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Class 6 Mathematics Chapter 11 of 14

Chapter 11 — Algebra

Open the lesson Play with this chapter — pictures, sound and practice.

Overview

This unit introduces algebra to Class 6 students. It explains how letters and symbols stand for numbers, how to write simple expressions, and how to form and solve basic equations. The unit covers terms like variables, constants, coefficients, and like terms. Students will learn how to simplify algebraic expressions using addition, subtraction and multiplication by a number, and how to use algebraic rules to evaluate expressions for given values. The chapter also teaches how to translate simple word problems into algebraic expressions and equations, and how to solve one-step equations. Learning algebra helps students think logically and solve problems in a step-by-step way. It is the foundation for higher mathematics, including geometry, arithmetic with unknowns, and later topics like percentages and ratios. Practising algebra strengthens reasoning, improves ability to generalise patterns, and enables solving practical problems such as finding an unknown number, sharing quantities equally, and working with patterns. This unit gives students tools to describe relationships, work with simple formulas, and build confidence with symbols that represent numbers.

Learning Objectives

  • Recognise and use variables, constants, coefficients and algebraic expressions.
  • Translate simple verbal statements into algebraic expressions.
  • Simplify algebraic expressions by combining like terms.
  • Evaluate algebraic expressions for given values of variables.
  • Use arithmetic operations with algebraic expressions on whole numbers.
  • Form and solve one-step linear equations using addition and subtraction.
  • Solve basic word problems by forming algebraic equations and solving them.
  • Use algebraic thinking to identify patterns and generalise simple rules.

Topics in this chapter

14 topics · tap a topic title to jump straight to it.

🔣1

Introduction to Algebra: What is Algebra?

Algebra is a branch of mathematics that uses letters and symbols to represent numbers. These letters (like a, b, x) are called variables because their value can change. Algebra helps us write general rules and solve problems when some numbers are unknown. In everyday life you meet algebraic ideas when you leave a blank to fill in later, for example, 'I have x sweets' or 'She is y years older than me'.

In this topic we learn the basic language of algebra: variable, constant, term, expression and equation. A constant is a fixed number such as 3 or 10. A term is a number, a variable, or a product of numbers and variables, like 5, x, or 3x. An algebraic expression combines terms using + or − signs, for example 2x + 3. An equation states that two expressions are equal, for example x + 5 = 12.

We will practise writing expressions and reading them aloud. For example, 4 + x is read as 'four plus x' and 7x as 'seven times x' or 'seven x'. Understanding these simple ideas helps solve unknowns and makes future maths easier.

Algebra is like a language for patterns and rules. Once you know the symbols and how to use them, you can work with many problems in a short, clear way. The next topics will build from these definitions to operations and solving equations.

📌 Examples
  • If x represents the number of apples, 3 + x means 3 apples plus x apples.
  • If y = 5, then the expression 2y means 2 × 5 = 10.
  • 3a is a single term where 3 is multiplied by a.
  • The phrase 'a number less than 7' can be written as 7 − x.
🧮 Formulas
  1. Variable: a symbol (like x, y) representing a number.
  2. Constant: a fixed number (like 2, 5).
  3. Term: a single number or product of number and variable (e.g., 4, 3x).
📊 Visual ideas
Draw a box labelled x to show a variable that can change; next to it list possible values like 1, 2, 3.
Diagram showing a balance with 'expression' on left and 'value' on right to illustrate an equation.
🔣2

Writing Algebraic Expressions

Writing algebraic expressions means changing words into a short mathematical form using variables and numbers. This helps us express general ideas without specific numbers. For example, 'a number increased by 4' becomes x + 4 if x stands for the number. Practice with many phrases makes translation simple.

Common words and how to write them: 'sum of' or 'plus' becomes +, 'difference' or 'minus' becomes −, 'product' or 'times' becomes ×, 'divided by' becomes ÷. Often we write multiplication without the sign: 3 × x is written as 3x. If a phrase says 'twice a number', write 2x. 'Three more than a number' is x + 3, while 'three less than a number' is x − 3. When two variables are multiplied, write them together, for example xy.

