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Chapter 3 — Algebra

Class 7 · Mathematics

Overview

This unit on Algebra introduces students to the language and tools of algebraic thinking. It covers variables and constants, writing and simplifying algebraic expressions, laws of exponents for whole number powers, factors and multiples, algebraic identities, linear equations in one variable, ratios and proportions using algebra, simple inequalities, and problems based on real situations. The unit teaches how to translate words into algebra, manipulate expressions using rules, factorise common expressions, and solve equations step by step. Algebra matters because it gives methods to solve many kinds of problems efficiently and prepares students for higher mathematics and science. Learning algebra strengthens logical reasoning, helps in recognizing patterns, and gives a compact way to represent relationships. The unit emphasises clear working, checking answers, and interpreting solutions in context. By the end of the unit, students will be able to form equations from word problems, simplify expressions, apply identities, and solve linear equations and simple inequalities confidently. They will also learn to present results neatly and verify solutions, useful skills for examinations and everyday quantitative reasoning.

Learning Objectives

  • Identify and use variables, constants, coefficients and terms in algebraic expressions.
  • Translate verbal statements into algebraic expressions and simple equations.
  • Simplify algebraic expressions using like terms and the distributive property.
  • Apply laws of exponents for whole number powers to simplify expressions.
  • Factorise algebraic expressions using common factors and simple identities.
  • Solve linear equations in one variable, including those arising from word problems, and check solutions.
  • Understand and solve simple linear inequalities and represent solutions on a number line.
  • Use ratios, proportions and percentages in algebraic form to solve practical problems.

Topics in this chapter

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

🔣1

Introduction to Algebraic Language

What is algebra and why we use it
Algebra is the part of mathematics that uses letters and symbols to stand for numbers. When we do not know a number, or when a number can vary, we use a letter such as x, y or n. Using letters makes it possible to write general rules and to solve many problems in a single, systematic way. Algebra also helps describe relationships: if one value depends on another, we can write that relation compactly.

Variables, constants and expressions
A variable is a symbol that represents a number which can change. A constant is a fixed number. An algebraic expression is made up of numbers, variables and mathematical operations. For example, 3x + 5 is an expression with variable x and constant 5. The expression gives a rule for producing a number when a particular value of x is chosen.

Parts of an expression
Each part separated by a + or − sign is called a term. In 4a − 7b + 9, there are three terms: 4a, −7b and 9. The coefficient is the numerical factor of a term that contains a variable; in 4a the coefficient is 4. The constant term is a stand-alone number like 9. Learning these names helps when following steps to simplify or solve.

Writing simple expressions from words
Many statements in everyday life can be written in algebraic form. “A number increased by 6” is x + 6. “Twice a number” is 2x. “Three times a number decreased by 4” is 3x − 4. Practising translation of words into algebraic expressions is an important first skill: it allows conversion of a real situation into a mathematical one that can be manipulated and solved.

Using brackets and clear notation
Brackets ( ) are used to show which operations should be done first and to keep the meaning clear, for example 2(x + 3) means double the whole quantity x + 3. Always write expressions neatly, use equal signs correctly, and show each step when working through problems. Good notation reduces errors and makes checking easier.

📌 Examples
  • Write an expression for: 7 less than a number x → x - 7
  • Identify parts: In 5a + 3, coefficient of a is 5 and constant is 3
  • Write the phrase 'three times a number increased by 4' as 3x + 4
🧮 Formulas
  1. Variable: a letter representing a number
  2. Coefficient: numerical factor of a term
  3. Term: part of an expression separated by + or −
📊 Visual ideas
A labelled diagram showing a term split into coefficient and variable, e.g., 6x with arrow to 6 (coefficient) and x (variable)
🔢2

Like and Unlike Terms; Simplifying Expressions

Understanding like and unlike terms
Like terms are terms that have exactly the same variable part raised to the same power. For example 3x and −7x are like terms because both contain x to the first power. Also 2x^2 and x^2 are like terms; constants such as 5 and −2 are like terms to each other. Unlike terms cannot be added directly, for example 3x and 4y are unlike because one has x and the other y. Being able to identify like terms is essential when simplifying expressions.

Combining like terms
To simplify an expression, collect like terms together and add or subtract their coefficients. Keep the variable part unchanged; only the numerical coefficients combine. For instance, 4x + 3x − 2 = (4 + 3)x − 2 = 7x − 2. Similarly, 5x^2 − 2x + 3x^2 = (5 + 3)x^2 − 2x = 8x^2 − 2x. Always reorder terms so like terms are adjacent – this helps avoid mistakes.

