aftab as a person. Stating that meaning makes the expression understandable and tells us what kind of answer to expect.

What are terms, factors and coefficients?

A term is a part of an expression that is added to the other parts. Subtraction can be read as addition of a negative term. In the expression 5m + 3, let m represent a number. The two terms are 5m and 3.

The notation 5m means 5 × m, where × means multiplication. Quantities multiplied together are called factors. Thus, 5 and m are factors of 5m. The numerical factor 5 is the numerical coefficient, usually shortened here to coefficient.

How do signs belong to terms?

In 7p − 3q, let p represent the score for a correct quiz answer and q the penalty for an incorrect answer. The signed terms are 7p and −3q. Their coefficients are 7 and −3. Keep the minus sign with its term.

An integral coefficient is a coefficient that is an integer. Integers include zero, positive whole numbers and their negatives. Coefficients in the expressions added and subtracted here are integers. A constant term has no variable, as with 3 in 5m + 3.

Definition: Like terms have the same variable part, including the same powers of those variables. Unlike terms have different variable parts. A power records repeated multiplication of the same factor.

For the linear terms used here, compare the letters: 5c, c and 10c are like terms, where c represents the price of a pencil. The unwritten coefficient of c is 1. But 18c and 11d are unlike terms when d represents the price of an eraser.

Different numerical coefficients do not make terms unlike. Before combining terms, compare their variable parts. Then calculate with the coefficients and keep the common variable part unchanged.

How do we translate words into algebraic expressions?

First identify the quantity that is unknown or allowed to vary. Choose a letter for it and state its meaning. Next identify the operation described in words. “More than” indicates addition, “less than” indicates subtraction, and “times” indicates multiplication.

Let x represent a number in the following translations. The position of the words matters: subtracting something from a number is different from subtracting that number from something else. Read the complete phrase before writing the expression.

How can an expression use two variables?

A coconut costs ₹35 and one kilogram of jaggery costs ₹60. The symbol ₹ denotes rupees and kg denotes kilograms. For this problem, let c be the number of coconuts and j the quantity of jaggery in kilograms.

The coconut cost is 35c rupees and the jaggery cost is 60j rupees. Adding the two costs gives 35c + 60j rupees. Both variable meanings belong to this problem; the letter c has a different stated meaning in the pencil example.

Worked example 2. Find the cost of 10 coconuts at ₹35 each and 5 kg of jaggery at ₹60 per kg.

Answer: Substitute c = 10 and j = 5 into 35c + 60j. The cost is 35 × 10 + 60 × 5 = 350 + 300 = ₹650.

A two-variable expression retains both quantities until their values are given. Do not combine the coconut count and the jaggery quantity into one count: each has its own price and its own role in the calculation.

How do we evaluate an expression by substitution?

Evaluation means finding an expression’s numerical value. Substitution means replacing a variable by its given value. Replace every occurrence of that variable consistently, then perform the indicated arithmetic operations. Multiplication must be completed before the terms are added or subtracted.

In algebraic notation, the multiplication sign is often omitted between a number and a letter. For example, 7k means 7 × k, where k represents a number. If k = 4, the expression has value 7 × 4 = 28.

Worked example 3. Evaluate 5m + 3 when the number m is 2.

Answer: Replace m by 2 to obtain 5 × 2 + 3. Multiplication gives 10, and adding 3 gives 13. Thus the value of the expression is 13.

Why must brackets and signs be preserved?

Brackets group quantities that must be treated together. If a variable is replaced by a negative number, brackets help preserve its sign. For a number a = −4, the expression 10 − a becomes 10 − (−4), whose value is 14.

The expression 5u, where u represents a number, means five times that number. The expression 5 + u means five more than that number. They give different values for most values of u. At u = 2, they give 10 and 7 respectively.

Note: Substitution changes letters into numbers; it does not change multiplication into joining digits. Nor does it allow a subtraction sign or a bracket to be dropped.

Two equivalent expressions have the same value for every permitted choice of their variables. By contrast, obtaining one numerical value after substitution answers a particular evaluation question. Keep the distinction between simplifying a general expression and evaluating it for given values.