Be careful with word order. 'Three less than a number' is x − 3, but 'a number less than three' is 3 − x. These give different results. Read the phrase slowly and decide which quantity comes first. Use brackets when needed: 'the sum of a number and 5, multiplied by 2' is 2(x + 5).

Writing expressions is a skill that improves with practice. Start with small sentences and convert them to algebra. Later you will use these expressions to form equations and solve problems. The next topic shows how to simplify expressions by combining like terms.

📌 Examples
  • Phrase: 'five more than a number' → Expression: x + 5
  • Phrase: 'twice a number plus three' → Expression: 2x + 3
  • 'Seven less than a number' → 7 − x
  • 'Three times the sum of a number and two' → 3(x + 2)
🧮 Formulas
  1. Sum of a and b: a + b
  2. Product of a and b: ab or a × b
  3. Twice a number x: 2x
📊 Visual ideas
Step diagram showing 'words' → 'math symbols' → 'expression' with arrows linking each step.
Box model for phrase '3 less than x': a box labeled x and an arrow subtracting 3.
🔢3

Terms, Like Terms and Unlike Terms

A term is a single part of an algebraic expression. It may be a number like 5, a variable like x, or a combination such as 3x or 2ab. Terms are the pieces separated by + or − signs in an expression. For example in 4x + 3 + 2y, the terms are 4x, 3 and 2y.

Like terms are terms that have the same variable part. This means the letters and their powers are identical. For instance, 3x and 5x are like terms because both have x; 2ab and 7ab are like terms because both have the variables a and b together. Like terms can be combined by adding or subtracting their coefficients (the numbers in front). For example 3x + 5x = (3 + 5)x = 8x.

Constants (numbers without variables) are like terms with each other: 4 and 7 are like terms and 4 + 7 = 11. Unlike terms have different variable parts and cannot be combined directly. For example 4x and 3y are unlike terms because one has x and the other has y. Also 3x and 3x^2 (x squared) are unlike because the powers differ; however at Class 6 we mainly work with single-letter variables without powers.

To simplify an expression, first group like terms together and then add or subtract their coefficients. Write the result with the variable part once and the combined coefficient before it. If an expression has both positive and negative terms, be careful with signs: 5x − 2x = 3x and 4 + (−3) = 1. When many terms are present, it helps to line up similar terms in columns (x-terms, y-terms, constants) and then combine vertically.

Practice identifying like and unlike terms in many expressions. This skill is essential for simplifying expressions, evaluating them after substitution and solving equations. It reduces mistakes and makes algebra work clearer and faster.

📌 Examples
  • 3x + 5x = 8x
  • 4 + 7 = 11 (constants are like terms)
  • 5x + 3y cannot be combined because x and y are different variables
  • 2a + 4a − a = (2 + 4 − 1)a = 5a
🧮 Formulas
  1. Combine like terms: ax + bx = (a + b)x
  2. Constants: a + b = (sum of numbers)
📊 Visual ideas
Draw groups of terms in circles: circle 1 with x-terms (3x, 2x), circle 2 with y-terms (4y), circle 3 with constants (5, −1).
Bar model showing 3x and 5x joining to make 8x.
➕4

Addition and Subtraction of Algebraic Expressions

Adding and subtracting algebraic expressions follows the same idea as with numbers, but first we must identify like terms. When adding, combine terms that have the same variable part. When subtracting, change the sign of each term in the expression being subtracted and then add. Careful attention to signs and grouping makes the process safe and simple.

Start by writing the expressions clearly, one under the other, aligning like terms if possible. For example, to add (3x + 2) and (5x + 4), place 3x under 5x and 2 under 4, then add columns: 3x + 5x = 8x and 2 + 4 = 6, giving 8x + 6. Always check for like terms of different variables too: (2x + 3y) + (5x − y) combines x-terms to 7x and y-terms to 2y, so the result is 7x + 2y.

For subtraction, take care with brackets. Example: (7x + 5) − (2x + 3). Remove the brackets by changing signs of the second group: 7x + 5 − 2x − 3. Now combine like terms: (7x − 2x) + (5 − 3) = 5x + 2. If there are more expressions, continue the same way: change signs for any expressions preceded by −, then group like terms and add coefficients.