Using the distributive law to remove brackets
The distributive law a(b + c) = ab + ac is used to remove brackets. For example 2(x + 3) = 2x + 6. When terms inside brackets are multiplied by an outside number, multiply each term inside the bracket. If the bracket has a negative sign in front, remember to change the signs of the terms inside when removing the bracket: −(x + 5) = −x − 5.

Step-by-step simplification
1) Remove brackets by using the distributive law. 2) Rearrange terms so that like terms are together. 3) Add or subtract coefficients of like terms. 4) Write the simplified expression in standard form, usually with terms ordered by degrees (highest power first) if required. Example: Simplify 3(x + 2) + 4x − 5 = 3x + 6 + 4x − 5 = 7x + 1.

Common errors to avoid
Do not combine unlike terms (such as x and x^2). Do not cancel terms that are added or subtracted inside sums unless they are common factors. Verify each simplification by expanding or substituting a sample value for variables to check that the original and simplified forms match numerically.

📌 Examples
  • Simplify: 3x + 5x − 2 = 8x − 2
  • Simplify: 2(a + 3) + 4a = 2a + 6 + 4a = 6a + 6
  • Simplify: −(2x − 5) + 3x = −2x + 5 + 3x = x + 5
🧮 Formulas
  1. Distributive law: k(p + q) = kp + kq
  2. Combining like terms: ax + bx = (a + b)x
📊 Visual ideas
A simple flow chart showing steps: Remove brackets → Group like terms → Add/Subtract coefficients
🔢3

Basic Identities and Their Use

What are algebraic identities?
Algebraic identities are equalities that hold true for all possible values of the variables involved. They are important because they allow quick expansion and factorisation when you recognise the form of an expression. Using identities saves time and reduces work in both simplification and solving problems.

Key basic identities
At this class level the most useful identities are the square of a sum and the square of a difference. They are: (a + b)^2 = a^2 + 2ab + b^2 and (a − b)^2 = a^2 − 2ab + b^2. Another frequently used identity is the difference of squares: a^2 − b^2 = (a − b)(a + b). Recognising these shapes in an expression helps either to expand a binomial square quickly or to factorise a quadratic-like expression.

How to apply identities
To expand (x + 5)^2 without multiplying term-by-term, use (a + b)^2: x^2 + 2·x·5 + 25 = x^2 + 10x + 25. To factorise x^2 + 10x + 25, observe it matches a^2 + 2ab + b^2 with a = x and b = 5, so the factorised form is (x + 5)^2. For difference of squares, rewrite x^2 − 9 as x^2 − 3^2 and then factor as (x − 3)(x + 3).

Recognising patterns
When you see a quadratic expression a^2 ± 2ab + b^2, check if the middle term equals twice the product of the square roots of the first and last terms: that is the test for a perfect square. For difference of squares, check that the first and last terms are perfect squares and the middle term is missing; then apply the identity directly. These patterns occur often in algebraic manipulation and in solving equations.

Practice and verification
Always expand your factorisation to verify it gives back the original expression. Practise several examples to build speed and pattern recognition. Learning these identities thoroughly lays the groundwork for factorisation techniques used in higher classes.

📌 Examples
  • Expand: (x + 3)^2 = x^2 + 6x + 9
  • Factorise: x^2 + 12x + 36 = (x + 6)^2
  • Factorise difference: x^2 - 16 = (x - 4)(x + 4)
🧮 Formulas
  1. (a + b)^2 = a^2 + 2ab + b^2
  2. (a - b)^2 = a^2 - 2ab + b^2
  3. a^2 - b^2 = (a - b)(a + b)
📊 Visual ideas
Draw a simple square labelled (a + b)^2, split into four parts a^2, ab, ab, b^2 to visualise the identity
🔢4

Monomials, Binomials and Polynomials

Definitions and basic ideas
A monomial is an algebraic expression consisting of a single term, for example 7x^2 or −3ab. A binomial has exactly two terms, like x + 5 or 2a − b. A polynomial is a sum of one or more monomials with non-negative whole number exponents, for example 3x^2 − x + 2. Classifying expressions helps decide what operations to use and how to simplify or factor them.

Degree of a term and polynomial
The degree of a single-variable term is the exponent of the variable: degree of 5x^3 is 3. For multi-variable terms the degree is the sum of the exponents, e.g., degree of 2a^2b is 3 (2 + 1). The degree of a polynomial is the highest degree among its terms. The degree guides the behaviour of polynomials and can determine the methods we apply for solving equations formed from them.

Operations on polynomials
Addition and subtraction are performed by combining like terms. For multiplication, use the distributive property: multiply each term of the first polynomial by each term of the second, then combine like terms. For example, (x + 2)(x + 3) expands to x^2 + 3x + 2x + 6 = x^2 + 5x + 6. Multiplication increases the degree: degree of product equals the sum of degrees.