How do we add and simplify like terms?

To simplify an expression is to write an equivalent expression in a simpler form. When adding algebraic expressions, collect like terms and add their coefficients. The variable part tells us what is being counted; the coefficient tells us how many of that quantity we have.

Property: Swapping and grouping terms preserves their sum

The order in which terms are added can be changed, and terms can be grouped conveniently, without changing the sum. Keep each sign attached to its term when rearranging. This allows terms with the same variable part to be placed together.

Let l and b represent a rectangle’s length and breadth, and p its perimeter, all in the same length unit. The perimeter is the total distance around its boundary. Then p = l + b + l + b = 2l + 2b.

Draw a rectangle. Label its opposite lengths l and its opposite breadths b. Write l + b + l + b alongside it, then group equal side lengths to obtain 2l + 2b.

Property: The distributive property combines like terms

The distributive property says that a multiple of a sum equals the sum of the corresponding multiples. It also explains combining like terms: 5c + 3c + 10c = (5 + 3 + 10)c = 18c.

Worked example 4. A shop sells pencils and erasers in the quantities below. Let c be the price per pencil and d the price per eraser, in the same currency. Find its total receipts over the three days.

Item	Day 1	Day 2	Day 3
Pencils, price c	5	3	10
Erasers, price d	4	6	1

Answer: Pencil receipts are 5c + 3c + 10c = 18c. Eraser receipts are 4d + 6d + d = 11d. The total is 18c + 11d, which retains the two unlike terms.

Combining unlike terms would lose the information that the prices can be different. Addition still gives a valid expression, but it cannot merge unlike terms into a single like term.

How do we subtract one algebraic expression from another?

In “subtract the second expression from the first”, begin with the first expression. Put the expression being subtracted inside brackets. A minus sign outside brackets applies to the whole expression inside them, not just its first term.

Property: Subtracting a bracket changes every term’s sign

To subtract an expression, add the negative of each of its terms. Positive terms inside the subtracted bracket become negative; negative terms become positive. After removing the brackets in this way, collect like terms and calculate their coefficients.

Worked example 5. A shop initially charges ₹40 per chair and ₹75 per table for a day’s rental. It returns ₹6 per chair and ₹10 per table afterwards. Let x and y be the numbers of chairs and tables rented. Find the final amount paid.

Answer: The initial payment is 40x + 75y rupees. The refund is 6x + 10y rupees. Subtracting gives (40x + 75y) − (6x + 10y) = 40x + 75y − 6x − 10y = 34x + 65y rupees.

The refund for tables must also be subtracted. Leaving +10y after removing the bracket would add that refund to the payment instead of taking it away. The context therefore provides a useful check on the signs.

How does subtraction work with negative terms?

In a quiz, let p represent the score for a correct answer and q the penalty for an incorrect answer. Charu’s total is 21p − 9q and Krishita’s total is 23p − 7q. Their score difference is found by subtracting Charu’s expression from Krishita’s.

The difference is (23p − 7q) − (21p − 9q). Remove the brackets to obtain 23p − 7q − 21p + 9q. Collecting terms gives 2p + 2q. The final +9q comes from subtracting the negative term −9q.

Note: A sign change caused by removing a subtracted bracket must be completed before like terms are combined. Write this intermediate line when it helps make the subtraction clear.

How do patterns and scores use algebraic expressions?

A pattern rule describes a repeated relationship. Algebra lets us state that relationship without writing a separate arithmetic calculation for every case. The letter must represent the quantity that controls the pattern, such as the number of repeated shapes.

Let n be the number of separate L shapes made from matchsticks. Each L needs two matchsticks. The total number of matchsticks is therefore 2n. This expression uses multiplication because the same number of sticks is needed for each L.

The drawing shows one L, then two Ls side by side, then three Ls side by side. Each L has one upright matchstick and one horizontal matchstick.

See Fig. 4.2 in your NCERT textbook

The diagram shows how a repeated count becomes an expression. To use the expression for a particular arrangement, substitute its number of Ls for n. The letter represents the number of shapes, while the expression represents the number of matchsticks.