Working with negative coefficients is common: (3x − 4) + (−2x + 7) = (3x − 2x) + (−4 + 7) = x + 3. Keep variable terms first and constants last to make the result tidy. Practise with several examples, including multiple variables and negative signs, until aligning and combining become routine. This prepares you for simplifying before evaluating or solving equations.

📌 Examples
  • (2x + 3) + (4x + 5) = 6x + 8
  • (6y + 2) − (3y + 4) = 3y − 2
  • 3x + (−2x + 7) = x + 7
  • (x + 5) + (2x − 3) + (3x + 1) = 6x + 3
🧮 Formulas
  1. (ax + b) + (cx + d) = (a + c)x + (b + d)
  2. (ax + b) − (cx + d) = (a − c)x + (b − d)
📊 Visual ideas
Vertical alignment of terms showing like terms in the same column for easy addition.
Number line to show combining constants (optional visual for constants).
🔢5

Multiplication of an Algebraic Expression by a Number

Multiplying an algebraic expression by a number means applying that number to every term inside the expression. The distributive idea is used: a number outside brackets multiplies each term inside. This rule is simple and powerful for expanding expressions and for later solving equations that have brackets.

Write the multiplier once and apply it to each term. For example, 3(x + 4) means 3 × x and 3 × 4. So 3(x + 4) = 3x + 12. If there are more terms, multiply each: 2(a + b + 3) = 2a + 2b + 6. When the multiplier is negative, keep the negative sign: −2(x − 5) = −2x + 10. Always be careful to multiply signs correctly: multiplying by a negative changes the sign of each term inside.

Multiplying a single term such as 4x by 3 gives 12x because you multiply the coefficients and keep the variable unchanged: 3 × 4x = 12x. If two variables are multiplied, write them together without a sign: x × y = xy. In this class we mainly multiply expressions by whole numbers, so follow these steps: distribute the multiplier, simplify each product, then combine like terms if any appear after expansion.

After multiplication, check if any like terms can be combined. For example 2(2x + 3x + 1) = 4x + 6x + 2 = 10x + 2. Also practise multiplying expressions that already have negative terms or multiple terms to avoid sign errors. Using brackets to show the original structure helps avoid mistakes. This skill is used when solving equations that have brackets, and when forming expressions from word problems.

📌 Examples
  • 3(x + 4) = 3x + 12
  • 2(2x + 3x + 1) = 4x + 6x + 2 = 10x + 2
  • −3(2y − 5) = −6y + 15
  • 4 × 5x = 20x
🧮 Formulas
  1. Distributive rule: k(a + b) = ka + kb
  2. k(ax) = (k × a)x
📊 Visual ideas
Box model showing 3 groups of (x + 4) each; combine to show 3x + 12.
Diagram of a bracket expanded into separate terms with arrows from multiplier to each term.
🔣6

Evaluating Algebraic Expressions

Evaluating an algebraic expression means finding its numerical value when the variables are given specific numbers. This requires careful substitution of the values into the expression and following the correct order of operations. The rule BODMAS (Brackets, Orders, Division/Multiplication, Addition/Subtraction) helps decide the order.

To evaluate, follow these steps: (1) Substitute the given values for each variable. (2) Work inside brackets first if any. (3) Multiply or divide as needed. (4) Add or subtract the remaining numbers. For example, evaluate 3(x + 2) when x = 4. Substitute x = 4 to get 3(4 + 2) = 3(6) = 18. If the expression is 2x + 3y and x = 2, y = 5, replace each variable: 2(2) + 3(5) = 4 + 15 = 19.

Be careful with negative numbers and with fractional results. If x = −2, then 4x = 4 × (−2) = −8. If a variable appears more than once substitute the same value each time. When expressions have several operations, write each step clearly to avoid mistakes: do multiplication before addition, simplify sums of like terms when possible, and keep checking signs. Using parentheses around substituted negative numbers prevents mistakes, for example 3(−2) is clearly negative.