Standard form and arranging terms
Polynomials are often written in descending powers of the variable: highest degree first. This standard form makes comparison simple and helps when using identities and factorisation. For example 4x^3 + x^2 − 2x + 5 is arranged by decreasing powers of x.

Why this matters
Understanding types of polynomials and degree is necessary for graphing, solving equations and for later study of algebraic divisions and factoring techniques. Many practical problems reduce to polynomial equations, so these basic definitions are building blocks for future topics.

📌 Examples
  • Identify: 5x is a monomial; x^2 + 3x is a binomial; 2x^3 - x + 1 is a polynomial of degree 3
  • Multiply: (x + 2)(x + 3) = x^2 + 5x + 6
  • Degree: degree of 6x^4 + x^3 − 2 is 4
🧮 Formulas
  1. Degree of polynomial: highest exponent of variable
  2. Multiplying binomials: (a + b)(c + d) = ac + ad + bc + bd
📊 Visual ideas
A labelled list showing monomial, binomial, trinomial and general polynomial examples with degrees
🔢5

Factors and Factorisation by Common Factor

Meaning of factorisation
Factorisation is the process of writing an algebraic expression as a product of simpler expressions called factors. Just like 12 = 3 × 4, algebraic expressions can be factored to show their building blocks. Factorisation is inverse to expansion and useful when simplifying fractions, solving equations and reducing expressions.

Common factor method explained
The simplest factorisation is to take out the greatest common factor (GCF) from all terms. Steps: (1) find the largest numerical factor that divides all coefficients; (2) find common variable powers in all terms (the smallest power present); (3) write the GCF outside a bracket and divide each term by the GCF to fill the bracket. For example 12x^2 + 8x has GCF 4x, so 12x^2 + 8x = 4x(3x + 2).

How to find GCF
List prime factors of numerical coefficients and common variable powers. For numbers, take the product of common prime factors; for variables, take the lowest power that appears in each term. If a term lacks a variable, its power is zero so variable cannot be part of GCF beyond that. Keep signs in mind: if all terms are negative, factor out a negative GCF to make the bracket simpler.

Examples and checks
If expression is 15a^2b + 10ab^2, numerical GCF is 5 and common variable power is ab, so factorisation is 5ab(3a + 2b). Always expand the factorised form to check you get the original expression. Factorising before solving equations often makes solution simpler because it reveals roots or simplifies fractions.

When to attempt GCF first
Always look for a common factor before trying other techniques like identities or grouping. Taking out the GCF can simplify later steps or make other factorisation methods applicable. This is a standard first step in most factorisation problems.

📌 Examples
  • Factorise: 6x^2 + 9x = 3x(2x + 3)
  • Factorise: 4a^3b + 6a^2b = 2a^2b(2a + 3)
  • Factorise: -8x + 12 = -4(2x - 3)
🧮 Formulas
  1. Common factor: ax + ay = a(x + y)
  2. GCF includes lowest power of each common variable and GCF of coefficients
📊 Visual ideas
A two-row table showing each term, its prime factors, and common factors leading to GCF
🔢6

Factorisation Using Identities and Grouping

Factorisation using identities
When an expression matches a standard identity, factorisation is quick. The perfect square identity a^2 + 2ab + b^2 becomes (a + b)^2. The difference of squares a^2 − b^2 becomes (a − b)(a + b). Recognising these shapes lets you factor expressions such as x^2 + 6x + 9 = (x + 3)^2 or x^2 − 25 = (x − 5)(x + 5). Look at the first and last terms for perfect squares and check whether the middle term equals twice their product for a perfect square identity.

Factorisation by grouping
For expressions with four terms, grouping helps. Group the first two terms and the last two terms (or try other pairings) and factor out the common factor from each pair. If the resulting grouped terms share a common binomial, factor that out. Example: x^3 + x^2 + 2x + 2 = x^2(x + 1) + 2(x + 1) = (x + 1)(x^2 + 2). Always try to rearrange terms if the first grouping does not work, because a different arrangement may produce a common binomial.

Combining methods
Sometimes you first take out a common factor then apply an identity or grouping on the remaining expression. For example 2x^2 + 8x + 8 = 2(x^2 + 4x + 4) = 2(x + 2)^2. Combining methods is a powerful tactic for many algebraic problems.

Tips and verification
Practice pattern recognition and try expanding the factorised form to verify it gives back the original expression. If factorisation seems difficult, check for GCF first, then try identities, then attempt grouping. This sequence usually leads to a successful factorisation for typical class problems.