How can scores from several rounds be combined?

In a quiz, p represents the score for a correct answer and q the penalty for an incorrect answer. The expression 7p − 3q means the score from seven correct answers with the penalty for three incorrect answers subtracted.

Worked example 6. Charu’s scores in three rounds are 7p − 3q, 8p − 4q and 6p − 2q. Find an expression for her total score.

Answer: Add the three expressions: (7p − 3q) + (8p − 4q) + (6p − 2q). Grouping like terms gives (7 + 8 + 6)p − (3 + 4 + 2)q = 21p − 9q.

If p = 4 and q = 1, the first-round expression becomes 7 × 4 − 3 × 1 = 25. This is an evaluation of one round; 21p − 9q describes the combined score across all three rounds.

What is a simple linear equation in one variable?

An equation states that two expressions have equal values. The expression before the equals sign is the left-hand side, abbreviated LHS. The expression after it is the right-hand side, abbreviated RHS.

A solution of an equation is a value of the unknown that makes both sides equal. Solving means finding such a value. In a simple linear equation in one variable, there is one unknown, occurring to the first power after simplification.

The first power means the variable itself rather than repeated multiplication by itself. The equation 2n + 1 = 99 is linear in n. Here n represents a position number in a matchstick sequence whose rule is 2n + 1 matchsticks.

How is an expression different from an equation?

The expression 2n + 1 tells us how to calculate a count from n. The equation 2n + 1 = 99 supplies an extra condition: the count must equal 99. That condition allows us to find which position has the required number of sticks.

Worked example 7. A matchstick sequence has 2n + 1 sticks at position n. Find the position with exactly 99 sticks.

Answer: Form 2n + 1 = 99. Subtract 1 from both sides to get 2n = 98. Divide both sides by 2 to get n = 49. Check: 2 × 49 + 1 = 99.

Trial and error means trying possible values and checking the result. It can be inefficient. For the equation above, a systematic approach finds the position directly by undoing the operations that produced the total.

Check a proposed solution in the original equation. A correct calculation in a later line cannot repair an earlier change that accidentally altered the equation. The original equality is the condition the answer must satisfy.

How do inverse operations solve a two-step equation?

Inverse operations undo each other. Addition and subtraction are inverse operations; multiplication and division by a non-zero number are inverse operations. The symbol ÷ means division. Choose operations that leave the unknown by itself while preserving equality.

Property: Equal operations preserve an equation’s balance

Add or subtract the same number on both sides. Both sides may also be multiplied or divided by the same non-zero number. The non-zero condition matters: division by zero is not defined, and multiplication by zero would lose the original restriction on the unknown.

Worked example 8. Solve 5x − 4 = 7, where x is the unknown number.

Answer: Add 4 to both sides: 5x − 4 + 4 = 7 + 4, so 5x = 11. Divide both sides by 5: x = 11/5. The notation 11/5 means 11 divided by 5.

Check: 5 × (11/5) − 4 = 11 − 4 = 7. This equals the original right-hand side, so the solution is correct.

The expression on the left multiplies x by 5 and then subtracts 4. To recover x, undo the subtraction first, then undo the multiplication. An equation with integral coefficients can have a fractional solution, as this example shows.

How do we handle a negative constant?

Worked example 9. Solve 11y + (−5) = 61, where y is the unknown number.

Answer: Subtract −5 from both sides, giving 11y = 61 − (−5) = 66. Divide by 11 to obtain y = 6. Checking gives 11 × 6 + (−5) = 66 − 5 = 61.

A shortened solution can omit the repeated operation on the left once its cancellation is understood. However, each new line must still mean the same thing as performing the stated operation on both sides. Explain the operation instead of treating a sign change as a guess.

How do we form equations from contextual problems?

A contextual problem describes quantities in a situation. Begin by deciding what is required. Define one unknown, express the remaining quantities using it, and identify the stated equality. Solve the resulting equation and interpret the answer using the original quantities and units.