Evaluating expressions is a frequent task in algebra and is useful for testing formulas, checking answers to equations and solving real problems. Practise with simple and slightly longer expressions so the substitution process becomes automatic. Always write the substituted expression before simplifying and then compute step by step to arrive at the final numeric value.

📌 Examples
  • Evaluate 4x − 1 when x = 3 → 4(3) − 1 = 12 − 1 = 11
  • Evaluate 2x + 3y when x = 2, y = 1 → 4 + 3 = 7
  • Evaluate 3(x + 2) when x = 1 → 3(1 + 2) = 9
  • If x = −1, then 5x = −5
🧮 Formulas
  1. Evaluate by substitution: replace variable with number and compute.
  2. Follow order: Brackets → Multiplication/Division → Addition/Subtraction
📊 Visual ideas
Flowchart: Expression → Substitute values → Simplify step-by-step → Final value.
Small table with variable values and evaluated results for different choices of x.
🟰7

Simple Equations and Balance Method

An equation is a mathematical sentence with an equal sign showing two expressions are equal, for example x + 5 = 12. The goal is to find the value of the variable that makes the equation true. The balance method helps: treat the equation like a balance scale. Whatever you do to one side must be done to the other so the two sides remain equal.

To solve x + 5 = 12, remove 5 from the left to leave x alone. Subtract 5 from both sides: x + 5 − 5 = 12 − 5, so x = 7. Always do the same operation on both sides. If the equation is x − 3 = 9, add 3 to both sides: x − 3 + 3 = 9 + 3, so x = 12.

When the unknown has a coefficient, use inverse operations. For 3x = 15, divide both sides by 3: x = 15 ÷ 3 = 5. For −2x = 8, divide both sides by −2 to get x = −4. Work step by step and check your answer by substituting back into the original equation.

Use balance pictures to show removing or adding the same amount from both sides. Equations may have variables on both sides; bring variable terms to one side and constants to the other. This simple idea forms the basis of solving linear equations in higher classes.

📌 Examples
  • x + 5 = 12 → subtract 5: x = 7
  • x − 4 = 10 → add 4: x = 14
  • 3x = 21 → divide by 3: x = 7
  • 2x + 3 = 11 → subtract 3 then divide by 2: 2x = 8 → x = 4
🧮 Formulas
  1. If ax + b = c, then ax = c − b and x = (c − b)/a
  2. Do same operation on both sides to keep equality
📊 Visual ideas
Balance scale drawing showing x + 3 on left and 8 on right; remove 3 from both sides to find x = 5.
Step diagram: Equation → Operation on both sides → Simplified equation → Solution
🟰8

Solving One-Step Equations

One-step equations need only a single inverse operation to find the variable. These equations are simple and teach the idea of undoing an operation. The two main types are those requiring addition or subtraction, and those requiring multiplication or division. Recognising which operation is used on the variable helps choose the correct inverse.

If the equation is x + a = b, remove a by subtracting a from both sides: x = b − a. If the equation is x − a = b, add a to both sides: x = b + a. For equations like ax = b, divide both sides by a: x = b/a. For example, x + 6 = 13 → subtract 6 → x = 7. 4x = 20 → divide by 4 → x = 5. When negative numbers are present, follow the same rule: −x = 4 → multiply both sides by −1 → x = −4.

Always perform the same operation on both sides to keep the equation balanced. Write each step clearly, showing the operation applied to both sides. After finding the value, substitute it back into the original equation to verify. This check helps catch mistakes such as sign errors or wrong arithmetic. Also remember that when dividing, division must be exact or leave the answer as a fraction if needed; for Class 6 examples usually give whole-number answers.

Practice many one-step problems so you can recognise the inverse quickly and solve accurately. One-step equations are the foundation for two-step and multi-step equations learned later.

