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Showing posts with the label algebra

Harder Proportion – Direct and Inverse Relationships | Pearson Edexcel International GCSE Maths

These two examples use proportional relationships involving square roots. In each case, the proportional statement is first converted into an equation containing a constant of proportionality, k . The given values are then used to find k before the required value is calculated. Question 1 – Direct Proportion and Square Roots The time t seconds taken for a stone to fall is directly proportional to the square root of the distance d metres. When t = 4.6 , d = 25 . (a) Express t in terms of d . (b) Find the time taken when d = 42.25 . (a) Express t in terms of d t ∝ √d ⇒ t = k√d when t = 4.6, d = 25 ⇒ 4.6 = k√25 ⇒ 4.6 = 5k ⇒ k = 4.6 / 5 ∴ k = 0.92 t = 0.92√d ...

Kinematics – Differentiation, Velocity and Maximum Height | Pearson Edexcel International GCSE Maths

Kinematics uses differentiation to connect displacement, velocity and acceleration. In this example, a stone is projected vertically upwards and its displacement is given as a function of time. Differentiation allows us to find its velocity and determine when it reaches its maximum height. The Displacement Function A stone is projected vertically upwards from the ground. After t seconds, its height above the ground, s metres, is given by: s(t) = 15t − 4.9t² 0 ≤ t ≤ 4 Question (a) – Find ds/dt Differentiate the displacement function with respect to time. Working s(t) = 15t − 4.9t² ds/dt = 15 − 4.9 × 2 × t = 15 − 9.8t Answer: ds/dt = 15 − 9.8t Question (b) – Velocity at...

Transforming Graphs – Questions Answered | Pearson Edexcel International GCSE Maths

These examples look at transformations of the graph y = f(x) . The original curve has a minimum point at (2, −1) . By examining how the function changes, we can determine how the coordinates of this minimum point are transformed. Original Minimum Point The curve y = f(x) has a minimum point at: (2, −1) Question (a)(i) – y = f(x + 2) Find the coordinates of the minimum point after the transformation y = f(x + 2) . Working y = f(x + 2) * Translation (−2, 0) Answer: (0, −1) Answer: (0, −1) Question (a)(ii) – y = 3f(x) Find the coordinates of the minimum point after the transformation y = 3f(x) . Working y = 3f(...

Pressure, Force and Area – Pearson Edexcel International GCSE Maths

Pressure, force and area are connected by a simple formula. Once the relationship is understood, the formula can be rearranged depending on which quantity needs to be found. Pressure, Force and Area The basic relationship is: To find pressure: P = F / A To find force: F = P × A To find area: A = F / P What the Symbols Mean P represents pressure. F represents force. A represents area. Units * Pressure is measured in pascals (Pa), where 1 Pa = 1 N/m². * Force is measured in newtons (N). * Area (A) is measured in square metres (m²). Understanding the Formula Pressure describes h...

Ratio Problems – Pearson Edexcel International GCSE (9–1) Mathematics A Higher Tier

These ratio problems are based on questions from the Revise Pearson Edexcel International GCSE (9–1) Mathematics A – Higher Tier Revision Guide . The solutions below use an algebraic ratio method in which each part of the ratio is represented as a multiple of x . Question 1 – Sharing Money in a Ratio Andre, Becky and Makito share money in the ratio 3 : 6 : 7 . Andre and Becky receive £207 altogether. Work out how much Makito receives. Working A : B : M = 3x : 6x : 7x 3x + 6x = 207 9x = 207 x = 207/9 7 × (207/9) = 161 Makito receives £161. Answer: £161 Question 2(a) – Ages in a Ratio Amir and Petra's ages are in the ratio 3 : 7 . Amir is 9 years old . Work out Petra's age. Wo...

Deriving the Addition and Subtraction Rules for Fractions from First Principles

Fractions | Algebra | First Principles The familiar rules for adding and subtracting fractions can be derived directly from the properties of real numbers. The key idea is that multiplying a number by 1 does not change its value. By writing 1 in a suitable fractional form, both fractions can be expressed with the same denominator. Assumptions Let a, b, c, d ∈ ℝ, with b ≠ 0 and d ≠ 0. Addition of Two Fractions Begin with two fractions whose denominators are not necessarily equal: a b + c d = a b · 1 + c d · 1 = a b ...