1. Identify the unknown and state precisely what the chosen letter represents.
2. Translate the given relationships into expressions without losing any fixed quantity.
3. Use the stated total or equality to form an equation.
4. Solve, check the original conditions, and state the answer in context.

How can two unknown quantities be expressed using one variable?

Worked example 10. Ramesh and Suresh have 60 marbles altogether. Ramesh has 30 more marbles than Suresh. How many does each have?

Answer: Let y be Suresh’s number of marbles. Ramesh then has y + 30. Their total gives y + (y + 30) = 60, or 2y + 30 = 60.

Subtract 30 from both sides to obtain 2y = 30. Divide by 2: y = 15. Suresh has 15 marbles and Ramesh has 45. Their total is 60, and their difference is 30.

There are two quantities to find, but the difference lets us describe both with one variable. Writing a separate unknown for each is unnecessary here. After finding Suresh’s quantity, remember to calculate Ramesh’s quantity as well.

How do fixed and variable charges form an equation?

Worked example 11. A taxi charges a fixed fee of ₹800 per day plus ₹20 for each kilometre travelled. For a day’s travel, the total cost is ₹2200. Find the distance travelled.

Answer: Let x represent the distance in kilometres. Then 800 + 20x = 2200. Subtract 800 to get 20x = 1400. Divide by 20 to obtain x = 70. The distance is 70 kilometres.

Check the charge by substituting the distance: 800 + 20 × 70 = 2200 rupees. The fixed fee is paid once for the day, while the distance charge depends on the number of kilometres travelled.

What are simple inequalities and how are their solutions found?

An inequality compares quantities without requiring equality. The symbol < means “is less than”, and > means “is greater than”. Thus 3 < 5 and 7 > 5 are true numerical comparisons.

These are strict inequalities: the quantities are unequal. In x < 5, where x represents a number, the value 5 itself is excluded. In y > 2, where y represents a number, the value 2 itself is excluded.

What does a solution of an inequality mean?

A solution of an inequality is an allowed value of its variable that makes the comparison true. The solution set is the collection of all such values. State the allowed kind of number, because it affects which answers can be included.

Natural numbers here are the counting numbers beginning with 1. Whole numbers also include zero. A count of rice packets cannot be negative or fractional. A purely numerical inequality, without that counting restriction, can allow other kinds of values.

Worked example 12. Ravi has ₹200. Rice costs ₹30 per packet of 1 kg. Let x be a whole-number count of packets, including zero. Find the packet counts satisfying 30x < 200.

Answer: The allowed counts are 0, 1, 2, 3, 4, 5 and 6. At 6 packets, the cost is ₹180, which is less than ₹200. At 7 packets, the cost is ₹210, which is greater than ₹200.

The collection can be written as {0, 1, 2, 3, 4, 5, 6}; braces enclose the members of a set. If at least one packet must be purchased, exclude zero and the permitted counts are 1 through 6.

How can a simple calculation help?

Dividing both sides of 30x < 200 by the same positive number, 30, preserves the comparison. It gives x < 20/3. For whole-number packet counts, this again allows zero through six. The restriction to whole numbers is essential when translating that bound into purchases.

Do not replace the inequality sign with an equals sign. The task is to find values that make the comparison true, not a value that makes the two sides equal. Check the largest allowed count and the next count to confirm the boundary in this example.

Note: Multiplying or dividing an inequality by a positive number preserves its direction. This statement includes the condition “positive”; do not apply it without checking that condition.

Glossary

• Algebra — The use of letters and operations to express numerical relationships and solve problems.
• Variable — A symbol representing a number whose value can vary within a stated context.
• Constant — A quantity with a fixed value in the expression or situation being considered.
• Unknown — A quantity whose value must be determined from the information given in a problem.
• Algebraic expression — A combination of numbers and variables connected through arithmetic operations.
• Term — A part of an expression added to other parts, with its sign retained.
• Numerical coefficient — The numerical factor multiplying the variable part of an algebraic term.
• Like terms — Terms having exactly the same variable part, including the powers of their variables.
• Unlike terms — Terms whose variable parts differ and cannot be combined as one like term.
• Substitution — Replacing a variable in an expression by its given numerical value.
• Equation — A statement that the expressions on either side of an equals sign have equal values.
• Linear equation — An equation in which the variable occurs to the first power after simplification.
• Inverse operations — Operations that undo one another, such as addition and subtraction.
• Inequality — A comparison between quantities using a less-than or greater-than relation, possibly allowing equality.
• Solution set — The collection of all allowed values that make a stated equation or inequality true.