📌 Examples
  • x + 6 = 13 → x = 13 − 6 = 7
  • x − 2 = 9 → x = 9 + 2 = 11
  • 4x = 20 → x = 20 ÷ 4 = 5
  • −3x = 9 → x = 9 ÷ (−3) = −3
🧮 Formulas
  1. x + a = b → x = b − a
  2. ax = b → x = b/a
📊 Visual ideas
Simple flow: Equation → One inverse step → Solution, with examples written on arrows.
Balance image showing division of both pans by the same number.
🟰9

Solving Two-Step Equations (Introductory)

Two-step equations involve two inverse operations to isolate the variable. A common form is ax + b = c. The correct method reverses the order of operations: first remove the constant b by addition or subtraction, then undo multiplication or division by the coefficient a. Practise the two-step process carefully and write each step clearly.

For example, solve 3x + 4 = 19. First subtract 4 from both sides to get 3x = 15. Next divide both sides by 3 to find x = 5. Similarly, for 2x − 3 = 9 add 3 to both sides to get 2x = 12, then divide by 2 to get x = 6. When fractions appear in the equation, apply inverse operations in the same order: for x/2 + 3 = 8, subtract 3 to get x/2 = 5 then multiply by 2 to get x = 10.

If variables appear on both sides, collect variable terms together by adding or subtracting. Example: 3x + 2 = x + 10. Subtract x from both sides: 2x + 2 = 10. Then subtract 2: 2x = 8. Divide by 2: x = 4. Always check the found value by substitution into the original equation to ensure both sides are equal. This prevents mistakes in sign or arithmetic.

Use balance pictures to visualise removing the same amount and then dividing both pans equally. Practise with positive and negative numbers and simple fractional steps so the two-step approach becomes routine. Two-step equations prepare students for longer equations and word problems that follow.

📌 Examples
  • 3x + 4 = 19 → 3x = 15 → x = 5
  • 2x − 5 = 7 → 2x = 12 → x = 6
  • x/2 + 3 = 8 → x/2 = 5 → x = 10
  • 4x − 2 = 10 → 4x = 12 → x = 3
🧮 Formulas
  1. ax + b = c → ax = c − b → x = (c − b)/a
📊 Visual ideas
Two-step flow: ax + b = c → subtract b → ax = c − b → divide by a → x.
Balance drawing showing removing same amount then dividing both pans.
🔣10

Using Algebra in Word Problems

Word problems use algebra to model real-life situations. Solving them needs careful reading, choosing a variable, writing an expression or equation, solving it, and then interpreting the answer. Each step must be clear to avoid mistakes when translating words into symbols.

Start by reading the problem twice to understand what is asked. Identify the unknown and give it a variable name like x. Look for key words: 'sum' or 'more than' means addition, 'less than' means subtraction, 'times' means multiplication, and 'shared equally' means division. Write an expression for each sentence and then join them to form an equation. For example: 'A number increased by 7 is 15' becomes x + 7 = 15. Solve the equation and check the result in the context.

Use diagrams or bar models for comparison problems. For example, if one child has 4 more marbles than another and total is known, draw two bars with a gap of 4 and write expressions for each bar. For consecutive numbers, represent them as x, x + 1, x + 2 and form equations for sums. Always check that the answer makes sense: ages should not be negative and counts should be whole numbers if the problem requires so.

Work through money, age, sharing and simple measurement problems. After solving, write the final answer in words and units (e.g., 'The number is 8' or 'She has 12 rupees'). Practising many word problems develops skill in forming correct equations and strengthens understanding of algebra as a tool for solving everyday questions.

📌 Examples
  • 'A number plus 6 equals 14.' → x + 6 = 14 → x = 8
  • 'If twice a number is 10, find the number.' → 2x = 10 → x = 5
  • 'Ravi has 3 more marbles than Sita. If Sita has x marbles, Ravi has x + 3.'
  • 'Sum of two consecutive numbers is 15. Let x and x+1. x + x + 1 = 15 → 2x + 1 = 15 → x = 7'
📊 Visual ideas
Bar model showing two quantities with a difference or sum to set up the equation.
Simple pictograph where boxes represent unknown x to visualise the equation.
🔣11

Patterns and Rules with Algebra

Algebra helps describe patterns with a rule. When a pattern repeats, we can write a formula to get any term. For example, consider sequence 2, 4, 6, 8 ... Each term increases by 2. If the first term is 2 and position is n, the nth term is 2n. We use algebra to find rules for many patterns.