The Associative, Commutative and Distributive Laws

The associative, commutative and distributive laws are three of the most important structural rules in algebra. They explain how expressions may be grouped, reordered, expanded and simplified without changing their mathematical value. These laws are used throughout arithmetic, algebra, factorisation, equation solving and mathematical proof. Associative Law The associative law describes how terms may be grouped when the same operation is repeated. If an operation is associative, changing the placement of the brackets does not change the final value of the expression. For addition: a + (b + c) = (a + b) + c For example: 1 + (2 + 3) = (1 + 2) + 3 The associative law also applies to multiplication: a × (b × c) = (a × b) × c For example: 2 × (3 × 4) = (2 × 3) × 4 Subtraction is not associative because changing the grouping can change the result. a − (b − c) ≠ (a − b) − c For example: 1 − (2 − 3) ≠ (1 − 2) − 3 Commutative Law The commutative law describe...

Proofs of the Base-10 Logarithm Laws

These workings derive the laws of base-10 logarithms from exponent laws by converting between exponential form and logarithmic form. Assume a > 0 , b > 0 , and n ≠ 0 . Product Rule log(ab) = log a + log b Let 10 x = a Let 10 y = b Therefore: log 10 a = x log 10 b = y 10 x 10 y = ab 10 x+y = ab Therefore: log 10 (ab) = x + y Substituting: log 10 (ab) = log 10 a + log 10 b Therefore: log(ab) = log a + log b Quotient Rule log(a / b) = log a - log b Let 10 x = a Let 10 y = b Therefore: log 10 a = x log 10 b = y 10 x / 10 y = a / b 10 x-y = a / b Therefore: log 10 (a / b) = x - y Substituting: log 10 (a / b) = log 10 a - log 10 b Therefore: log(a / b) = log a - log b Power Rule log(a n ) = n log a Let log(a n ) = x Therefore: 10 x = a n Taking the n-th root of both sides: (10 x ) 1/n = (a n ) 1/n 10 x/n = a Therefore: log 10 a = x / n n log 10 a = x Theref...

Algebraic Proof Toolkit for Edexcel International GCSE (Higher): Standard Forms That Make Proofs Easy

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Proving that the sum of three consecutive integers is divisible by 3. Algebraic proof questions in Edexcel International GCSE (Higher) often look difficult because they are written in words. The quickest way to handle them is to translate the words into standard algebraic forms that guarantee the number property you need (even, odd, multiple, consecutive, square, etc.). Once the translation is correct, the rest of the proof is usually straightforward simplification, factoring, and a clear final statement. This post gives a compact “toolkit” of the most common forms, presented in a table you can reuse, plus a small set of extras and techniques that frequently appear in Higher-tier proof questions. The core principle In an algebraic proof, represent the numbers so the required property is built in. For example: If a number is even, write it as 2n for some integer n. If a number is a multiple of 3, write it as 3n for some integer n. The phrase “for some integer n” matters...

Why Completing the Square Matters for Vertex Form and the Turning Point

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A quadratic function and its turning point. Link to graph:  https://www.desmos.com/calculator/fktyfs12st A quadratic function is any function of the form f(x) = ax² + bx + c with a ≠ 0 . Its graph is a parabola, and every parabola has exactly one turning point (also called the vertex ). Completing the square is fundamental because it rewrites the quadratic as a shifted square , which makes the turning point immediately visible. Vertex form: the turning point is built in The vertex form of a quadratic is: f(x) = a(x − h)² + k This form is powerful because it exposes two facts at once: (x − h)² ≥ 0 for all real x (a square is never negative). (x − h)² = 0 happens exactly when x = h . So: If a > 0 , then a(x − h)² ≥ 0 , so the smallest possible value of f(x) is k , achieved at x = h (a minimum). If a < 0 , then a(x − h)² ≤ 0 , so the largest possible value of f(x) is k , achieved at x = h (a maximum). Therefore, in vertex form, the turning ...