📌 Examples
  • Factorise: x^2 - 16 = (x - 4)(x + 4)
  • Factorise by grouping: x^3 + x^2 + 2x + 2 = x^2(x + 1) + 2(x + 1) = (x + 1)(x^2 + 2)
  • Use identity: 4y^2 + 12y + 9 = (2y + 3)^2
🧮 Formulas
  1. Difference of squares: a^2 - b^2 = (a - b)(a + b)
  2. Perfect square: a^2 + 2ab + b^2 = (a + b)^2
📊 Visual ideas
Sketch of two rectangles representing (a + b)^2 split into a^2, 2ab and b^2 to visualise identities
🔋7

Laws of Exponents for Whole Number Powers

Understanding exponents
An exponent (also called a power) tells how many times a number, called the base, is multiplied by itself. For example x^3 means x × x × x. Exponents provide a compact way to write repeated multiplication and they follow consistent rules that help simplify expressions quickly when bases are the same or when a power is raised to another power.

Basic laws and how to use them
The key laws for whole number exponents are: product rule x^m × x^n = x^(m+n), quotient rule x^m ÷ x^n = x^(m−n) for x ≠ 0, power of a power (x^m)^n = x^(mn), and power of a product (ab)^n = a^n b^n. These laws follow from the repeated multiplication idea and are valid for whole number exponents. For example x^2 × x^3 means x multiplied 2 times and 3 times, so total of 5 multiplications giving x^5.

Applying rules carefully
When simplifying, make sure bases are the same before applying the product or quotient rule. For instance, x^2 × y^3 cannot be combined into a single power of x or y. Use the power of a product rule if both factors share the same exponent: (2x)^3 = 8x^3. Be careful with subtraction of exponents: x^3 ÷ x^2 = x^(3−2) = x. Negative exponents and fractional exponents are not required at this level, but be careful not to add exponents when bases differ.

Worked approach to simplification

  • Write each expression with explicit exponents.
  • Use rules in a sequence that reduces complexity: remove parentheses with power-of-power first, then use product or quotient rules.
  • Simplify numerical coefficients separately from variable powers.

Common mistakes and checks
Students sometimes add exponents when multiplying terms with different bases — this is incorrect. Always re-evaluate by rewriting as repeated multiplication when unsure. Try sample values for the base and confirm both sides give the same number to check a simplification.

📌 Examples
  • Simplify: x^2 × x^4 = x^6
  • Simplify: (2x^3)^2 = 4x^6
  • Simplify: x^5 ÷ x^2 = x^3
🧮 Formulas
  1. x^m × x^n = x^(m+n)
  2. x^m ÷ x^n = x^(m-n) for x ≠ 0
  3. (x^m)^n = x^(mn)
  4. (ab)^n = a^n b^n
📊 Visual ideas
A step diagram showing x^2 × x^3 combining exponents to x^5
8

Algebraic Fractions and Simplification

What are algebraic fractions?
An algebraic fraction has algebraic expressions in the numerator, denominator, or both. Examples include (2x + 4)/x and (x^2 − 9)/(x + 3). We treat algebraic fractions much like numerical fractions: simplify by cancelling common factors after factorisation, add or subtract by finding common denominators, and multiply or divide by multiplying numerators and denominators or inverting the divisor respectively.

Simplifying procedure
Step 1: Factorise numerator and denominator completely. Step 2: Cancel any common factors that appear in both numerator and denominator, provided those factors are not zero. Step 3: State restrictions on variable values that make any denominator zero (these values are excluded). For example (6x^2 + 9x)/(3x) = (3x(2x + 3))/(3x) = 2x + 3, but x ≠ 0 because original denominator 3x cannot be zero.

Adding and subtracting
To add or subtract algebraic fractions, find a common denominator (often the least common multiple of denominators). Convert each fraction to an equivalent fraction with that common denominator, add/subtract numerators and simplify. Be careful: do not cancel terms across addition or subtraction unless they form a common factor. For example (x + 2)/(x + 3) cannot be simplified further without factorising, while x(x + 2)/(x + 2) can be simplified to x (for x ≠ −2).

Multiplying and dividing
Multiply numerators and multiply denominators, then simplify by cancelling common factors. For division, multiply by the reciprocal of the divisor and then simplify. Always check variable restrictions arising from original denominators before and after simplification to avoid invalid solutions when solving equations involving fractions.

Practical tip
Always perform factorisation before attempting to cancel factors. Cancelling without factorisation is a common source of errors. Use substitution of a sample non-excluded value to verify simplified results if unsure.