Look for how numbers change from term to term. If they add the same amount each time, the pattern is arithmetic and can be written as a × n + b. For example sequence 3, 6, 9, 12 increases by 3; nth term = 3n. For patterns that multiply, such as 2, 4, 8, 16, the rule is 2^n for powers — this appears later in higher classes.

Use algebraic expressions to describe pattern stages, shapes or steps. For example, a pattern that needs 3 matches for one shape, 6 for two shapes, 9 for three shapes can be written as 3n. Testing values confirms the rule: put n = 1, 2, 3 and get 3, 6, 9. Finding patterns builds algebraic thinking and helps to generalise results for any term without listing all earlier terms.

Practice by creating tables of n and term value, then try to find a formula connecting them. This skill is useful for problem solving and later work in sequences and series.

📌 Examples
  • Sequence 5, 8, 11, 14 → nth term = 3n + 2 (check n = 1 gives 5).
  • Squares: 1, 4, 9, 16 → nth term = n^2 (introduction only).
  • If pattern adds 4 each time and first term is 2 → nth term = 4n − 2.
🧮 Formulas
  1. Arithmetic pattern with difference d and first term a: nth term = a + (n − 1)d
  2. If term increases by k each time and first term is A, nth term = A + (n − 1)k
📊 Visual ideas
Table with two columns n and term value, and a straight line sketch for arithmetic sequences.
Dot pattern images showing shape growth and an algebraic expression beside each stage.
🔢12

Simple Identities and Properties

Some algebraic statements and rules always hold true for any number you choose; these are called properties or identities. At Class 6 level we use simple, useful properties such as commutative, associative and distributive laws and basic identities like a + 0 = a and a × 1 = a. Knowing these helps to rearrange and simplify expressions safely.

The commutative property says the order does not matter for addition and multiplication: a + b = b + a and a × b = b × a. For example, 3 + x = x + 3. The associative property tells us that grouping does not matter for addition: (a + b) + c = a + (b + c). This helps when adding three or more terms. The distributive property connects multiplication and addition: k(a + b) = ka + kb. This is used to expand brackets and to simplify calculations.

Simple identities help when solving equations. For instance, (x + y) − y = x shows that adding and then subtracting the same value returns the original. Also a − a = 0 and a × 0 = 0 are basic facts to remember. Using these identities can reduce work: for example, if you see x + 0, you can immediately replace it by x.

Practice applying these properties with numbers and simple algebraic expressions. Use the distributive law to expand brackets before combining like terms. Recognising when to use commutative or associative laws makes calculation faster and prevents unnecessary errors. These properties form the rules you will apply in all algebra work.

📌 Examples
  • Commutative: 3 + x = x + 3
  • Associative: (2 + 3) + x = 2 + (3 + x)
  • Distributive: 2(x + 3) = 2x + 6
  • Identity: x + 0 = x
🧮 Formulas
  1. Commutative: a + b = b + a, a × b = b × a
  2. Associative: (a + b) + c = a + (b + c)
  3. Distributive: k(a + b) = ka + kb
  4. Identity: a + 0 = a, a × 1 = a
📊 Visual ideas
Illustration with boxes to show commutative property by swapping positions of terms.
Diagram showing distribution of 3 across (x + 4) with arrows to 3x and 12.
➗13

Introduction to Simple Algebraic Fractions (Optional)

An algebraic fraction contains variables in numerator and/or denominator, such as x/2 or 3x/4. At Class 6 level we focus on simple cases where the denominator is a whole number, and we treat variables like numbers when dividing. These fractions are often used in equations and evaluation problems.

x/2 means one-half of x. To evaluate x/2 for a given x, substitute the value and divide. Example: if x = 10, x/2 = 5. When a coefficient appears, compute the product first if needed: 3x/4 with x = 8 gives 3×8/4 = 24/4 = 6. When solving equations such as x/3 = 4, use the inverse operation: multiply both sides by 3 to get x = 12. If the equation has a coefficient, for example (2x)/5 = 6, first multiply by 5: 2x = 30, then divide by 2 to get x = 15.