Common errors and misconceptions

• Misconception: A letter stands for the name of an object. Correct: It represents a numerical quantity. Define whether it means a count, a price, a length or another quantity before using it.
• Misconception: 5u and 5 + u mean the same operation. Correct: The first means five times u; the second means five more than u. They give different results for most values of u.
• Misconception: Terms with different coefficients must be unlike. Correct: Compare the variable parts. The terms 5c, c and 10c are like terms and combine by adding their coefficients.
• Misconception: All terms in a total can be merged into one term. Correct: Unlike terms remain separate. The expression 18c + 11d retains the separate contributions of pencils and erasers.
• Misconception: Subtracting a bracket changes only its first sign. Correct: Every term is subtracted. In (40x + 75y) − (6x + 10y), both 6x and 10y must be taken away.
• Misconception: An equation can be solved by changing just one side. Correct: Perform the same valid operation on both sides and check the resulting value in the original equation.
• Misconception: The boundary value belongs to a strict inequality’s solution set. Correct: The signs < and > exclude equality. Also respect any restriction to whole numbers or natural numbers.

Exam-style questions with model answers

Q1. In 5m + 3, m represents a number. State the coefficient of m and the constant term. [2 marks]

1. The coefficient of m is 5, because 5m means that the variable m is multiplied by 5.
2. The constant term is 3, because this term contains no variable and has a fixed value.

Q2. Evaluate 5m + 3 when m = 2. Show substitution, calculation and the final value. [3 marks]

1. Substitute the given number 2 for m. The expression becomes 5 × 2 + 3; the coefficient still multiplies the substituted value.
2. Carry out the multiplication first. Since 5 × 2 = 10, the remaining arithmetic expression is 10 + 3.
3. Add the constant term to obtain 13. Therefore, when m = 2, the value of the complete expression 5m + 3 is 13.

Q3. Pencils cost c each and erasers cost d each in the same currency. A shop sells 5, 3 and 10 pencils, and 4, 6 and 1 erasers, on three successive days. Form and simplify its total receipts, explaining why two terms remain. [4 marks]

1. The receipts from pencils are 5c + 3c + 10c. Each term uses the same price c, multiplied by the number of pencils sold.
2. These are like terms, so their sum is (5 + 3 + 10)c = 18c.
3. The receipts from erasers are 4d + 6d + d = (4 + 6 + 1)d = 11d.
4. Total receipts are 18c + 11d. These terms have different variable parts, so they remain separate in the simplified expression.

Q4. A shop initially charges ₹40 per chair and ₹75 per table for a day’s rental. It later refunds ₹6 per chair and ₹10 per table. For x chairs and y tables, form and simplify the final payment, showing the sign changes. [5 marks]

1. The initial chair payment is 40x rupees and the initial table payment is 75y rupees. Together they give the initial payment 40x + 75y.
2. The chair refund is 6x rupees and the table refund is 10y rupees. The complete refund is therefore 6x + 10y.
3. Subtract the entire refund from the initial payment. The required expression is (40x + 75y) − (6x + 10y).
4. Remove the subtracted bracket by changing both signs inside it. This gives 40x + 75y − 6x − 10y.
5. Collect like terms: (40 − 6)x + (75 − 10)y = 34x + 65y. This is the final amount paid, in rupees.

Q5. Solve 5x − 4 = 7, where x is an unknown number, and verify your answer. [3 marks]

1. Add 4 to both sides to undo the subtraction: 5x − 4 + 4 = 7 + 4. This gives 5x = 11.
2. Divide both sides by 5 to undo the multiplication. The result is x = 11/5, a fractional solution.
3. Substitute into the original equation: 5 × (11/5) − 4 = 11 − 4 = 7. Both sides equal 7, so the solution is verified.