📌 Examples
  • Simplify: (6x^2 + 9x)/(3x) = (3x(2x + 3))/(3x) = 2x + 3, x ≠ 0
  • Add: 1/x + 2/x = 3/x (for x ≠ 0)
  • Multiply: (x^2 - 1)/(x + 1) × (x + 1)/(2x) = (x - 1)/(2x) for x ≠ -1, x ≠ 0
🧮 Formulas
  1. Cancel common factor: (a·c)/(b·c) = a/b for c ≠ 0
  2. LCM used as common denominator when adding fractions
📊 Visual ideas
A two-column table showing factorisation of numerator and denominator and cancellation steps
🟰9

Setting Up and Solving Linear Equations in One Variable

Translating a situation into an equation
Many real problems can be converted into a linear equation in one variable. Begin by reading the problem carefully and choosing a letter for the unknown. Express other quantities in terms of that letter. Use given relationships to write an equality. For example, if “a number increased by 5 equals 12” you set x + 5 = 12. Clear definition of the variable and neat notation are essential to avoid errors.

Solving linear equations step-by-step
Use inverse operations to isolate the variable. Keep the equation balanced by performing the same operation on both sides. Typical steps are: remove brackets using the distributive law, collect like terms and simplify, move variable terms to one side, move constants to the other side, and finally divide by the coefficient of the variable. For example solving 3x + 5 = 20 gives 3x = 15 and then x = 5. Work from outermost operations inward, reversing each operation in turn.

Equations with variable on both sides
If the variable appears on both sides, bring all variable terms to one side by adding or subtracting and constants to the other. Example: 4x + 2 = 2x + 10 leads to 2x = 8, so x = 4. Always simplify first to avoid unnecessary algebraic steps.

Equations with fractions and special checks
When fractions appear, multiply both sides by the least common multiple of denominators to remove fractions. Before doing so, note any value that would make a denominator zero and exclude it. After solving, substitute the value back into the original equation to check for correctness and to ensure no invalid solution was introduced during manipulation.

Interpretation and answer presentation
After finding a value, interpret the solution in the problem context (for example ages should be non-negative). Present the solution clearly with steps shown and the final answer boxed. Checking by substitution builds confidence and prevents small algebraic mistakes from leading to wrong conclusions.

📌 Examples
  • Solve: 3x + 5 = 20 → 3x = 15 → x = 5
  • Word problem: A number plus 7 is 12. Find the number. x + 7 = 12 → x = 5
  • Solve: 4x + 2 = 2x + 10 → 2x = 8 → x = 4 (check: 16 + 2 = 8 + 10 → 18 = 18)
🧮 Formulas
  1. Balance method: If ax + b = c then ax = c - b and x = (c - b)/a
📊 Visual ideas
A flow chart showing steps: Translate → Simplify → Solve → Check
10

Equations Involving Brackets and Fractions

Handling brackets in equations
Brackets indicate that the operations inside should be considered together. Use distributive law to remove brackets: a(b + c) = ab + ac. In equations, distribute carefully and keep track of signs, especially when a negative sign precedes the bracket: −(x + 2) = −x − 2. After removing brackets, collect like terms and isolate the variable as with any linear equation. Removing brackets simplifies the structure and makes solving straightforward.

Equations with fractional terms
When variables appear in denominators, clear fractions by multiplying both sides of the equation by the least common multiple (LCM) of denominators. For example, to solve (x/3) + 2 = 5 multiply both sides by 3 to get x + 6 = 15. Before multiplying, note any values that make denominators zero and exclude these from possible solutions. Multiplying through removes denominators and reduces the equation to an ordinary linear one.

Combined cases and careful checking
In equations that have both brackets and fractions, it is often easiest to clear fractions first by multiplying through by the LCM, then expand any resulting brackets. Alternatively, expand first then clear denominators — either method works if done carefully. After solving, substitute the value back into the original equation with brackets and fractions to confirm it satisfies the equation and does not make any denominator zero.

Common pitfalls
A common mistake is canceling terms across addition or subtraction rather than cancelling common factors. Another is forgetting to reverse an inequality sign when multiplying or dividing by a negative number — this applies to inequalities rather than equations, but similar sign-care is needed around brackets. Always perform algebraic manipulations stepwise and check final values in the original equation.

📌 Examples
  • Solve: 2(x - 3) = 8 → 2x - 6 = 8 → 2x = 14 → x = 7
  • Solve: (x/3) + 2 = 5 → multiply by 3: x + 6 = 15 → x = 9
  • Solve: (2x + 1)/3 = x - 1 → multiply by 3: 2x + 1 = 3x - 3 → x = 4
🧮 Formulas
  1. Distributive law: a(b + c) = ab + ac
  2. To clear fractions multiply both sides by LCM of denominators
📊 Visual ideas
A stepwise diagram showing clearing denominators then simplifying to isolate x
🧴11

Simple Inequalities and Their Solution

What is an inequality?
An inequality compares two expressions and tells us that one is less than, greater than, or equal in a non-exact way. Symbols used are <, >, ≤ and ≥. Solving an inequality means finding all values of the variable that make the statement true. Solutions are usually ranges of values, not a single number.