Quadratic Functions in Vertex Form (A Clear Guide for Everyone)

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Parabolas in sight: The Clifton Suspension Bridge, Bristol, United Kingdom. A quadratic function is a function whose graph is a parabola (a U-shaped curve). One of the most useful ways to write a quadratic is in vertex form , because it shows the parabola’s turning point immediately. 1) The vertex form A quadratic function in vertex form is written as: f(x) = a(x - h) 2 + k This form is especially helpful because the values h and k tell you the vertex directly. 2) The vertex (turning point) The vertex is the point where the parabola changes direction. In vertex form: Vertex = (h, k) If the parabola opens up , the vertex is the lowest point (a minimum). If the parabola opens down , the vertex is the highest point (a maximum). 3) What the number a does The number a controls two key things: the direction the parabola opens, and how wide or narrow it is. a > 0 means the parabola opens up (U-shape). a < 0 means the parabola opens dow...

Rules of Logarithms

This article presents the rules of logarithms using complete, line-by-line derivations. Every identity is built directly from its exponential origin, without shortcuts, matching the structure of formal handwritten algebra. 1. Definition We begin with fundamental exponent facts: a⁰ = 1 ⇒ logₐ(1) = 0 a¹ = a ⇒ logₐ(a) = 1 Say: aᵐ = p Then, by definition: logₐ(p) = m Raise both sides of aᵐ = p to the power 1/m (with m ≠ 0 ): p^(1/m) = a Therefore: logₚ(a) = 1/m Since m = logₐ(p) , we obtain: logₐ(p) = 1 / logₚ(a) 2. Product Rule — Full Derivation Say: aᵐ = p and aⁿ = q Multiply: aᵐ · aⁿ = p · q Using index addition: a^(m+n) = p · q Taking logarithms: logₐ(p · q) = m + n Substitute: logₐ(p · q) = logₐ(p) + logₐ(q) 3. Quotient Rule — Full Derivation Say: aᵐ = p and aⁿ = q Divide: aᵐ / aⁿ = p / q Index subtraction gives: a^(m−n) = p / q Taking logarithms: logₐ(p / q) = m − n So: log...

Arithmetic and Geometric Sequences

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Arithmetic and Geometric Sequences Arithmetic and geometric sequences are two fundamental types of numerical progressions. They describe how quantities grow or shrink by addition or by multiplication, and they form the foundation for topics such as series, summation formulas, and exponential growth. 1. Arithmetic Sequence An arithmetic sequence is a list of numbers in which each term differs from the previous one by a fixed amount called the common difference d . a, a + d, a + 2d, a + 3d, … , a + (n − 1)d a – first term d – common difference The n th term, denoted T n , is given by: T n = a + (n − 1)d Each new term is obtained by adding d to the previous term. The difference between consecutive terms remains constant: T k+1 − T k = d Example: If a = 4 and d = 3, the sequence is 4, 7, 10, 13, 16, … The 20th term is T 20 = 4 + (20 − 1)×3 = 61. 2. Geometric Sequence A geometric sequence is a list of numbers where each term is fo...

What Is an Isomorphism?

What Is an Isomorphism? In mathematics, an isomorphism is a function that shows two mathematical objects have the same structure. Although the objects may look different, an isomorphism demonstrates that they behave in exactly the same way with respect to the operations that define them. If such a map exists, the objects are called isomorphic . An isomorphism tells us that two systems are essentially the same, differing only by a relabelling of their elements. The Basic Idea An isomorphism is a function: f : A → B that must be: Injective — different elements of A map to different elements of B. Surjective — every element of B comes from some element of A. Together these mean f is bijective , and so it has an inverse: f⁻¹ : B → A No information is lost moving from A to B or back. Preserving Structure Bijectivity alone is not enough. An isomorphism must also preserve structure. For groups, this means: f(a ⋆ b) = f(a) ∘ f(b) for all a, b ∈ A ,...

Proof by Induction: The Sum of Squares Formula

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Proof by Induction: The Sum of Squares Formula Theorem. For any integer n ≥ 1, ∑ i=1 n i² = (1/6)·n·(n+1)·(2n+1) Proof (by mathematical induction) 1. Basis Step For n = 1: LHS = 1² = 1 RHS = (1/6)·1·(1+1)·(2·1+1) = (1/6)·1·2·3 = 1 Therefore, the formula holds for n = 1. 2. Inductive Hypothesis Assume the statement is true for some integer k ≥ 1: ∑(i=1→k) i² = (1/6)·k·(k+1)·(2k+1) 3. Inductive Step We must show it is true for n = k + 1: ∑(i=1→k+1) i² = [∑(i=1→k) i²] + (k+1)² = (1/6)·k·(k+1)·(2k+1) + (k+1)² = (1/6)(k+1)[k(2k+1) + 6(k+1)] = (1/6)(k+1)[2k² + 7k + 6] = (1/6)(k+1)(k+2)(2k+3) This matches the same form with k replaced by k+1: (1/6)·(k+1)·((k+1)+1)·(2(k+1)+1) Hence, the formula holds for n = k + 1. 4. Conclusion Since the statement is true for n = 1 (basis step), and true for n = k + 1 whenever it is true for n = k (inductive step), it follows by the principle of mathematical induction that ∑(i=1→n) i² = (1/6)·n·(n+1)·(2n+1) for al...