Handle negative values carefully: if x = −6, then x/2 = −3. If an expression has several fractional terms, find common denominators when adding or subtracting (this appears later), but for now practise substitution and simple solving. Avoid putting variables in the denominator when forming equations in Class 6; instead rearrange to keep denominators as numbers and use multiplication to clear them.

Working with algebraic fractions builds comfort with variables in division and prepares students for more advanced fraction algebra later. Practise evaluating, simplifying by canceling common factors when possible (for example 3x/3 = x) and solving straightforward fractional equations by reversing operations carefully.

📌 Examples
  • If x = 12, then x/3 = 4
  • Solve x/2 = 6 → x = 12
  • Evaluate 3x/4 for x = 8 → 3×8/4 = 6
  • Solve (2x)/5 = 6 → 2x = 30 → x = 15
🧮 Formulas
  1. If x/a = b then x = a × b
  2. (kx)/a = b → kx = ab → x = ab/k
📊 Visual ideas
Simple fraction bar showing x divided into equal parts to visualise x/3.
Step diagram solving x/3 = 4: multiply both sides by 3 → x = 12.
👑14

Checking Answers and Common Mistakes

Checking answers is an important habit in algebra. After solving an equation or evaluating an expression, substitute the found value back into the original expression or equation. If both sides of an equation match after substitution, the solution is correct. This simple check catches many mistakes such as sign errors or arithmetic slips.

Common mistakes include combining unlike terms (for example adding 3x and 4), forgetting to change signs when removing brackets in subtraction, and applying operations to only one side of an equation. To avoid errors, write each step clearly and show the operation being done to both sides, for example write x + 5 − 5 = 12 − 5. Use parentheses when substituting negative numbers, e.g., 3(−2) to show the negative sign belongs to the number.

Another frequent error is wrong order of operations. Remember BODMAS and apply it when evaluating expressions. When solving equations with two steps, do the inverse operations in reverse order of BODMAS: first remove addition/subtraction then do multiplication/division. For fractional steps, clear denominators by multiplying both sides, but do this carefully and then simplify. If your solution gives a value that does not fit the word problem (for example a negative age), re-check the setup and calculations.

Write down a short check after finishing: substitute the value and calculate both sides. If the check passes, state the final answer clearly with units. Practising these checks will reduce careless mistakes and build confidence in your algebra solutions.

📌 Examples
  • Solve x + 4 = 10 → x = 6. Check: 6 + 4 = 10 (correct).
  • Wrong: 3x + 2 = 11 → x = 11 − 2 = 9 (then divide by 3) is incorrect order; correct: 3x = 9 → x = 3.
  • When subtracting bracket: (x + 3) − (2x − 1) → x + 3 − 2x + 1 (change signs).
  • If x/2 = 5 → multiply both sides by 2: x = 10 and check: 10/2 = 5.
📊 Visual ideas
Checklist image: Solve → Substitute → Simplify → Verify equals true.
Flow showing how an error in one step can be traced back by checking substitution.

Key Concepts

Variable
A symbol (like x or y) used to represent an unknown or changeable number.
Constant
A fixed number that does not change in an expression or equation.
Term
A single element in an expression, such as a number, variable, or product like 3x.
Expression
A combination of terms connected by + or − that shows a value but has no equal sign.
Equation
A mathematical sentence stating that two expressions are equal, using =.
Coefficient
The numerical factor multiplied by a variable in a term, e.g., 5 in 5x.
Like terms
Terms that have the same variable part and can be combined by adding or subtracting coefficients.
Unlike terms
Terms with different variable parts that cannot be combined directly.
Distributive property
A rule that allows multiplication over addition: k(a + b) = ka + kb.
Evaluate
To find the value of an expression by substituting numbers for variables.
Balance method
Solving equations by doing the same operation on both sides to keep equality.
One-step equation
An equation that can be solved using a single inverse operation.
Two-step equation
An equation that needs two inverse operations to isolate the variable.
Identity
An algebraic statement that is true for all values of the variable.