Q6. Ramesh and Suresh have 60 marbles altogether. Ramesh has 30 more marbles than Suresh. Use one variable to find both quantities and check both conditions. [5 marks]

1. Let y represent the number of marbles Suresh has. Since Ramesh has 30 more, his number of marbles is y + 30.
2. Use the total of 60 to form y + (y + 30) = 60. Combining the two like terms gives 2y + 30 = 60.
3. Subtract 30 from both sides of the equation. This leaves 2y = 30, with twice Suresh’s number on the left.
4. Divide both sides by 2 to obtain y = 15. Suresh has 15 marbles, and Ramesh has 15 + 30 = 45 marbles.
5. Check both statements: 15 + 45 = 60 confirms the total, and 45 − 15 = 30 confirms that Ramesh has 30 more.

Q7. A taxi charges ₹800 for a day plus ₹20 per kilometre travelled. The total charge for that day is ₹2200. Form an equation, find the distance and check it. [4 marks]

1. Let x be the distance travelled in kilometres. The distance charge is 20x rupees, so 800 + 20x = 2200.
2. Subtract the fixed fee of 800 from both sides. The remaining distance charge satisfies 20x = 1400.
3. Divide both sides by 20. This gives x = 70, so the taxi travelled 70 kilometres.
4. Check the complete charge: 800 + 20 × 70 = 800 + 1400 = 2200 rupees, matching the stated total.

Q8. Solve 30x < 200 when x is a whole number, including zero. Give the complete solution set and verify its upper boundary. [4 marks]

1. Divide both sides by the positive number 30. The inequality keeps its direction and becomes x < 20/3.
2. The complete set of whole-number solutions is {0, 1, 2, 3, 4, 5, 6}. Negative numbers and fractions are excluded by the question.
3. At the largest listed value, x = 6, the left-hand side is 30 × 6 = 180, which is less than 200.
4. At the next whole number, x = 7, it becomes 210, which exceeds 200. Larger whole numbers also exceed the bound.

Key takeaways

• A variable represents a numerical quantity. State its meaning and unit before using it to describe a relationship.
• Terms carry their signs. A numerical coefficient is the number multiplying a term’s variable part.
• Like terms have matching variable parts; combine their coefficients while preserving the common variable part.
• Substitution replaces variables by given numbers. Preserve brackets and complete multiplication before adding or subtracting terms.
• Subtracting an expression means subtracting every term inside its bracket, including any negative terms.
• Solve a two-step equation using inverse operations on both sides, then check the original equality.
• A contextual answer must satisfy every stated relationship, including totals, differences, prices and the units of quantities.
• A strict inequality excludes equality at its boundary. Its solutions must also obey the stated restrictions on allowed numbers.

Test yourself

Shabnam is 3 years older than Aftab. If her age is 20 years, how old is Aftab?

Aftab’s age is 20 − 3 = 17 years, because he is three years younger than Shabnam.

What does 7k mean, and what is its value when k = 4?

It means seven multiplied by the number k. At k = 4, its value is 7 × 4 = 28.

Why are 5c and 10c like terms?

Both have the same variable part, c. Their different coefficients do not prevent them from being like terms.

If a = −4, what is the value of 10 − a?

It is 10 − (−4) = 14. Subtracting a negative number adds its positive counterpart.

Why does 18c + 11d retain two terms?

The terms have different variable parts, c and d. They cannot be combined as a single like term.

What is the solution of 11y + (−5) = 61?

Add 5 to both sides to obtain 11y = 66, then divide by 11 to find y = 6.

A sequence uses 2n + 1 matchsticks at position n. Which position uses 99 sticks?

Solve 2n + 1 = 99. Subtract 1, then divide by 2, to obtain position n = 49.

Does x = 5 satisfy x < 5, where x represents a number?

No. The strict less-than sign excludes equality, so the boundary value 5 is not a solution.