The Method of Differences — A Clean Proof of the Sum of Cubes

The Method of Differences — A Clean Proof of the Sum of Cubes The method of differences is a remarkably elegant tool for evaluating finite sums. When each term of a series can be written in the form f(r+1) − f(r) , the sum “collapses” — all interior terms cancel, leaving only a boundary expression. This behaviour is called a telescoping sum . 1) Telescoping Sums Assume the general term u r can be written as: u r = f(r+1) − f(r). Then the finite sum from r = 1 to r = n becomes: Σ u r = Σ ( f(r+1) − f(r) ). To see what happens, write out a few terms: u₁ = f(2) − f(1) u₂ = f(3) − f(2) u₃ = f(4) − f(3) ⋮ uₙ = f(n+1) − f(n) When these are added, everything cancels except the first and last pieces: Σ u r = f(n+1) − f(1). This is the essence of the method: interior structure disappears, leaving just the difference between the final and initial states. 2) A Classic Application — The Sum of Cubes We will use this technique to prove the well-known ...

Injective, Surjective, and Bijective Functions

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Injective, Surjective, and Bijective Functions In mathematics, a function describes how elements of one set are assigned to elements of another. Three important properties capture how completely and uniquely a function connects its domain to its codomain: injective , surjective , and bijective . These properties tell us whether different inputs can share the same output and whether every possible output is used. Injective (One-to-One) A function is injective if different inputs always produce different outputs. No two distinct elements in the domain are allowed to map to the same result in the codomain. Formally, a function f : A → B is injective if: f(a₁) = f(a₂) ⇒ a₁ = a₂ Equivalently, if a₁ ≠ a₂ , then f(a₁) ≠ f(a₂) . Example: f(x) = e x from ℝ → ℝ is injective. Different inputs produce different outputs, but not every real number is reached, so it is not surjective. Surjective (Onto) A function is surjective if every element of the codomain is reach...

What Is a Group in Mathematics?

What Is a Group in Mathematics? In abstract algebra, a group is a set G together with a binary operation (written as * ). The pair (G, *) is called a group when the operation satisfies the four conditions below. Closure: for all g₁, g₂ ∈ G , the product g₁ * g₂ is still in G . Identity: there exists an element e ∈ G such that for all g ∈ G , e * g = g * e = g . This element e is called the identity. Inverses: for each g ∈ G there exists an element g⁻¹ ∈ G such that g * g⁻¹ = g⁻¹ * g = e . Associativity: for all g₁, g₂, g₃ ∈ G , g₁ * (g₂ * g₃) = (g₁ * g₂) * g₃ . These four conditions are exactly what is required for (G, *) to be a group. Side note: Commutativity An operation is commutative if swapping the elements does not change the result: a * b = b * a . Commutativity is not required for a group to exist. It is important not to confuse commutativity with associativity . These are distinct i...

Essential Elements of Algebra Problem Solving

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Essential Elements of Algebra Problem Solving Solving equations in algebra is about preserving truth while working step by step toward the unknown. Although equations can look complicated, most of the time we are simply applying a small set of rules with care. If these rules are followed, the solution you reach will be valid; if they are broken, the result becomes unreliable. This post introduces three essential principles that underpin almost all algebraic manipulation. Follow them consistently, and you will have a strong foundation for solving equations with confidence. 1) Start with a True Statement Everything in algebra begins with a statement that is already true. If the starting point is false, no amount of manipulation can produce a trustworthy conclusion. When you write an equation such as: 2x + 3 = 13 you are asserting that doubling a number and then adding three gives thirteen. This is your initial truth. All further steps must preserve this truth. If y...