End-of-Chapter Trial Paper & Test Questions

Topic-wise questions to test your understanding of every concept in this chapter.

  1. Write an expression for 'twice a number increased by 5' / 'किसी संख्या का दुगना बढ़ाकर 5 किया गया'
    Show answer

    English: 2x + 5 where x is the number. / हिंदी: यदि संख्या x है तो व्यंजक होगा 2x + 5।

  2. Simplify: 3x + 5x − 2 + 7 / 3x + 5x − 2 + 7 सरल कीजिए
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    English: Combine like terms: (3x + 5x) + (−2 + 7) = 8x + 5. / हिंदी: सामान पद मिलाएँ: (3x + 5x) + (−2 + 7) = 8x + 5।

  3. Evaluate 4x − 3 when x = 4 / x = 4 होने पर 4x − 3 का मान ज्ञात कीजिए
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    English: 4(4) − 3 = 16 − 3 = 13. / हिंदी: 4×4 − 3 = 16 − 3 = 13।

  4. Solve x + 7 = 15 / x + 7 = 15 हल कीजिए
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    English: Subtract 7 from both sides: x = 15 − 7 = 8. / हिंदी: दोनों ओर से 7 घटाएँ: x = 15 − 7 = 8।

  5. Form an equation: 'Three times a number is 27'. Find the number. / 'किसी संख्या का तीन गुना 27 है' से समीकरण बनाकर संख्या ज्ञात कीजिए
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    English: Let x be the number. 3x = 27 → x = 27 ÷ 3 = 9. / हिंदी: संख्या मान लेते हैं x. 3x = 27 → x = 27 ÷ 3 = 9।

  6. A number divided by 4 gives 6. Write equation and solve / किसी संख्या को 4 से भाग करने पर 6 मिलता है। समीकरण लिखें और हल करें
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    English: Let x be the number. x/4 = 6 → x = 6 × 4 = 24. / हिंदी: संख्या = x लें. x/4 = 6 → x = 6 × 4 = 24।

  7. Translate: 'Seven less than a number is 12' and solve / 'किसी संख्या से सात कम 12 है' का अनुवाद करें और हल कीजिए
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    English: Let x be the number. x − 7 = 12 → x = 12 + 7 = 19. / हिंदी: x − 7 = 12 → x = 19।

  8. If 5x + 2 = 22, find x / यदि 5x + 2 = 22 हो तो x निकालिए
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    English: Subtract 2: 5x = 20. Divide by 5: x = 4. / हिंदी: 2 घटाएँ: 5x = 20. 5 से भाग करें: x = 4।

  9. Which are like terms in 4a + 5b − 2a + 3? / 4a + 5b − 2a + 3 में सामान पद कौन से हैं?
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    English: Like terms are 4a and −2a (both have a); constants 3 stands alone; 5b is unlike these. / हिंदी: 4a और −2a सामान पद हैं; 3 एक स्थिरांक है; 5b अलग है।

  10. Find nth term: sequence 2, 4, 6, 8. Write formula for nth term / श्रेणी 2, 4, 6, 8 के लिए nवाँ पद लिखें
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    English: Terms increase by 2, first term 2. nth term = 2n. / हिंदी: हर बार 2 जुड़ता है, पहले पद 2 है। nवाँ पद = 2n।

  11. Solve: 2(x + 3) = 16 / 2(x + 3) = 16 हल कीजिए
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    English: Divide both sides by 2: x + 3 = 8. Subtract 3: x = 5. / हिंदी: दोनों ओर 2 से भाग करें: x + 3 = 8. 3 घटाएँ: x = 5।

  12. Check the solution: If x = 4 for equation 3x − 5 = 7, does it satisfy? / समीकरण 3x − 5 = 7 के लिए x = 4 समाधान जाँचें, क्या वह सच है?
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    English: Substitute x = 4: 3(4) − 5 = 12 − 5 = 7, which equals right side. So it satisfies. / हिंदी: x = 4 रखें: 3×4 − 5 = 12 − 5 = 7, जो दाईं ओर के बराबर है। अतः समाधान सही है।

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