Solving inequalities — operations that keep the sign
You can add or subtract the same number from both sides of an inequality without changing the direction of the inequality. Similarly, multiplying or dividing by a positive number leaves the inequality sign unchanged. For example, from x + 3 < 8 we get x < 5 by subtracting 3 from both sides.

Multiplying or dividing by a negative and reversing sign
If you multiply or divide both sides of an inequality by a negative number, you must reverse the direction of the inequality sign. For instance, from −2x < 6 dividing by −2 gives x > −3 (note the change from < to >). This rule follows from the properties of order on the number line and is a common source of errors if forgotten.

Writing and representing solutions
Write solutions as inequalities, for example x ≤ 4, or in interval notation like (−∞, 4]. Draw a number line to represent the solution visually: use a filled (closed) dot for ≤ or ≥ and an open dot for < or >, and shade the appropriate direction. When solving compound inequalities (with AND or OR), treat each part and then find the intersection or union of solutions accordingly.

Applications and checks
Inequalities appear in questions about minimum or maximum requirements, budgets and measurements. After solving, test a value from the solution range in the original inequality to verify correctness. Always show the steps and indicate any boundary values clearly when presenting answers for exams.

📌 Examples
  • Solve: 2x + 3 < 11 → 2x < 8 → x < 4
  • Solve: -3x ≥ 9 → divide by -3 → x ≤ -3 (inequality sign reverses)
  • Represent x ≤ 2 on number line with filled dot at 2 and shading to left
📊 Visual ideas
Number line showing solution x < 4 with open circle at 4 and arrow to left
🔣12

Ratio, Proportion and Algebraic Methods

Understanding ratio and proportion
A ratio is a way to compare two quantities, written as a:b or a/b. A proportion states that two ratios are equal: a/b = c/d. Algebra is useful when one or more parts of a ratio problem are unknown because it allows us to introduce variables and solve systematically. Ratios are common in mixture problems, sharing amounts, and scaling.

Setting up algebraic models
When dividing a quantity in a given ratio, let the common multiplier be k. If a quantity is to be split in ratio 2:5, write the parts as 2k and 5k; the total gives an equation to solve for k. For proportions with unknowns, cross-multiplication converts the equality into a simple equation: if a/b = c/d then ad = bc. This is often the quickest route to the unknown.

Solving ratio problems stepwise

  • Assign variables to the unknown parts, using k to scale ratio numbers.
  • Express totals or relationships as equations involving k or the variables.
  • Solve for k or the variable using algebraic steps and check results.

Examples of applications
Divide money, lengths, or time in given ratios; convert recipes by scaling; or solve mixture problems by equating ratios. Proportions also solve problems involving similarity in geometry (scale factors) and rates like speed and density when quantities maintain a constant ratio.

Practical tips and checking
Always reduce ratios to simplest whole-number form before solving. After obtaining numerical parts, check that they satisfy the original ratio and any totals or other conditions. Watch for integer requirements in some problems and interpret answers in context, stating units where appropriate.

📌 Examples
  • Divide 84 in ratio 2:5 → parts 2k and 5k; 7k = 84 → k = 12 → parts 24 and 60
  • Solve proportion: x/6 = 4/9 → 9x = 24 → x = 8/3
🧮 Formulas
  1. If a/b = c/d then ad = bc (cross-multiplication)
  2. If total = k(sum of ratio parts) then each part = k × corresponding ratio number
📊 Visual ideas
Bar model showing a quantity divided into parts in given ratio, labelled with k × numbers
📏13

Word Problems: Ages, Money and Measurement

Reading and planning
Word problems require careful reading to pick out what is known and what is required. Underline key facts and decide the unknown to be represented by a variable. Write expressions for related quantities in terms of that variable. Clear labelling of the variable is essential: for example let x be the present age of A, then express ages after or before years as x + n or x − n.

Common types and translations
Typical topics are age problems (current age, age difference, ages after or before some years), money problems (costs, totals, shares), and simple measurement relations (perimeter, area relations as simple algebraic relations). Translate phrases into algebraic form: “is” becomes “=”, “sum” becomes “+”, “difference” becomes “−”, “times” becomes multiplication, and “ratio” becomes fraction or colon. Use these translations to form equations.

Solving structure

  • Assign a variable to the unknown quantity and write expressions for other quantities.
  • Form an equation using given relationships and totals.
  • Solve the equation stepwise and check the solution by substitution into the original verbal statement.
  • Interpret the result: ensure it meets context conditions (ages positive, money non-negative, integer part counts if required).

Examples and checks
If the sum of ages of A and B is 30 and A is 4 years older than B, let B = x so A = x + 4; then x + x + 4 = 30 gives x = 13 and A = 17. For money problems, set up cost equations with the correct multiplication for quantities and prices. Always verify the final numeric answers back in the original sentences to ensure correctness and to show work clearly in exams.

Presentation and units
Answer with correct units (years, rupees, metres). Show steps: variable definition, equation, algebraic work, and final interpretation. This makes the solution easy to follow and reduces careless mistakes.

📌 Examples
  • Age: Sum of ages of A and B is 30. If A is 4 years older than B, find ages. Let B = x; A = x + 4; x + x + 4 = 30 → 2x = 26 → x = 13 → B = 13, A = 17
  • Money: A buys 3 pens and 2 books for Rs 350; pen costs p, book costs b. 3p + 2b = 350. If p = 40 find b: 120 + 2b = 350 → 2b = 230 → b = 115
📊 Visual ideas
Simple bar or timeline for age problems showing present and future/past ages
🔢14

Revision: Putting It All Together

Why revision matters
Revision helps to link vocabulary and techniques learned across the unit so that when you see a new problem you can choose the correct method quickly. Algebra builds in layers: knowing how to simplify expressions, factorise, use identities, apply exponent laws, and solve equations and inequalities allows you to tackle a wide range of problems confidently. Revision also reveals common mistakes and gives practice in avoiding them.

Organising revision
Make a checklist of core skills: translating words to algebraic expressions, combining like terms, removing brackets, factorisation techniques (GCF, identities, grouping), exponent rules, simplifying algebraic fractions, solving linear equations (including brackets/fractions), and solving simple inequalities. For each skill, practise representative questions until the steps become routine. Timed practice helps prepare for exams and improves accuracy.

Common errors to watch for
Avoid combining unlike terms (e.g., x and x^2), cancelling terms across sums, forgetting sign changes when distributing negative signs, and not reversing inequality direction when multiplying or dividing by a negative. Also always check for excluded values in algebraic fractions and verify back-substitution into original equations especially when denominators were cleared during solution.

Exam technique and presentation
Read questions carefully and underline key data. State the chosen variable clearly. Show steps systematically: write the equation, simplify, solve, and then check. Box or write the final answer clearly with units. If stuck, attempt a simpler form of the question or substitute a trial value to gain insight: sometimes numerical checks help spot algebraic mistakes.

Practice mix and self-check
Solve mixed sets of questions that require different techniques so you learn to recognise which method to apply. After solving, check answers by substitution, and where possible, perform the reverse operation (e.g., expand a factorised result) to confirm correctness. Regular, focused practice is the best way to consolidate the whole unit.

📌 Examples
  • Mixed: Simplify and solve: 2(x + 3) - x = 7 → 2x + 6 - x = 7 → x = 1
  • Check: Solve (x/2) + (x/3) = 5 → multiply by 6: 3x + 2x = 30 → 5x = 30 → x = 6
📊 Visual ideas
A checklist diagram students can draw: Read → Define variable → Form equation → Solve → Check

Key Concepts

Variable
A symbol, usually a letter, that represents a number which can change.
Constant
A fixed numerical value that does not change within a given expression.
Coefficient
The numerical factor multiplying a variable in a term.
Term
A single part of an expression separated by + or −.
Like terms
Terms that have the same variable raised to the same power.
Distributive law
A rule stating a(b + c) = ab + ac to remove brackets.
Factorisation
Expressing an algebraic expression as a product of its factors.
Greatest common factor (GCF)
The largest expression that divides all terms of an algebraic expression.
Identity
An equality true for all values of the variable, such as (a + b)^2 = a^2 + 2ab + b^2.
Monomial/Binomial/Polynomial
Expressions with one, two, or many terms respectively.
Degree
The highest exponent of the variable in a polynomial.
Exponent
A small number showing how many times a base is multiplied by itself.
Algebraic fraction
A fraction whose numerator or denominator (or both) are algebraic expressions.
Linear equation
An equation of the first degree in one variable that gives a straight-line relationship.
Inequality
A statement that compares two expressions using <, >, ≤, or ≥.
Ratio
A comparison of two quantities expressed as a:b or a/b.
Proportion
An equation stating that two ratios are equal, e.g., a/b = c/d.

Practice Questions

  1. Write an algebraic expression for: 5 less than a number n. / किसी संख्या n से 5 घटाने के लिए बीजगणितीय व्यंजक लिखिए।
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    The expression for 5 less than a number n is n - 5. / किसी संख्या n में से 5 घटाने का व्यंजक n - 5 है।

  2. Simplify: 4x + 3x - 7. / 4x + 3x - 7 को सरल कीजिए।
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    Combine like terms 4x and 3x to get 7x, so the simplified form is 7x - 7. / समान पदों 4x और 3x को जोड़ने पर 7x मिलता है, अत: सरल रूप 7x - 7 है।

  3. Expand using identity: (x + 4)^2. / किसी पहचान का उपयोग करके (x + 4)^2 का विस्तार कीजिए।
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    Using (a + b)^2 = a^2 + 2ab + b^2 with a = x and b = 4 gives x^2 + 8x + 16. / (a + b)^2 = a^2 + 2ab + b^2 में a = x और b = 4 लगाने पर x^2 + 8x + 16 मिलता है।

  4. Factorise: 9y^2 - 16. / 9y^2 - 16 को गुणनखंडों में विभाजित कीजिए।
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    Recognise it as a difference of squares: (3y)^2 - 4^2, so factorised form is (3y - 4)(3y + 4). / इसे वर्गों के अंतर के रूप में पहचानें: (3y)^2 - 4^2, अत: गुणनखंड (3y - 4)(3y + 4) है।

  5. Solve the equation: 5x - 7 = 18. / समीकरण हल कीजिए: 5x - 7 = 18।
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    Add 7 to both sides: 5x = 25, then divide by 5 to get x = 5. Check: 5(5) - 7 = 25 - 7 = 18. / दोनों ओर 7 जोड़ें: 5x = 25, फिर 5 से भाग करने पर x = 5 मिलता है। जाँच: 5(5) - 7 = 25 - 7 = 18।

  6. Solve and check: (x/4) + 3 = 8. / हल कीजिए और जाँच कीजिए: (x/4) + 3 = 8।
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    Subtract 3: x/4 = 5, multiply by 4: x = 20. Check by substitution: (20/4) + 3 = 5 + 3 = 8, so solution is correct. / 3 घटाएँ: x/4 = 5, फिर 4 से गुणा करें: x = 20. जाँच: (20/4) + 3 = 5 + 3 = 8, अत: हल सही है।

  7. A number is divided in the ratio 3:5 and the larger part is 40. Find the number. / किसी संख्या को 3:5 के अनुपात में बाँटा जाता है और बड़ा भाग 40 है। संख्या ज्ञात कीजिए।
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    Let parts be 3k and 5k with 5k = 40, so k = 8. Total = 3k + 5k = 8k = 8×8 = 64. So the number is 64. / भाग 3k और 5k मान लें और 5k = 40 से k = 8 मिलता है। कुल = 8k = 8×8 = 64। अत: संख्या 64 है।

  8. Solve the inequality: 3x - 4 < 11. / असमानता हल कीजिए: 3x - 4 < 11।
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    Add 4: 3x < 15, divide by 3: x < 5. On a number line use an open circle at 5 and shade to the left. / दोनों ओर 4 जोड़ें: 3x < 15, 3 से भाग करने पर x < 5 मिलता है। संख्या रेखा पर 5 पर खुला वृत्त बनाकर बाएँ की ओर छायांकन करें।

  9. Simplify the algebraic fraction: (8x^2 + 4x)/(4x). / बीजगणितीय भिन्न सरल कीजिए: (8x^2 + 4x)/(4x)।
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    Factor numerator: 4x(2x + 1) over 4x cancels to give 2x + 1, with restriction x ≠ 0 because denominator 4x cannot be zero. / अंश को गुणनखंड में लिखें: 4x(2x + 1)/(4x) = 2x + 1, जहाँ x ≠ 0 क्योंकि हरकत में हर बार हरकत (denominator) 4x शून्य नहीं हो सकता।

  10. Translate and solve: Seven added to twice a number gives 21. / अनुवाद कीजिए और हल कीजिए: किसी संख्या के दो गुने में सात जोड़ने पर 21 मिलता है।
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    Let number be x. Then 2x + 7 = 21, so 2x = 14 and x = 7. Check: 2×7 + 7 = 14 + 7 = 21. / संख्या x मान लें। तब 2x + 7 = 21, अत: 2x = 14 और x = 7। जाँच: 2×7 + 7 = 14 + 7 = 21।

  11. Factorise by grouping: x^3 + x^2 + 3x + 3. / समूहकरण करके गुणनखंड करें: x^3 + x^2 + 3x + 3।
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    Group as (x^3 + x^2) + (3x + 3) = x^2(x + 1) + 3(x + 1) = (x + 1)(x^2 + 3). / समूह करें: (x^3 + x^2) + (3x + 3) = x^2(x + 1) + 3(x + 1) = (x + 1)(x^2 + 3)।

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